UCLA Basic Exam Problems and Solutions Brent Woodhouse ...bwoodhouse729/Basic Exam... · UCLA Basic Exam Problems and Solutions Brent Woodhouse This document covers nearly all ...

UCLA Basic Exam Problems and Solutions Brent Woodhouse

This document covers nearly all problems on the UCLA Basic Exam from Fall 2001 to Spring 2013.Problems are listed by category and by exam and linked below. Relevant definitions are listed at the startof most categories below. The linear algebra section in particular starts with many standard theorems. Icannot guarantee this material is completely accurate, but it should at least help you along the way. Goodluck!

Problems listed by category:

AnalysisCountabilityMetric space topologyTopology on realsFixed pointInverse and Implicit Function TheoremsInfinite sequences and seriesPartial derivativesDifferentiationRiemann integrationTaylor SeriesJacobianLagrange MultipliersMiscellaneous

Linear AlgebraRecurring ProblemsOther Problems

Problems listed by exam:

Fall 2001: 1 2 3 4 5 6 7 8 9 10Winter 2002: 1 2 3 4 5 6 7 8 9 10 11Spring 2002: 1 2 3 4 5 6 7 8 9 10 11Fall 2002: 1 2 3 4 5 6 7 8 9 10Spring 2003: 1 2 3 4 5 6 7 8 9 10Fall 2003: 1 2 3 4 5 6 7 8 9 10Spring 2004: 1 2 3 4 5 6 7 8 9 10Fall 2004: 1 2 3 4 5 6 7 8 9 10Spring 2005: 1 2 3 4 5 6 7 8 9 10 11 12 13Fall 2005: 1 2 3 4 5 6 7 8 9 10Winter 2006: 1 2 3 4 5 6 7 8 9 10Spring 2006: 1 2 3 4 5 6 7 8 9 10Spring 2007: 1 2 3 4 5 6 7 8 9 10 11 12Fall 2007: 1 2 3 4 5 6 7 8 9 10 11 12Spring 2008: 1 2 3 4 5 6 7 8 9 10 11 12Fall 2008: 1 2 3 4 5 6 7 8 9 10 11 12Spring 2009: 1 2 3 4 5 6 7 8 9 10 11 12Fall 2009: 1 2 3 4 5 6 7 8 9 10 11 12Spring 2010: 1 2 3 4 5 6 7 8 9 10 11 12Fall 2010: 1 2 3 4 5 6 7 8 9 10 11 12Spring 2011: 1 2 3 4 5 6 7 8 9 10 11 12Fall 2011: 1 2 3 4 5 6 7 8 9 10 11 12Spring 2012: 1 2 3 4 5 6 7 8 9 10 11 12Fall 2012: 1 2 3 4 5 6 7 8 9 10 11 12Spring 2013: 1 2 3 4 5 6 7 8 9 10 11 12

1


Analysis

Countability

A set S is countable if there exists a one-to-one map f : S → N.

Fall 2001 # 4. Let S be the set of all sequences (x1, x2, . . .) such that for all n,

xn ∈ 0, 1.

Prove that there does not exist a one-to-one mapping from the set N = 1, 2, . . . onto the set S.

Suppose for the sake of contradiction that there exists f : N → S such that f is a bijection (one-to-oneand onto). Define the sequence (xn)∞n=0 so that for all natural numbers n, xn = 0 if (f(n))n = 1 and xn = 1if (f(n))n = 0. Then (xn)∞n=1 ∈ S. Hence there exists some natural number M such that

(xn)∞n=1 = f(M).

But this meansxM = (f(M))M ,

which contradicts the construction of xM . Thus no such f exists.

Fall 2003 # 1. Prove that R is uncountable. If you like to use the Baire category theorem, you have toprove it.

Suppose for the sake of contradiction that the real numbers are countable, so there exists a sequence(rn)∞n=1 such that rn : n ≥ 1 = R. Then we can choose a closed interval [a1, b1] such that r1 /∈ [a1, b1].Next, choose a subinterval [a2, b2] ⊂ [a1, b1] such that r2 /∈ [a2, b2]. Repeating this procedure inductively, weselect a decreasing sequence of intervals ([an, bn])∞n=1 such that for all n, rn /∈ [an, bn].

Now the an form an increasing sequence, the bn form a decreasing sequence, and for all n we have an ≤ bn.In fact, for any natural numbers n,m with n ≤ m, an ≤ am ≤ bm. Letting m→∞,

an ≤ limm→∞

bm = infm∈N

bm.

Then taking the limit as n→∞,supn∈N

an = limn→∞

an ≤ infm∈N

bm.

In particular, there exists some x ∈ [supn∈N an, infn∈N bn]. Then for all n we have x ∈ [an, bn], so x 6= rnby construction of an, bn. But since x ∈ R = rn : n ≥ 1, x = rn for some n, a contradiction. Hence R isuncountable.

Fall 2005 #1. A real number α is said to be algebraic if for some finite set of integers a0, . . . , an, not all 0,

a0 + a1α+ · · ·+ anαn = 0.

Prove that the set of algebraic real numbers is countable.

We assume that a countable union of countable sets is countable.Let Snn∈N be a sequence of countable sets. Define

S =⋃n∈N

Sn.

For all n ∈ N, let Fn denote the set of all 1-1 maps from Sn to N. Since Sn is countable, Fn is non-empty.

2


Using the axiom of countable choice, there exists a sequence fnn∈N such that fn ∈ Fn for all n ∈ N.Let φ : S → N→ N be the mapping defined by

φ(x) = (n, fn(x)),

where n is the smallest natural number such that x ∈ Sn. (Clearly Sn : x ∈ Sn is non-empty, so the Well-Ordering Principle ensures that such an n exists.) Since each fn is 1-1, φ must be 1-1. By the FundamentalTheorem of Arithmetic, g(n,m) = 2n3m is an injection from N×N→ N. Hence g φ is an injection from Sinto N, so S is countable.

Clearly the set of integers is countable. Since each corresponds to a finite selection of integers, the setof polynomials of degree n with integer coefficients is countable for each n. The set Z[x] of all polynomialswith integer coefficients is the union over all n of sets of polynomials of degree n. As a countable union ofcountable sets, Z[x] is thus countable.

Note that the set A of algebraic real numbers is

A =⋃

p∈Z[x]

x ∈ R : p(x) = 0.

Now each p ∈ Z[x] has some finite degree n, and then can only have at most n real roots. Thus the setx ∈ R : p(x) = 0 is finite for each p ∈ Z[x]. Hence as a countable union of finite sets, A is countable, asdesired.

Fall 2005 #5. Prove carefully that R2 is not a (countable) union of sets Si, i = 1, 2, . . . with each Si beinga subset of some straight line Li in R2.

Suppose for the sake of contradiction that R2 is a countable union of sets Si, i = 1, 2, . . . with each Sibeing a subset of some straight line Li in R2. Since R is uncountable, there exists some x ∈ R such that theline Lx = (x, y) : y ∈ R is not equal to any Si. (Otherwise (x, 0) : x ∈ R would be countable, implyingthat R was countable.) Now each line Si intersects Si at either zero or one point. Thus the set⋃

i∈NLx ∩ Si

is countable. Since R2 is a countable union of the Si however,⋃i∈N

Lx ∩ Si = Lx,

so Lx is countable. But this implies that R is countable, a contradiction.

Spring 2008 #7. Let a(x) be a function on R such that

(i) a(x) ≥ 0 for all x, and

(ii) There exists M <∞ such that for all finite F ⊂ R,∑F

a(x) ≤M.

Prove x : a(x) > 0 is countable.

Define

Sn =

x : f(x) >

1

n

.

3


Fix n and suppose for the sake of contradiction that #(Sn) > Mn. Then there exists a subset F of Sn ofcardinality Mn+ 1. By property (ii), ∑

F

a(x) ≤M.

However, ∑F

a(x) ≥∑F

1

n=Mn+ 1

n> M,

a contradiction. Thus Sn is finite for each n.We have

x : a(x) > 0 =⋃n∈N

x : a(x) >

1

n

=⋃n∈N

Sn,

which is a countable union of finite sets, and thus countable. (Apparently assumption (i) is not needed.)

Fall 2011 #3. Prove that the set of real numbers can be written as the union of uncountably many pairwisedisjoint subsets, each of which is uncountable.

Define the map f : (0, 1)× (0, 1)→ (0, 1) so that

f(x, y) = 0.x0y0x1y1 . . .

where x = 0.x0x1 . . . and y = 0.y0y1 . . .. Here we replace any infinite chain of 9’s in x or y by incrementingthe digit preceding the chain and replacing the chain of 9’s by a chain of 0’s, then evaluate f . Then this mapis well-defined and actually an injection. Consider the set S of all vertical lines in (0, 1)× (0, 1). There areuncountably many, and their union is (0, 1)× (0, 1). Also, each vertical line is an uncountable set of points.Now consider f(S). Note the only decimals that f misses form a countable set. Since f is an injection, foreach line L ∈ S, f(S) is uncountable. Also, the images of distinct lines of S under f are disjoint. Thus

(0, 1) = f(S) =⋃L∈S

f(L) ∪⋃

(countable collection of points)

can be written as a union of uncountably many pairwise disjoint subsets, each of which is uncountable.Let g be a bijection between (0, 1) and R (tan(π(x− 1/2)), for instance). Composing g with f above, we

can write the set of real numbers as the desired union.

4


Metric space topology

X is compact if every open cover of X has a finite subcover.X is complete if every Cauchy sequence of elements in X converges to some element of X.X is connected if for every pair of non-empty open sets A and B with A ∪B = X, A ∩B 6= ∅.X is sequentially compact if every sequence of elements in X has a convergent subsequence.X is totally bounded if for all ε > 0, there exists a finite open cover of X using balls of radius ε.X is separable if there is a dense subset of X that is countable. (A dense subset S ⊂ X is one which

satisfies S = X.)A base of open sets for X is a family B of open subsets of X such that every open subset of X is the

union of sets in B.X is second-countable if there is a base of open sets of X that is at most countable.An accumulation point of a sequence (xn)∞n=1 is a point x such that for each neighborhood B of x there

are infinitely many natural numbers i such that xi ∈ B.A homeomorphism is a bijection f such that both f and f−1 are continuous.

Spring 2009 #4; Spring 2005 #13. Spring 2013 #3. Let (X, d) be an arbitrary metric space.

(a) Give a definition of compactness of X involving open covers.

(b) Define completeness of X.

(c) Define connectedness of X.

(d) Is the set of rational numbers Q (with the usual metric) connected? Justify your answer.

(e) Suppose X is complete. Show that X is compact in the sense of part (a) if and only if for every r > 0,X can be covered by finitely many balls of radius r. (X is totally bounded.)

(a) X is compact if every open cover of X has a finite subcover.

(b) X is complete if every Cauchy sequence of elements in X converges to some element of X.

(c) X is connected if for every pair of non-empty open sets A and B with A ∪B = X, A ∩B 6= ∅.

(d) No, the set of rational numbers (with the usual metric) is not connected. Let α be an irrationalnumber. Consider the open sets S = Q ∩ (−∞, α) and T = Q ∩ (α,∞). Clearly S and T are non-empty,have union Q, and S ∩ T = ∅. Let ε > 0. For any s ∈ S, B(s, α − s) ⊂ S, hence S is open. Likewise, forany t ∈ T , B(t, t− α) ⊂ T , hence T is open. (Here the balls are with respect to the usual metric restrictedto Q × Q.) Thus we have exhibited two non-empty disjoint open subsets of Q with union Q, so Q is notconnected.

(e) Suppose X is complete. We show that X is compact if and only if X is totally bounded in multiplesteps:

Step 1: If X is compact, then X is sequentially compact.

Find a point to converge to: Let yj∞j=1 be a sequence in X. Suppose for the sake of contradictionthat for each x ∈ X, there exists ε = ε(x) > 0 such that only finitely many terms of the sequence yj liein B(x, ε(x)). Note that the set of open balls B(x, ε(x)) : x ∈ X forms an open cover of X. Since X iscompact, there is a finite subcover

X = B(x1, ε(x1)) ∪ · · · ∪B(xm, ε(xm)).

5


Since yj belongs to B(xi, ε(xi)) for only finitely many indices j, we conclude that yj belongs to X for onlyfinitely many indices j, contradicting that yi ∈ X for all i. Hence there exists x ∈ X such that for eachε > 0, B(x, ε) contains infinitely many terms of the sequence yj.

Construct a convergent subsequence: Choose j1 so that yj1 ∈ B(x, 1). Inductively choose jn+1 so thatjn+1 > jn and yjn+1

∈ B(x; 1/(n + 1)). Then yjn∞n=1 is a subsequence of yj that converges to x. ThusX is sequentially compact.

Step 2: If X is sequentially compact, then X is totally bounded.

Select points which are spread out to form a sequence; at some point you must stop. Let ε > 0. Lety1 ∈ X. If X = B(y1, ε), we are finished. Otherwise, let y2 be any point in X \ B(y1, ε). As long as⋃nj=1B(yj , ε) 6= X, select

yn+1 ∈ X \ (

n⋃j=1

B(yj , ε)).

Suppose for the sake of contradiction that this procedure does not terminate. Then the points y1, y2, . . .satisfy

d(yk, yj) ≥ εfor all 1 ≤ j < k. It follows that yj∞j=1 has no convergent subsequence, contradicting the sequentialcompactness of X. Thus the procedure does terminate, so there exists N with

X =

N⋃j=1

B(yj , ε).

Hence X is totally bounded.

Step 3: If X is totally bounded, then X is sequentially compact.

Construct a sequence of subsequences using the pigeonhole principle, then diagonalize. Let xj∞j=1 be asequence in X. Rewrite the sequence in the form x1j∞j=1. By induction, we construct sequences xkj∞j=1,k ≥ 2, with the properties

(i) xkj∞j=1 is a subsequence of xk−1,j∞j=1, k ≥ 2.(ii) xkj∞j=1 is contained in a ball of radius 1/k, k ≥ 2.Suppose k ≥ 2 and we already have the sequences xij∞j=1 for i < k. Let B1, . . . Bn be a finite number

of open balls of radius 1/k that cover X. Since there are infinitely many indices j and only finitely manyballs, there must exist at least one ball, say Bm such that xk−1,j ∈ Bm for infinitely many j ≥ 1. Now letxk1 be the first of the xk−1,j ’s that belong to Bm, let xk2 be the second, etc. Then xkj∞j=1 has properties(i) and (ii).

Now set yn = xnn, so that yn∞n=1 is a subsequence of xj∞j=1. Also note that yn∞n=k is a subsequenceof xkj∞j=1 for each k. Thus by construction of the xkj , for any n,m ≥ k,

d(yn, ym) < 2/k.

Thus yn∞n=1 is a Cauchy sequence, so yn∞n=1 converges, and X is sequentially compact.

Step 4: If X is totally bounded, then X is separable.

Direct approach. Let n be a positive integer. Then there exist xn1, . . . , xnm such that the open balls withcenters at the xnj and radii 1/n cover X. The family xnj : 1 ≤ j ≤ mn, 1 ≤ n < ∞ is then a countablesubset of X. For each x ∈ X and each integer n, there is an xnj such that d(xnj , x) < 1/n. Consequentlythe xnj are dense in X.

Step 5: If X is separable, then X is second-countable.

6


Direct approach, similar to Step 4. Let xj∞j=1 be a dense sequence in X. Consider the family of opensets

B = B(xj , 1/n) : j ≥ 1, n ≥ 1.Let U be an open set of X and let x ∈ U . For some n ≥ 1, we have B(x, 2/n) ⊂ U . Choose j so thatd(xj , x) < 1/n. Then x ∈ B(xj , 1/n), and the triangle inequality shows that B(xj , 1/n) ⊂ B(x, 2/n) ⊂ U .Thus each x ∈ U has some associated Vx ∈ B such that x ∈ Vx and Vx ⊂ U . It follows that

U =⋃x∈U

Vx

represents U as a union of sets in B. Thus B is a base of open sets. Since B is countable, X is second-countable.

Step 6: If X is second-countable, then every open cover of X has a countable subcover. (Lindelof’sTheorem).

Pick elements of the base which are inside sets from the open cover. They cover X. Let Uαα∈A be anopen cover of X, where A is some index set. Let B be a countable base of open sets. Let C be the subset ofB consisting of those sets V ∈ B such that V ⊂ Uα for some α. We claim that C is a cover of X. Indeed, ifx ∈ X, then there is some index α such that x ∈ Uα. Since B is a base and Uα is open, there exists V ∈ Bsuch that x ∈ V and V ⊂ Uα. In particular, V ∈ C, so C covers X.

For each V ∈ C, select one index α = α(V ) such that V ⊂ Uα(V ). Then the sets Uα(V ) : V ∈ C coverX. Since B is countable, so is C, so that the Uα(V )’s form a countable subcover of X.

Step 7: If X is sequentially compact and every open cover of X has a countable subcover, then everyopen cover of X has a finite subcover.

Argue by contradiction. Make a sequence, get subsequence, use completeness of X, and that a set isclosed. Let Un∞n=1 be a sequence of open subsets of X that cover X. Suppose for the sake of contradictionthat for all positive integers m,

X 6= U1 ∪ · · · ∪ Um.For each m, let xm be any point in X \ (

⋃mj=1 Uj). Since X is sequentially compact, the sequence (xm)∞m=1

has a subsequence which converges. Since X is complete, this subsequence converges to some x ∈ X. Nowxj ∈ X \ (

⋃mj=1 Uj) for all j ≥ m. Thus since X \ (

⋃mj=1 Uj) is closed, x ∈ X \ (

⋃mj=1 Uj). But this is true

for all m, hence

x ∈ X \ (

∞⋃j=1

Uj) = ∅,

a contradiction. Thus Un∞n=1 has a finite subcover.

Note: Tao’s Analysis II, Ch. 12 also has an argument that sequential compactness implies compactness.

Fall 2004 #4. Suppose that (M,ρ) is a metric space, x, y ∈M , and that xn is a sequence in this metricspace such that xn → x. Prove that ρ(xn, y)→ ρ(x, y).

Let ε > 0. Since xn → x, there exists N such that for all n ≥ N , ρ(xn, x) ≤ ε. By the reverse triangleinequality, for any n ≥ N ,

|ρ(xn, y)− ρ(y, x)| ≤ ρ(xn, x) ≤ ε.Thus ρ(xn, y)→ ρ(x, y).

Fall 2002 #1. Let K be a compact subset and F be a closed subset in the metric space X. SupposeK ∩ F = ∅. Prove that

0 < infd(x, y) : x ∈ K, y ∈ F.

7


Suppose for the sake of contradiction that infd(x, y) : x ∈ K, y ∈ F = 0. Then for each n, there existxn ∈ K and yn ∈ F such that d(xn, yn) < 1/n. Since K is compact, K is sequentially compact and complete,so there exists a subsequence xnj

∞j=1 that approaches some x ∈ K. It follows that ynj→ x as j → ∞.

Since F is closed, x ∈ F . Hence x ∈ K ∩ F , a contradiction. Thus

0 < infd(x, y) : x ∈ K, y ∈ F.

Fall 2008 #4. (a) Suppose that K and F are subsets of R2 with K closed and bounded and F closed.Prove that if K ∩ F = ∅, then d(K,F ) > 0. Recall that

d(K,F ) = infd(x, y) : x ∈ K, y ∈ F.

Fall 2010 #1. Also show the converse, that if K ⊂ X is compact and

infx∈K,y∈F

d(x, y) > 0,

then K ∩ F = ∅.

(b) Is (a) true if K is just closed? Prove your assertion.

(a) This is the above exercise with X = R2, since R2 is complete with respect to the standard metric.

For the converse, suppose for the sake of contradiction that K ∩ F 6= ∅. Then there exists x ∈ K ∩ F ,and 0 = d(x, x) ∈ d(x, y) : x ∈ K, y ∈ F, so d(K,F ) = 0, a contradiction. Thus K ∩ F = ∅.

No, part (a) is no longer true if K is just closed. Let

K = (n, 0) : n ∈ N and F = (n+1

2n, 0) : n ∈ N \ 0.

Note that K and F contain all their limit points, so they are closed. However, there are points in F and Kwith distance 1

2n for each n ∈ N, so d(K,F ) = 0.

Spring 2002 #3. Suppose that X is a compact metric space (in the covering sense of the word compact).Prove that every sequence xn : xn ∈ X, n = 1, 2, 3 . . . has a convergent subsequence. [Prove this directly.Do not just quote a theorem.]

Let xn∞n=1 be a sequence in X. Let ε > 0. Suppose for the sake of contradiction that there does notexist x ∈ X such that i ∈ N \ 0 : xi ∈ B(x, ε) is infinite. Clearly

⋃x∈X B(x, ε) covers X, so since X is

compact, there exists a finite cover

X = B(x1, ε) ∪ · · · ∪B(xN , ε).

Now

i ∈ N \ 0 : xi ∈ X =

N⋃j=1

i ∈ N \ 0 : xi ∈ B(xj , ε).

Clearly the left hand side is an infinite set, but by assumption, each set in the union on the right hand sideis finite, so the union on the right hand side is finite, a contradiction. Hence there exists some x ∈ X suchthat i ∈ N \ 0 : xi ∈ B(x, ε) is infinite.

Select xn1∈ B(x, 1). Then for each k ≥ 1, inductively select xnk+1

> xnkwith xnk+1

∈ B(x, 1/(k + 1)).It follows that (xnk

)∞k=1 converges to x, hence xn∞n=1 has a convergent subsequence.

Spring 2005 #12. Let (X, d) be a metric space. Prove that the following are equivalent:

8


(a) There is a countable dense set.

(b) There is a countable basis for the topology.

Recall that a collection of open sets U is called a basis if every open set can be written as a union ofelements of U .

Let xn∞n=1 be a countable dense set in X. Let B be a basis for the topology. It follows that

B(xn, 1/m), n ≥ 1,m ≥ 1

is a countable basis for the topology. To see this, let U be an open set in X and let x ∈ U . For somem ≥ 1, we have B(x, 2/m) ⊂ U . Choose n so that d(xn, x) < 1/m. Then x ∈ B(xn, 1/m), and the triangleinequality implies B(xn, 1/m) ⊂ B(x, 2/m) ⊂ U . Thus each x ∈ U has some associated Vx ∈ B such thatx ∈ Vx and Vx ⊂ U . It follows that

U =⋃x∈U

Vx

represents U as a union of sets in B. Thus B is a countable basis for the topology.Conversely, suppose Bn∞n=1 is a countable basis for the topology. Choose xn ∈ Bn for each n ≥ 1. It

follows that xn is dense in X, so there is a countable dense set in X.

Spring 2005 #6. Let X be the set of all infinite sequences σn∞n=1 of 1’s and 0’s endowed with the metric

dist(σn∞n=1, σ′n∞n=1) =

∞∑n=1

1

2n|σn − σ′n|.

Give a direct proof that every infinite subset of X has an accumulation point.

An accumulation point of a sequence (xn)∞n=1 is a point x such that for each neighborhood B of x thereare infinitely many natural numbers i such that xi ∈ B. Let S0 be an infinite subset of S. Now eitherinfinitely many sequences in S0 start with 0 or infinitely many sequences in S0 start with 1. Let x1 be 0 ifinfinitely many sequences in S0 start with 0 and otherwise, let x1 = 1. Note that if x1 = 1, then infinitelymany sequences in S0 start with 1. Now for each n ≥ 0, inductively let Sn+1 be the set of sequences in Snwhose n-th digit is xn. By construction Sn+1 is infinite. Define xn+1 to be 0 if infinitely many sequences inSn+1 have n+ 1-st digit 0, and xn+1 = 1 otherwise.

Thus we form a sequence (xn)∞n=1 such that for each N ≥ 1, there exists σn∞n=1 ∈ S such that σn = snfor all n ≤ N . Hence

dist(σn∞n=1, xn∞n=1) =

∞∑n=1

1

2n|σn − xn| =

∑n=N+1

1

2n|σn − xn| ≤

∑n=N+1

1

2n≤ 1

2N.

Since N is arbitrary, this shows that xn∞n=1 is an accumulation point of S.

Spring 2005 #7. Let X,Y be two topological spaces. We say that a continuous function f : X → Y isproper if f−1(K) is compact for any compact set K ⊂ Y .

(a) Give an example of a function that is proper but not a homeomorphism.

(b) Give an example of a function that is continuous but not proper.

(c) Suppose f : R→ R is C1 (that is, has a continuous derivative) and for all x ∈ R,

|f ′(x)| ≥ 1.

9


Show that f is proper.

(a) Pick any continuous function which is not a bijection. For example, f : R→ R given by f(x) = 0 iscontinuous and proper, but f is clearly not a bijection, so f is not a homeomorphism.

(b) Let X be a non-compact metric space (R for example) and let Y = 0. Then the constant functionf : X → Y is continuous, but f−1(0) = X is not compact, and 0 is compact. Hence f is not proper.

(c) Let K ⊂ Y be a compact set. Let Uαα∈I be an open cover of f−1(K). It follows that f(Uα)α∈Iis an open cover of f(f−1(K)) = K. Since K is compact, there exists a finite subcover f(Un)Nn=1 of K.

Let x ∈ f−1(K). Then since f(Un)Nn=1 covers K, f(x) ∈ f(Uj) for some 1 ≤ j ≤ n. Hence there existsy ∈ Uj such that f(x) = f(y).

By the mean value theorem (valid since f is C1), there exists c ∈ (x, y) such that

f(x)− f(y) = f ′(c)(x− y).

Thus by the given property of f ,

0 = |f(x)− f(y)| = |f ′(c)||x− y| ≥ |x− y|.

Thus x = y, so x ∈ Uj . Thus UnNn=1 is a finite cover of f−1(K), so f−1(K) is compact, and f is proper.

Spring 2008 #6. Let Y be a complete countable metric space. Prove there is y ∈ Y such that y is open.

Suppose for the sake of contradiction that y has non-empty interior for each y ∈ Y . Then y isnowhere dense, and Y =

⋃y∈Y y is a countable union of closed nowhere-dense sets since Y is countable.

But this contradicts the Baire Category Theorem, since Y is a complete metric space. Hence there existssome y ∈ Y such that y has non-empty interior. It follows that y is an interior point of y, hence y isopen.

Spring 2010 #8. Let (X, d) be a complete metric space and let K be a closed subset of X such that forany ε > 0, K can be covered by a finite number of sets Bε(x), where

Bε(x) = y ∈ X : d(x, y) < ε.

Prove that K is compact.

Follow the proof in Spring 2009 #4.

Fall 2012 #3; Fall 2011 #6; Spring 2008 #4; Winter 2006 #4. Let fn(x) be a sequence ofnon-negative continuous functions on a compact metric space X. Assume fn(x) ≥ fn+1(x) for all n and x,so that limn→∞ fn(x) = f(x) exists for every x ∈ X. Prove f is continuous if and only if fn converges to funiformly on X.

The forward direction is called Dini’s Theorem. Let ε > 0. For each n, let gn = fn − f , and defineEn := x ∈ X : gn(x) < ε. Each gn is continuous, so each En is open. Since fn is monotonicallydecreasing, gn is monotonically decreasing, so En ⊂ En+1 for all n ≥ 1. Since fn converges pointwise to f ,it follows that the collection En is an open cover of X. Since X is compact, there exist n1 < n2 < · · · < nKwith

X = En1∪ En2

∪ · · · ∪ EnK= EnK

.

Thus for any n ≥ nK and x ∈ X, x ∈ EnK, so

|fn(x)− f(x)| = fn(x)− f(x) ≤ fnK(x)− f(x) = gnK

(x) ≤ ε.

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Thus fn converges to f uniformly on X.Suppose fn converges to f uniformly on x. Let ε > 0. Select N such that for all n ≥ N and all x ∈ X,

|fn(x)− f(x)| ≤ ε/3.

Since fN is continuous, there exists δ > 0 such that if x, y ∈ X with d(x, y) ≤ ε, then

|fN (x)− fN (y)| ≤ ε/3.

It follows that for any x, y ∈ X with d(x, y) ≤ δ,

|f(x)− f(y)| ≤ |f(x)− fN (x)|+ |fN (x)− fN (y)|+ |fN (y)− f(y)| ≤ ε/3 + ε/3 + ε/3 = ε.

Thus f is continuous.

Fall 2012 #4. A subset K of a metric space (X, d) is called nowhere dense if K has empty interior, (i.e., ifU ⊂ K, U open in X imply U = ∅.) Prove the Baire theorem that if (X, d) is a complete metric space, then Xis not a countable union of closed nowhere-dense sets. Hint: Assume X = ∪nKn where each Kn is closed andnowhere dense. Show there is x1 ∈ X and 0 < δ1 < 1/2 such that B1 = B(x1, δ1) = y ∈ X : d(y, x) < δ1satisfies B1 ∩K1 = ∅ and there is x2 ∈ X and 0 < δ2 <

δ12 such that B2 = B(x2, δ2) satisfies B2 ⊂ B1 and

B2 ∩K2 = ∅. Then continue by induction to find a sequence xn in X that converges to x ∈ X \⋃∞n=1Kn.

Let Kn, n ≥ 1 be closed nowhere-dense sets such that X =⋃n≥1Kn. Choose some x1 ∈ X \K1. Since

K1 is closed, X \ K1 is open, so there exists 0 < δ1 < 1/2 such that B1 := B(x1, δ1) ⊂ X \ K1. ThenB1 ∩K1 = ∅. Suppose inductively that n ≥ 2 and we have selected Bn = B(xn, δn) such that Bn ⊂ Bn−1

and Bn ∩Kn = ∅. Select xn+1 ∈ Bn \Kn+1. (Since Kn+1 is nowhere dense, it cannot contain Bn, so somesuch xn+1 exists.) Now Kn+1 is closed, so Bn \Kn+1 is open, hence there exists 0 < δn+1 < δn/2 such thatB(xn+1, δn+1) ⊂ (Bn \Kn+1). Then Bn+1 ⊂ Bn and Bn+1 ∩Kn+1 = ∅, completing the induction.

Choosing countably many xn and Bn in this way requires the axiom of countable choice. It follows thatδn ≤ δ1/2

n−1, so the sequence (xn)∞n=1 is a Cauchy sequence. Since (X, d) is complete, (X, d) converges tosome x ∈ X. Because Bn+1 ⊂ Bn for each n, it follows that x ∈ Bn for each n ≥ 1. Since Bn ∩Kn = ∅, wemust have x /∈ Kn for each n ≥ 1, hence

x ∈ X \∞⋃n=1

Kn = ∅,

a contradiction. Hence X is not a countable union of closed nowhere-dense sets.

Winter 2002 #1. Spring 2012 #1. Let Ω denote the set of all closed subsets of [0, 1] and let ρ : Ω×Ω→[0, 1] be defined by

ρ(A,B) := maxsupx∈A

infy∈B|x− y|, sup

y∈Binfx∈A|x− y|.

Show that (Ω, ρ) is a metric space.

Clearly ρ is non-negative and symmetric.Suppose A and B are closed subsets of [0, 1] with ρ(A,B) = 0. Then

supx∈A

infy∈B|x− y| = 0,

which implies that for any x ∈ A,infy∈B|x− y| = 0.

Thus x is a limit point of B, and since B is closed, x ∈ B. Hence A ⊆ B.Likewise, ρ(A,B) = 0 implies supy∈B infx∈A |x−y| = 0; reasoning as above implies B ⊆ A. Thus A = B.

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Finally, we verify that ρ satisfies the triangle inequality on Ω. Let A,B,C be closed subsets of [0, 1]. Forany a ∈ A, b ∈ B, c ∈ C,

|a− b| ≤ |a− c|+ |c− b|.

Taking the infimum of both sides over all b ∈ B,

infb∈B|a− b| ≤ |a− c|+ inf

b∈B|c− b|.

It follows thatinfb∈B|a− b| ≤ |a− c|+ sup

c∈Cinfb∈B|c− b|,

Then taking the infimum over all c ∈ C,

infb∈B|a− b| ≤ inf

c∈C|a− c|+ sup

c∈Cinfb∈B|c− b|.

Finally, taking the supremum over all a ∈ A,

supa∈A

infb∈B|a− b| ≤ sup

a∈Ainfc∈C|a− c|+ sup

c∈Cinfb∈B|c− b|.

Thussupa∈A

infb∈B|a− b| ≤ ρ(A,C) + ρ(C,B).

By symmetry of A and B,supb∈B

infa∈A|a− b| ≤ ρ(A,C) + ρ(C,B).

Thusρ(A,B) ≤ ρ(A,C) + ρ(C,B).

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Topology on realsA family of functions is equicontinuous if for every ε > 0, there exists a δ > 0 such that if x1, x2 ∈ X

with d(x1, x2) < δ, then for any f ∈ F , d(f(x1), f(x2)) < ε.

Winter 2006 #6. Let −∞ < a < b <∞. Prove that a continuous function f : [a, b]→ R attains all valuesin [f(a), f(b)].

Let y ∈ [f(a), f(b)]. If y = f(a) or y = f(b), we are done. Otherwise, f(a) < y < f(b). Define

E := x ∈ [a, b] : f(x) < y.

Clearly E is a subset of [a, b] and is hence bounded. Also, a ∈ E, so E is non-empty. By the least upperbound principle,

c := sup(E)

is finite. Clearly c ∈ [a, b].Select N such that c− 1

N ≥ a. By definition of supremum, there must exist xn ∈ E with

c− 1

n≤ xn ≤ c

for all n ≥ N . Letting N →∞, it follows that limn→∞ xn = c. Since f is continuous at c, this implies

limn→∞

f(xn) = f(c).

But since xn ∈ E, f(xn) < y for every n. This implies f(c) ≤ y.Since f(c) ≤ y < f(b), c < b. Choose N such that c + 1

n < b for all n ≥ N . Then c + 1n /∈ E for all

n ≥ N , so

f(c+1

n) ≥ y

for all n ≥ N . Taking the limit as n→∞ and using the continuity of f ,

f(c) ≥ y.

Thus f(c) = y, hence f attains all values in [f(a), f(b)].

Fall 2004 #2. State and prove Rolle’s Theorem. (You can use without proof theorems about the maximaand minima of continuous or differentiable functions.)

Rolle’s Theorem: Let a < b, f be a continuous function on [a, b] which is differentiable on (a, b), andsuppose f(a) = f(b). Then there exists c ∈ (a, b) such that f ′(c) = 0.

Proof: Since f is continuous on the compact set [a, b], it attains its maximum and minimum on [a, b].If both the maximum and minimum occur at the endpoints a and b, then f(a) = f(b) implies that f isconstant, so taking c = (a+ b)/2 ∈ (a, b), f ′(c) = 0. Otherwise, there exists c ∈ (a, b) such that f(c) is eitherthe maximum or minimum of f on [a, b].

Suppose that f(c) is the maximum of f on [a, b] (the other case is similar). For every h > 0,

f(c+ h)− f(c)

h≤ 0,

thus letting h→ 0 from the right,f ′(c+) ≤ 0.

Likewise, for every h < 0,f(c+ h)− f(c)

h≥ 0,

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thus letting h→ 0 from the left,f ′(c−) ≥ 0.

Since f is differentiable at c, f ′(c) = f ′(c+) = f ′(c−), so f ′(c) = 0.

Spring 2011 #7. Prove that there is a real number x such that

x5 − 3x+ 1 = 0.

Let f(x) = x5 − 3x + 1. Clearly f is continuous. Also, f(−2) = −25 and f(0) = 1. Thus by theintermediate value theorem, there exists x ∈ [−2, 0] such that f(x) = 0.

Fall 2001 #1. Let K be a compact set of real numbers and let f(x) be a continuous real-valued functionon K. Prove there exists x0 ∈ K such that f(x) ≤ f(x0) for all x ∈ K.

Steps: Show f is bounded, so supremum of f on K exists. Find a sequence that converges to thesupremum. Use sequential compactness to get a subsequence converging to some d ∈ [a, b]. Use continuity.

Because K is compact, the Heine-Borel Theorem implies it is closed and bounded. By the Bolzano-Weierstrauss Theorem, K is also sequentially compact.

Suppose for the sake of contradiction that f is unbounded. Then for each natural number n, there existsxn ∈ K such that f(xn) > n. Since K is sequentially compact, there exists a subsequence of xn∞n=0

converging to some x ∈ R. Since K is closed, x ∈ K. Since f is continuous, f(xn) → f(x). But f(xn) isunbounded as n→∞, a contradiction. Hence f is bounded.

Thus by the least upper bound principle, M := sup(f(K)) exists. Select a sequence (xn)∞n=1 such that

M − 1

n≤ f(xn) ≤M

for each n. Thus (f(xn))∞n=1 converges to M . Since K is sequentially compact, there exists a subsequence(xnk

)∞k=1 which converges to some x0 ∈ R. Since K is closed, x0 ∈ K. Now (f(xnk))∞k=1 must converge to

M . By continuity of f , this implies f(x0) = M . Thus f attains its maximum on K.

Fall 2002 #2. Show why the Least Upper Bound Property (every set bounded above has a least upperbound) implies the Cauchy Completeness Property (every Cauchy sequence has a limit) of the real numbers.

Let (xn)∞n=1 be a Cauchy sequence of real numbers. We first show that (xn)∞n=1 is bounded. Fix ε > 0and let N be such that |xn − xm| < ε for n,m ≥ N . Let R = max(d(xN , x1), . . . , d(xN−1, xN ), ε). Then theentire sequence (xn)∞n=1 is contained in B(xn, 2R). Thus (xn)∞n=1 is bounded.

By the least upper bound property, we can define

zn := supxk : k ≥ n

for each n ≥ 1. Clearly (zn)∞n=1 is decreasing and bounded since (xn)∞n=1 is bounded. The least upper boundproperty implies the greatest lower bound property, thus we can define

x = infzn : n ≥ 1 = limn→∞

zn.

We show that (xn)∞n=1 → x. First, we exhibit a subsequence of (xn)∞n=1 which converges to x. Let j bea positive integer. Since x is the limit of the zn, there exists Nj such that for all n ≥ Nj ,

|zn − x| ≤1

2j.

Since zNj= supxk : k ≥ n, there exists nj ≥ Nj such that

|zNj− xnj

| ≤ 1

2j.

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Thus we obtain infinitely many distinct nj ; re-index the distinct nj to obtain the subsequence xn`∞`=1. By

construction,

|xn`− x| ≤ |xn`

− zN`|+ |zN`

− x| ≤ 1

2`+

1

2`=

1

`.

Thus (xn`)∞`=1 converges to x.

Now since (xn)∞n=1 is a Cauchy sequence, it follows that the whole sequence converges to x. Let ε > 0.There exists N such that for all ` ≥ N ,

|xn`− x| ≤ ε/2.

There also exists N ′ such that if j, k ≥ N ′,

|xj − xk| ≤ ε/2.

Hence for all n ≥ max(nN , N′),

|xn − x| ≤ |xn − xnN|+ |xnN

− x| ≤ ε/2 + ε/2 = ε.

Thus (xn)∞n=1 converges to x.

Winter 2002 #4. Prove that the set of irrational numbers in R is not a countable union of closed sets.

Suppose for the sake of contradiction that the set of irrational numbers I can be represented as

I =⋃n∈N

FN

where the FN are closed. Then

R = (⋃n∈N

Fn) ∪ (⋃r∈Qr).

Since R has non-empty interior, the Baire Category Theorem implies that one of the sets in the union onthe right hand side has non-empty interior. Clearly it is not r for some rational r, so some Fn must havenon-empty interior. Thus there exists x ∈ Fn such that B(x, ε) ⊂ Fn ⊂ I. But the rationals are dense in R,so some rational number is an element of B(x, ε) and thus an element of I, a contradiction.

Fall 2002 # 3; Spring 2002 #2 Show that the set Q of rational numbers in R is not expressible as theintersection of a countable collection of open subsets of R.

Fall 2012 #5. Use the Baire Category Theorem to prove this.

Suppose for the sake of contradiction that Q =⋂n∈N Un, where Un is open for each n. Clearly Q ⊂ Un

for each n, and since the rational numbers are dense in R, each Un is dense in R. For each rational numberr, X \ r is open and dense in R. Thus

∅ = I ∩Q =

⋂r∈Q

X \ r

∩ ⋂n∈N

Un

is a countable intersection of dense open sets. Applying the Baire Category Theorem however, we expectthat ∅ is dense in R, a contradiction.

Spring 2003 #3. Find a subset S of the real numbers R such that both (i) and (ii) hold for S:

(i) S is not the countable union of closed sets.

(ii) S is not the countable intersection of open sets.

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Let A be a subset of [0, 1] that is not a countable union of closed sets, and let B be a subset of [2, 3] thatis not a countable intersection of open sets. (Irrationals and rationals for instance, respectively.) We showthat S := A ∪B satisfies (i) and (ii).

Suppose for the sake of contradiction that S is the countable union of closed sets Fn∞n=1. Then

A = S ∩ [0, 1] =

( ∞⋃n=1

Fn

)∩ [0, 1] =

∞⋃n=1

(Fn ∩ [0, 1]).

Note that Fn ∩ [0, 1] is closed for each n, so A is a countable union of closed sets, a contradiction.Likewise, if S is the countable intersection of open sets, it follows that B is the countable intersection of

open sets, a contradiction. Hence S satisfies (i) and (ii).

Spring 2002 #1. Prove that the closed interval [0, 1] is connected.

Suppose there exist disjoint non-empty open sets A,B such that A ∪B = [0, 1]. Suppose without loss ofgenerality that 1 ∈ B. Clearly [0, 1] is bounded, thus A is bounded, so by the least upper bound principle,we can define

c = sup(A).

Since [0, 1] is closed, c ∈ [0, 1]. We show that c cannot be in either A or B. Suppose for the sake ofcontradiction that c ∈ A. Note that c < 1 since 1 ∈ B and A and B are disjoint. Since A is open, thereexists some ball of radius ε > 0 at c such that B(c, ε)∩[0, 1] ∈ A. But then c+min(ε, 1−c) ∈ A, contradictingthat c = sup(A).

Suppose for the sake of contradiction that c ∈ B. If c = 0, then A = 0, which is closed, a contradiction.Hence c > 0. Since B is open, there exists some ball of radius ε at c such that B(c, ε)∩ [0, 1] ∈ B. But thenc−min(ε, c) is an upper bound for A, contradicting that c = sup(A).

Winter 2002 #3. Prove that the open ball in R2

(x, y) ∈ R2 : x2 + y2 < 1

is connected. [You may assume that intervals in R are connected. You should not just quote other generalresults, but give a direct proof.]

Lemma: The image of a connected set under a continuous function is connected.Proof: Let S be connected and f be continuous. Suppose for the sake of contradiction that f(S) is

disconnected, so f(S) = A ∪ B, with A and B disjoint non-empty open sets. It follows that f−1(A)and f−1(B) are disjoint and non-empty. Since f is continuous, f−1(A) and f−1(B) are also open. ButS = f−1(A) ∪ f−1(B), so S is not connected, a contradiction.

Lemma: Let Sα ∈M ⊂ X be connected sets. Suppose ∩αSα 6= ∅. Then⋃Sα is connected.

Proof: Let S =⋃Sα = G ∪H, where G,H are non-empty disjoint open sets. Choose x0 ∈

⋂α Sα. Fix

α. Note Sα = (Sα ∩G)∪ (Sα ∩H) and x0 ∈ Sα. If x0 ∈ G, since Sα is connected, we get Sα ∩H = ∅. Sincethis holds for all α, S ∩H = ∅. Since H ⊂ ∪Sα, H = ∅, a contradiction.

Let θ ∈ [0, 2π) and define f : R→ R2 by fθ(t) = (t cos(θ), t sin(θ)). Then f is continuous, so by the firstlemma fθ([0, 1]) is open for each θ. We may write the open unit ball as

⋃θ∈[0,2π) fθ([0, 1]). Note also that

(0, 0) ∈ fθ([0, 1]) for each θ. Hence by the second lemma,⋃θ∈[0,2π) fθ([0, 1]) is connected, so the unit ball is

connected.

Winter 2002 #2. Prove that the unit interval [0, 1] is sequentially compact, i.e., that every infinite sequencehas a convergent subsequence. [Prove this directly. Do not just quote general theorems like Heine-Borel].

Let (xn)∞n=1 be an infinite sequence in [0, 1]. Clearly this sequence is bounded. Let I0 = [0, 1]. Let n0 = 0.If the left half of I0 contains infinitely many terms of (xn)∞n=2, set I1 = [0, 1/2]. Otherwise, the right half ofI0 must contain infinitely many terms of the sequence; set I1 = [1/2, 1]. Now assume inductively that k ≥ 1

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and we have chosen nj ∈ Ij for all j < k and constructed Ik of length 2−k such that infinitely many terms ofthe sequence (xn)∞n=1 lie in Ik. Select nk > nk−1 such that nk ∈ Ik. If the left half of Ik contains infinitelymany terms of (xn)∞0 , set Ik+1 to be the left half of Ik. Otherwise, the right half of Ik contains infinitelymany terms of (xn)∞n=0; set Ik+1 to be the right half of Ik. This completes the induction, so we have aninfinite subsequence (xnk

)∞k=0 such that xnk∈ Ik for each k. Since Ik+1 ⊂ Ik for each k, with the length of

Ik given by 2−k, (xnk)∞k=0 is a Cauchy sequence. Thus this subsequence converges to some x ∈ [0, 1], since

[0, 1] is closed.

Fall 2009 #1. (i) For each n ∈ N let fn : N → R be a function with |fn(m)| ≤ 1 for all m,n ∈ N.Prove that there is an infinite subsequence of distinct positive integers ni, such that for each m ∈ N, fni

(m)converges.

(ii) For ni as in (i), assume that in addition limm→∞ limi→∞ fni(m) exists and equals 0. Prove or disprove:

The same holds for the reverse double limit limi→∞ limm→∞ fni(m).

(i) Consider the set Ω of functions from N to R whose images lie in [−1, 1]. Define the norm

d(f, g) = supm∈N|f(m)− g(m)|

on Ω, so that (Ω, d) is a metric space. Note that d(f, g) ≤ 2 for any f, g ∈ Ω, so clearly (Ω, d) is totallybounded. It also follows easily that (Ω, d) is closed. Thus by a well-known theorem for metric spaces, Ω issequentially compact, so there exists a subsequence (fni

)∞i=0 of (fn)∞n=0 such that (fni)∞i=0 converges with

respect to d. This implies that (fni(m))∞i=1 converges with respect to the usual norm on R for all m ∈ N.

(ii) Consider fn(m) = 1 for n < m and fn(m) = 0 for n ≥ m. Then fn ∈ Ω, and

limm→∞

limi→∞

fni(m) = lim

m→∞0 = 0.

However,limi→∞

limm→∞

fni(m) = lim

i→∞1 = 1.

This serves as a counterexample to the given statement.

Spring 2002 #4; Spring 2003 #1; Spring 2009 #6. (a) Define uniform continuity of a functionF : X → R, X a metric space.

(b) Prove that a function f : (0, 1)→ R is the restriction to (0, 1) of a continuous function F : [0, 1]→ R ifand only if f is uniformly continuous on (0, 1).

(a) F : X → R is uniformly continuous if for all ε > 0, there exists δ > 0 such that whenever d(x, y) < δ,|F (x)− F (y)| ≤ ε.

(b) Suppose F : [0, 1] → R is continuous and f = F |(0,1). Show f is uniformly continuous. Letε > 0. For each x ∈ [0, 1], let δx > 0 be such that if |x − y| ≤ δx, then |F (x) − F (y)| ≤ ε. Note that⋃n∈Nx : δx < 1/(n + 1) = [0, 1]. Since [0, 1] is compact, there exists a finite subcover of [0, 1] consisting

of sets x : δx < 1/(n+ 1). Since these are decreasing as n increases, there exists some natural number Nsuch that x : δx < 1/N = [0, 1]. Thus f is uniformly continuous. (Faster: F is continuous on a compactset, so F is uniformly continuous. Thus f is uniformly continuous.)

Suppose f is uniformly continuous on (0, 1). Show there exists a continuous function F : [0, 1] → Rsuch that F |(0,1) = f . Define F : [0, 1] → R such that F (x) = f(x) for x ∈ (0, 1) and F (0) = f(0+),F (1) = f(1−). By construction, F is continuous at 0 and 1. Since f is uniformly continuous on (0, 1), F iscontinuous on (0, 1). Thus F is continuous on [0, 1]. We also have F |(0,1) = f by construction.

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Spring 2004 #2. Is f(x) =√x uniformly continuous on [0,∞)? Prove your assertion.

Spring 2006 #5. Prove that if 0 < α < 1, then F (x) = xα is unformly continuous on [0,∞).Fall 2008 #1. For which of the values a = 0, 1, 2 is the function f(t) = ta uniformly continuous on

[0,∞)? Prove your assertions.

Consider f(t) = tα for 0 < α < 1. Let ε > 0 and take δ = ε/α. If x, y ∈ [1,∞) and |x− y| < δ, then bythe mean value theorem, there exists c ∈ (x, y) such that

f(x)− f(y) = f ′(c)(x− y) = αcα−1(x− y) ≤ α(x− y) ≤ ε.

Thus f is uniformly continuous on [1,∞). Now xα is continuous on [0, 1], so since [0, 1] is compact, xα isuniformly continuous on [0, 1].

For each ε > 0, there exists δ1 > 0 such that if x, y ∈ [0, 1] and |x−y| ≤ δ1, then |f(x)−f(y)| ≤ ε/2. Therealso exists δ2 > 0 such that if x, y ∈ [1,∞) and |x− y| ≤ δ2, then |f(x)− f(y)| ≤ ε/2. Let δ = min(δ1, δ2).Suppose x, y ∈ [0,∞). If x, y ∈ [0, 1] or x, y ∈ [1,∞), clearly |f(x)− f(y)| ≤ ε. Otherwise, suppose withoutloss of generality that x ∈ [0, 1] and y ∈ [1,∞). Then if |x− y| ≤ δ, |x− 1| ≤ δ and |y − 1| ≤ δ, so

|f(x)− f(y)| ≤ |f(x)− f(1)|+ |f(1)− f(y)| ≤ ε/2 + ε/2 = ε.

Thus f is uniformly continuous on [0,∞).Clearly f(t) = ta is uniformly continuous on [0,∞) for a = 0, 1. But f(t) = t2 is not uniformly continuous

on [0,∞). Let δ > 0. Note that by selecting x = n, y = n+ δ for n ≥ 1/(2δ), we have |x− y| ≤ δ, but

|f(y)− f(x)| = 2nδ + δ2 ≥ 2nδ ≥ 1.

Thus f is not uniformly continuous on [0,∞).

Fall 2009 #6. Consider the function f(x, y) = sin3(xy) + y2|x| defined on the region S ⊂ R2 given by

S = (x, y) ∈ R2;x2010 + y2010 ≤ 1.

Define what it means for f to be uniformly continuous on S and prove that f is indeed uniformly continuous.(You can use any theorem you wish in the proof, as long as it is stated correctly and you justify properly whyit can be applied, e.g., if you are using a general theorem on continuous functions, show that the function inquestion is indeed continuous, and if you are using a metric property of a set explain why it has it.)

We say f is uniformly continuous on S if given ε > 0, there exists δ > 0 such that for any (x, y), (x′, y′) ∈ Swith |(x, y)− (x′, y′)|2 ≤ δ, we have

|f(x, y)− f(x′, y′)| ≤ ε.

Clearly the projections from R2 to 0×R and R×0 are continuous. Also, |x| and sin(x) are continuous.Then f(x, y) is the composition, addition, and product of continuous functions, hence it is continuous.

We show that S is a compact set, so since f is continuous on the compact set S, f is uniformly continuouson S. If |x| > 1 or |y| > 1, then x2010 + y2010 > 1, so (x, y) /∈ S. Thus S is bounded. Let ((xn, yn))∞n=1 be asequence of points in S which converges to some (x, y). By definition of S,

x2010n + y2010

n ≤ 1

for all n. Now g((x, y)) = x2010 + y2010 is a continuous function on R2 (iterating that a product or sum ofcontinuous functions is continuous). Thus (g((xn, yn)))∞n=1 approaches g(x, y). The terms in (g((xn, yn)))∞n=1

are inside the closed set [0, 1], hence g(x, y) ∈ [0, 1]. Hence x2010 + y2010 ≤ 1, so S is closed. As a closed andbounded set in R2, S is compact.

Spring 2012 #3. Prove the Bolzano-Weierstrass theorem in the following form: Each sequence (an)n∈N ofnumbers an in the closed interval [0, 1] has a convergent subsequence.

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Keep bisecting the range and continuing on in the region with infinitely many terms of the sequence.

Fall 2004 #6. The Bolzano-Weierstrauss Theorem in Rn states that if S is a bounded closed subset of Rnand (xn) is a sequence which takes values in S, then (xn) has a subsequence which converges to a point inS. Assume this statement known in case n = 1, and use it to prove the statement in case n = 2.

Suppose S is a bounded closed subset of R2 and ((xn, yn))∞n=1 is a sequence which takes values in S.Define Sx and Sy to be the projections of S into the first and second coordinates. Then (xn)∞n=1 is a sequencein Sx. By the Bolzano-Weierstrauss Theorem on R, (xn)∞n=1 has a subsequence (xn1)∞n=1 which converges tosome x ∈ Sx.

By the Bolzano-Weierstrauss Theorem in R, the sequence (yn1)∞n=1 in Sy has a subsequence (yn2)∞n=1

which converges to y ∈ Sy. It follows that (xn2)∞n=1 converges to x in Sx and

|(xn2, yn2)− (x, y)|2 = (|xn2 − a|2 + |yn2 − y|2)1/2.

Letting n→∞, the expression on the right approaches 0, hence

((xn2, yn2)∞n=1

is a subsequence of((xn, yn))∞n=1

which converges to (x, y). Since S is closed, (x, y) ∈ S, so we have shown the Bolzano-Weierstrauss Theoremholds in R2.

Spring 2007 #10. Suppose the functions fn(x) on R satisfy:

(i) 0 ≤ fn(x) ≤ 1 for all x ∈ R and n ≥ 1.

(ii) fn(x) is increasing in x for every n ≥ 1.

(iii) limn→∞ fn(x) = f(x) for each x ∈ R, where f is continuous on R.

(iv) limx→−∞ f(x) = 0 and limx→∞ f(x) = 1.

Show that fn(x)→ f(x) uniformly on R.

For any x, y ∈ R with x ≤ y, by condition (ii),

fn(x)− fn(y) ≤ 0

for all n ≥ 1. Taking the limit as n→∞ and using condition (iii)

f(x)− f(y) ≤ 0.

Thus f is increasing.Taking the limit as n→∞ in condition (i) and using condition (iii), we find

0 ≤ f(x) ≤ 1

for all x ∈ Rn.Since [a, b] is compact and f is continuous, f is uniformly continuous on [a, b]. Let ε > 0. Using condition

(iv), select a and b such that f(a) < ε/2 and f(b) > 1− ε/2. Then for any x, y < a, since f is increasing,

|f(x)− f(y)| ≤ f(a)− 0 ≤ ε/2 < ε.

Likewise, if x, y > b, then

|f(x)− f(y)| ≤ 1− f(b) ≤ 1− (1− ε/2) = ε/2 < ε.

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Now since [a, b] is compact and f is continuous, f is uniformly continuous on [a, b]. Thus there exists δ > 0such that if x, y ∈ [a, b] and |x− y| < δ, then |f(x)− f(y)| ≤ ε/2. Choose δ small enough to prohibit x < aand y > b. If x < a and a < y < b with |x− y| ≤ δ, then |y − a| ≤ δ, so

|f(x)− f(y)| ≤ |f(x)− f(a)|+ |f(a)− f(y)| ≤ ε/2 + ε/2 = ε.

Thus f is uniformly continuous on R.Let ε > 0. Select δ such that if x, y ∈ [a, b] and |x − y| ≤ δ, then |f(x) − f(y)| ≤ ε/2. Partition [a, b]

into a finite number of intervals of length less than δ. That is, choose a = x0 ≤ x1 ≤ · · · ≤ xn = b suchthat xi+1 − xi ≤ δ for all i. Since fn converges to f pointwise, there exists N such that for all n ≥ N ,|f(xi)− fn(xi)| ≤ ε/4 for all i. For any y ∈ [xi, xi+1] and any n ≥ N ,

|fn(y)− f(y)| ≤ |fn(y)− fn(xi)|+ |fn(xi)− f(xi)|+ |f(xi)− f(y)|

≤ (f(xi+1)− ε/4)− (f(xi)− ε/4) + |fn(xi)− f(xi)|+ |f(xi)− f(y)|

≤ ε+ ε/4 + ε/2 < 2ε.

Thus the fn converge to f uniformly on [a, b].By the choice of a, for any y < a, and any n ≥ N ,

|fn(y)− f(y)| ≤ fn(a)− 0 ≤ f(a) + ε/4 ≤ ε/2 + ε/4 < ε.

Likewise, for any y > b, and any n ≥ N ,

|fn(y)− f(y)| ≤ ε.

Thus (fn)∞n=1 converges uniformly to f on R.

Spring 2010 #7. Let fn be a sequence of real-valued functions on the line, and assume that thereis a B < ∞ such that |fn(x)| ≤ B for all n and x. Prove that there is a subsequence fnk

such thatlimk→∞ fnk

(r) exists for all rational numbers r.

Define (Ω, d) as in Fall 2009 #1, with Ω consisting of function from Q → R, and take d to be thesupremum over rational inputs. Then fn|Q ∈ Ω for each n. Then (Ω, d) is a metric space.

We see that Ω is totally bounded from the condition |fn(x)| ≤ B. Let (fm)∞m=1 be a sequence in Ω whichconverges to f with respect to d. Clearly |f(x)| ≤ B for all x ∈ Q, so Ω is closed. Thus Ω is sequentiallycompact, so there exists a subsequence fnk

∞k=1 such that this subsequence converges to some f ∈ Ω withrespect to d. It follows that we have pointwise convergence on all rationals, as desired.

Spring 2004 #4. Are there infinite compact subsets of Q? Prove your assertion.

Yes, 1/n : n ∈ N ∪ 0 is an infinite compact subset of Q. Clearly this set is bounded by 1, and itcontains its only limit point of 0, hence it is closed. Thus it is compact.

Fall 2008 #2. Suppose that A is a non-empty connected subset of R2.

(a) Prove that if A is open, then it is path connected.

(b) Is part (a) true if A is closed? Prove your assertion.

(a) Let a ∈ A. Define H to be the subset of points in A which can be joined to a by a path in A. LetK = A \H.

Let x ∈ H. Since A is open, there exists ε > 0 such that Bε(x) ⊂ A. Given any y ∈ Bε(x), there is astraight line path g in Bε(x) ⊂ A connecting x and y. Since x ∈ H, there is a path f in A joining a to x.

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Thus traversing f and then g forms a path from a to y. It follows that y ∈ H, hence Bε(x) ⊂ H. Thus H isopen.

Let x ∈ K. Since A is open, there exists ε > 0 such that Bε(x) ⊂ A. If any point in Bε(x) could bejoined to a by a path in A, then so could x, a contradiction. Hence Bε ⊂ K, so K is open.

Clearly H ∩K = ∅, H ∪K = A, and a ∈ H, so H is non-empty. Since A is connected, we must haveK = ∅, so H = A. Thus A is path connected.

(b) No. Consider the set

A = (x, sin(1

x)) : x ∈ (0, 1] ∪ (0, x) : x ∈ [−1, 1].

We show that A is closed and connected, but not path connected. For convenience, let B = (x, sin( 1x )) :

x ∈ (0, 1] and C = (0, x) : x ∈ [−1, 1]. Then C is the set of limit points of B which are not already inB. Since C is closed, A = B ∪ C is closed. Defining f : (0, 1] → R2 by f(x) = (x, sin( 1

x )), we see that f iscontinuous, thus since (0, 1] is connected B = f((0, 1]) is connected as well. Clearly C is connected.

Suppose for the sake of contradiction that A = B ∪ C is not connected. Then there exist disjoint non-empty open sets U, V with A ⊂ U ∪ V . Since B and C are connected, we can assume without loss ofgenerality that B ⊂ U and C ⊂ V . Now each element in C is a limit point of B, so every ball centered atsome (0, x) ∈ C must contain infinitely many elements of B. Since U is open, U and V are not disjoint, acontradiction. Hence A is connected.

Now any path in A connecting some point of B with some point of C is a continuous map f : [0, 1]→ Awith f(0) ∈ B. But such a map must have f([0, 1]) entirely contained within B. Thus there does not exista path between an element of B and an element of C.

Spring 2011 #11. Show that a connected subset A ⊆ R is arcwise connected (= path-connected).

Let A be a connected subset of R. It follows that A is an interval. Let x, y ∈ A. Then necessarily[x, y] ⊂ A. Define f : [0, 1] → [x, y] by f(t) = x + t(y − x). Clearly f is continuous, and f(0) = x andf(1) = y, so f is a path connecting x and y. Hence A is path-connected.

Spring 2004 #6. Let || || be any norm on Rn.

(a) Prove that there exists a constant d with ||x|| ≤ d||x||2 for all x ∈ Rn, and use this to show thatN(x) = ||x|| is continuous in the usual topology on Rn.

(b) Prove that there exists a constant c with ||x|| ≥ c||x||2. (Hint: use the fact that N is continuous on thesphere x : ||x||2 = 1).

(c) Show that if L is an n-dimensional subspace of an arbitrary normed vector space V , then L is closed.

(a) Define d =√∑

||ei||2. Write x =∑xiei. By the Cauchy-Schwarz inequality,

||x|| = ||∑

xiei|| ≤∑||xiei|| =

∑|xi|||ei||

= (|x1|, . . . , |xn|) · (||e1||, . . . , ||en||) ≤√∑

x2i ·√∑

||ei||2 = d||x||2.

Let ε > 0. For any x, y ∈ Rn with ||x− y||2 ≤ ε/(d+ 1), by the reverse triangle inequality,

||x|| − ||y|| ≤ ||x− y|| ≤ d||x− y||2 ≤ d(ε/(d+ 1)) ≤ ε.

By symmetry, we conclude|||x|| − ||y||| ≤ ε.

Thus N is continuous with respect to the usual topology on Rn.

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(b) It follows from (a) that N is continuous on the sphere T := x : ||x||2 = 1. Since T is a compactset, N achieves its minimum c on T . For any x ∈ Rn with x 6= 0, since x

||x||2 ∈ T ,

||x|| = ||x||2||x

||x||2|| ≥ c||x||2.

(c) Since V is n-dimensional, there exists an isomorphism T : Rn → V . Define || || : Rn → R by||x|| = ||Tx||V .

From parts (a) and (b), there exist constants c and d such that for any x ∈ Rn,

c||x||2 ≤ ||x|| ≤ d||x||2.

Let (yk)∞k=1 be a sequence of points in L that converges to some v ∈ V with respect to the norm || ||V ofV . In particular, (yk)∞k=1 is a Cauchy sequence with respect to || ||V . It follows that

c||T−1yi − T−1yj ||2 ≤ ||T−1yi − T−1yj || = ||yi − yj ||V ≤ d||T−1yi − T−1yj ||2.

Let ε > 0. There exists N such that for all i, j ≥ N , ||yi − yj ||V ≤ cε. It follows that

||T−1yi − T−1yj ||2 ≤ ε.

Hence (T−1yk)∞k=1 is a Cauchy sequence with respect to || ||2. Since this norm is complete, there exists someT−1y ∈ Rn such that (T−1(yn))∞n=1 converges to T−1(y). It follows that (yk)∞k=1 converges to y with respectto || ||V (using the other side of the inequality above). Hence L is closed.

Fall 2005 #8. For a real n × n matrix A, let TA : Rn → Rn be the associated linear mapping. Set||A|| = supx∈Rn||TAx|| : ||x|| = 1 (here ||x|| = usual euclidean norm, i.e.,

||(x1, . . . , xn)|| = (x21 + · · ·+ x2

n)1/2).

(a) Prove that ||A+B|| ≤ ||A||+ ||B||.

(b) Use part (a) to check that the set M of all n× n matrices is a metric space if the distance function d isdefined by

d(A,B) = ||B −A||.

(c) Prove that M is a complete metric space with this “distance function”. (Suggestion: The ij-th elementof A =< TAej , ei >, here ei = (0, . . . , 1, . . . , 0), with a 1 in the ith position.)

(a) By definition of the matrix norm, for all x ∈ Rn, it follows that

||TAx|| ≤ ||A||||x||.

Now let x ∈ Rn with ||x|| = 1. Then

||TA+Bx|| = ||TAx+ TBx|| ≤ ||TAx||+ ||TBx|| ≤ ||A|| · ||x||+ ||B|| · ||x||

= (||A||+ ||B||) · ||x|| ≤ ||A||+ ||B||.Thus taking the supremum over all such x,

||A+B|| ≤ ||A||+ ||B||.

(b) Clearly this distance function is non-negative and symmetric. If d(A,B) = 0, then ||B − A|| = 0. Inparticular, this implies that ||(TB − TA)(ei)|| = 0 for all i, so TB − TA = 0. This implies that A = B.

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Finally, for matrices A,B,C, by part (a),

d(A,B) = ||B −A|| ≤ ||(B − C) + (C −A)|| ≤ ||B − C||+ ||C −A|| = d(B,C) + d(C,A).

Thus d is a metric on real n× n matrices.

(c) Suppose (An)∞n=1 is a Cauchy sequence with respect to the matrix norm. Thus there exists N suchthat for all j, k ≥ N ,

||Aj −Ak|| ≤ ε.

This implies that for all j, k ≥ N ,

n∑`=1

(TAj− TAk

)2`i = ||(TAj

− TAk)ei|| ≤ ε.

Taking ε < 1, it follows that each entry of TAj − TAkis less than ε. Thus ((TAn)ij)

∞n=1 forms a Cauchy

sequence for each i, j. Define a matrix A by

Aij = limn→∞

(TAn)ij .

It follows that (An)∞n=1 converges to A with respect to the matrix norm. Thus the set of n by n matrices Mis a complete metric space with this distance function.

Spring 2005 #10. Consider the set of f : [0, 1]→ R that obey

|f(x)− f(y)| ≤ |x− y| and

∫ 1

0

f(x)dx = 1.

Show that this is a compact subset of C([0, 1]).

Call this set of functions S. The first condition clearly implies that S ⊂ C([0, 1]). We will show that Sis closed, bounded, and equicontinuous, thus by the Arzela-Ascoli theorem, S is compact.

Let (fn)∞n=1 be a sequence of functions in S converging to f . We want to show f ∈ S. Let ε > 0. SelectN such that for all n ≥ N ,

supx∈[0,1]

|f(x)− fn(x)| ≤ ε/2.

Then for any x, y ∈ [0, 1],

|f(x)− f(y)| ≤ |f(x)− fn(x)|+ |fn(x)− fn(y)|+ |fn(y)− f(y)| ≤ ε+ |x− y|.

Letting ε→ 0, we have|f(x)− f(y)| ≤ |x− y|.

Since (fn)∞n=1 converges to f in the supremum norm, they converge uniformly to f . Since the fn arecontinuous, it follows that f is continuous. Also, because we have uniform convergence,∫ 1

0

f(x)dx =

∫ 1

0

limn→∞

fn(x)dx = limn→∞

∫ 1

0

fn(x)dx = limn→∞

1 = 1.

Hence f ∈ S.Let f ∈ S. Suppose for the sake of contradiction that there exists y ∈ [0, 1] such that f(y) ≥ 3. Then for

any x ∈ [0, 1],|f(x)− f(y)| ≤ |x− y| ≤ 1,

hence f(x) ≥ 2. Then ∫ 1

0

f(x) ≥∫ 1

0

2 = 2,

23


a contradiction. Thus f(y) ≤ 3 for all y ∈ [0, 1]. Likewise, f(y) ≥ −3 for all y ∈ [0, 1], so f is uniformlybounded.

Finally, uniform equicontinuity on S follows immediately from the first condition of functions in S.

Spring 2006 #6. Let W be the subset of the space C[0, 1] of real-valued continuous functions on [0, 1]satisfying the conditions:

|f(x)− f(y)| < |x− y| and

∫ 1

0

f(x)2dx = 1.

Edit: The first condition must be strict.

(a) Prove that W is uniformly bounded, i.e., there exists M > 0 such that |f(x)| ≤ M for all x ∈ [0, 1].Hint: Show first that |f(0)| ≤ 2 for all f ∈W .

(b) Prove that W is a compact subset of C[0, 1] under the sup norm ||f ||∞ = supx∈[0,1] |f(x)|.

Essentially the same proof as the previous exercise. Note f2n converges uniformly to f2.

Fall 2007 #1. Let S be a subset of Rn with the distance function d(x, y) = ((x1−y1)2 + · · ·+(xn−yn)2)1/2

so that (S, d|S×S) is a metric space.

(a) Given y ∈ S, is E = x ∈ S : d(x, y) ≥ r a closed set in S?

(b) Is the set E in part (a) contained in the closure of x ∈ S : d(x, y) > r in S?

(a) Yes. Let x ∈ S be a limit point of E. Then there exists a sequence (xn)∞n=1 in E converging to x.We have

d|S×S(xn, y) = d(xn, y) ≥ r

for each n ≥ 1. Let ε > 0. Select N such that for all n ≥ N , d(xn, x) < ε. It follows that

d(xn, y) ≤ d(xn, x) + d(x, y)

thusd(x, y) ≥ d(xn, y)− d(xn, x) ≥ r − ε.

Since ε was arbitrary, d(x, y) ≥ r. Thus because we assumed x ∈ S, it follows that x ∈ E. Thus E is aclosed set in S.

(b) No. Let S = 0 ∪ 1 ∪ 2 ⊂ R, y = 0, and r = 1. Then

x ∈ S : d(x, y) > r = x ∈ S : d(x, 0) > 1 = 2.

The closure of this set in S is 2, but

E := x ∈ S : d(x, y) ≥ r = x ∈ S : d(x, 0) ≥ 1 = 1 ∪ 2,

which is not a subset of 2.

Spring 2007 #12. Let c0 be the normed space of real sequences x = (x1, x2, . . .) such that limk→∞ xk = 0with the supremum norm ||x|| = supk |xk|.

(a) Show that c0 is complete.

(b) Is the unit ball x ∈ c0 : ||x|| ≤ 1 compact? Prove your answer.

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(c) Is the set E = x ∈ c0 :∑k k|xk| ≤ 1 compact? Prove your answer.

(a) Let (x(m))∞m=0 be a Cauchy sequence of elements of c0. Then for any n,m, k ∈ N, we have

|x(m)k − x(n)

k | ≤ ||x(m) − x(n)||

so that x(m)k is a Cauchy sequence of real numbers for each k and thus converges. Define x so that

xk = limm→∞

x(m)k .

We show that x is in c0. Let ε > 0. Since x(m) is a Cauchy sequence, let N be an integer such thatm,n ≥ N implies

||x(m) − x(n)|| < ε/2.

Since x(N) ∈ c0, choose M such that k ≥M implies

|x(N)k | < ε/2.

Then for k ≥M , we have

|xk| ≤ |xk − x(N)k |+ |x(N)

k | ≤ ε/2 + ε/2 = ε.

Hence x ∈ c0.For m ≥ N and k ∈ N, we also have

|xk − x(m)k | ≤ |x(n)

k − x(m)k |+ |x(n)

k − xk|.

For large enough n, by the definition of xk, this is small. Since k was arbitrary,

||x− x(m)|| ≤ ε

for m ≥ N , so x(m) → x. Hence c0 is complete.

(b) No. Suppose for the sake of contradiction that x ∈ c0 : ||x|| ≤ 1 is compact. For each n ∈ N,

define the sequence x(n) by x(n)k = 1 if k ≤ n and 0 if k > n. Then each x(n) ∈ c0, and ||x(n)|| = 1,

so x(n) ∈ x ∈ c0 : ||x|| ≤ 1. By sequential compactness and completeness, there exists a subsequence(x(nk))∞k=1 which converges to some x ∈ x ∈ c0 : ||x|| ≤ 1. Thus as k →∞,

||x− x(nk)|| → 0.

In particular, for each j,

|xj − 1| = limk→∞

|xj − x(nk)j | = 0.

Hence xj = 1. But then limj→∞ xj = 1, so x /∈ c0, a contradiction.

(c) Yes. Let x(n) be a sequence in E. Note that |x(n)k | ≤ 1 for all k ∈ N, so for each k, x(n)

k has aconvergent subsequence. Thus we may choose a subsequence y(n,1)∞n=1 of x(n) and y1 ∈ R such that

limn→∞

y(n,1)1 = y1.

Continuing, as in the proof of the Arzela-Ascoli Theorem, we may construct subsequences y(n) of x(n) andnumbers y1, y2, . . . such that

limk→∞

y(n)k = yk

for all k. We claim y = (y1, y2, . . .) ∈ E and that limn→∞ y(n) = y.

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Fix an integer N . ThenN∑k=1

k|yk| = limn→∞

N∑k=1

k|y(n)k | ≤ 1,

since y(n)k ∈ E. Letting N →∞, it follows that y ∈ E.

Let ε > 0. Select N large enough so that 1N < ε. Then choose N1 ≥ N such that

||y(n)k − yk|| < ε

for k = 1, . . . , N and all n ≥ N1. This is valid since limn→∞ y(n)k = yk and 1, . . . , N is finite. Then for

n ≥ N1 and k ≥ N , we have that

|y(n)k − yk| ≤ |y(n)

k |+ |yk| ≤ 1/N + 1/N = 2/N < 2ε,

whereas for k < N we have that |y(n)k − yk| < ε by construction. Thus for n ≥ N we have that

||y(n) − y|| < 2ε

so that y(n) is a subsequence of x(n) which converges to y. Hence E is sequentially compact and thuscompact.

Spring 2012 #2. Recall that f : [a, b]→ R is convex if for all x, y ∈ [a, b] and α ∈ [0, 1], f(αx+(1−α)y) ≤αf(x) + (1− α)f(y). Let fn : [a, b]→ R be convex functions and suppose that f(x) := limn→∞ fn(x) existsat all x ∈ [a, b] and is continuous on [a, b]. Prove that fn → f uniformly.

Assume for the sake of contradiction that fn does not converge uniformly to f . Let ε > 0. Pass to asubsequence where |fn − f |sup ≥ ε. Then there exists a sequence (xn)∞n=1 with elements in [a, b] such that|fn(xn) − f(xn)| ≥ ε/2. By compactness of [a, b], there exists a subsequence (also denoted by (xn)∞n=1) of(xn)∞n=1 which converges to some x ∈ [a, b]. We may choose xn to be monotone and without loss of generality,assume they are non-decreasing.

Now we localize the problem to [a1, b1], where b1 = x. Since f is continuous, there exists c such that

|f(z)− c| < ε/20

for all z ∈ [a1, b1] (choose a1 such that this holds).We will use convexity of the fn to show pointwise convergence fails somewhere. We are given that

fn(a1)→ f(a1) and fn(b1)→ f(b1) as n→∞. Choose N large enough that

|fn(a1)− c| ≤ ε/10 and |fn(b1)− c| ≤ ε/10.

From the construction of xn, for all n,

fn(xn) ≤ f(xn)− ε/2.

Consider the line from (a1, c+ ε/10) to (b1, c− ε/4). This line intersects the horizontal line at c− ε/10at some point z ∈ (a1, b1). By convexity of fn, fn(z) must lie below this line on [a1, xn], so choosing n largeenough that z < xn,

fn(z) < c− ε/10.

Since this holds for each n and fn(z)→ f(z), taking n→∞ yields

f(z) ≤ c− ε/10.

But this contradicts |f(z)− c| < ε/20 above. Hence fn does converge uniformly to f .

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Fixed point

Spring 2008 #1. Let g ∈ C([a, b]), with a ≤ g(x) ≤ b for all x ∈ [a, b]. Prove the following:

(i) g has at least one fixed point p in the interval [a, b].

(ii) If there is a value γ < 1 such that

|g(x)− g(y)| ≤ γ|x− y|

for all x, y ∈ [a, b], then the fixed point is unique, and the iteration

xn+1 = g(xn)

converges to p for any initial guess x0 ∈ [a, b].

(i) Note that g(x)− x is continuous on [a, b], non-negative at x = a and non-positive at x = b. Thus bythe intermediate value theorem, there exists some p ∈ [a, b] such that g(p)− p = 0, or g(p) = p.

(ii) Suppose there exists γ < 1 such that

|g(x)− g(y)| ≤ γ|x− y|

for all x, y ∈ [a, b]. Suppose both p, q ∈ [a, b] were fixed points of g. Then if p 6= q,

|p− q| = |g(p)− g(q)| ≤ γ|p− q| < |p− q|,

a contradiction. Hence p = q, so the fixed point of g is unique.Let x0 ∈ [a, b] be an initial guess and consider the iterated sequence

xn+1 = g(xn)

defined for all n ≥ 0. We first show that (xn)∞n=0 converges. Follow the procedure below to show this is aCauchy sequence and converges to a fixed point of g. Since the fixed point of g is unique, it must be that(xn)∞n=0 converges to p.

Fall 2009 #2. (i) Let X be a complete metric space with respect to a distance function d. We say that amap T : X → X is a contraction if for some 0 < λ < 1 and all x, y ∈ X:

d(f(x), f(y)) ≤ λd(x, y).

Prove that if T is a contradiction then it has a fixed point, i.e., there is an x ∈ X such that T (x) = x.

(ii) Using (i) show that given a differentiable function f : R → R whose first derivative satisfies f ′(x) =

e−x2 − e−x4

there exists α ∈ R with f(α) = α.

(i) Let T : X → X be a contraction. We assume X is non-empty, so there exists x0 ∈ X. Inductivelydefine xn+1 = T (xn) for all n ≥ 0. A simple induction shows that for any natural number n,

d(xn+1, xn) = d(Tn+1(x0)), Tn(x0)) ≤ λn(T (x0), x0).

Thus for any natural numbers n,m,

d(xn, xn+m) ≤m−1∑i=0

d(xn+i, xn+i+1) ≤m−1∑i=0

λn+id(x0, T (x0))

27


= λnd(x0, T (x0))

m−1∑i=0

λi = λnd(x0, T (x0))1− λm

1− λ≤ λn

1− λd(x0, T (x0)).

Let ε > 0. Choose N so that λN

1−λd(x0, T (x0)) < ε. Then for any n,m ≥ N , we have shown

d(xn, xm) ≤ λn

1− λd(x0, T (x0)) ≤ λN

1− λd(x0, T (x0)) < ε.

Thus the sequence (xn)∞n=0 is a Cauchy sequence. Because X is complete, this sequence converges to somex∗ ∈ X.

Since T is Lipschitz continuous on X, T is continuous on X. Proof: Let δ = ε/λ.Recall for all n ≥ 0 we defined

xn+1 = T (xn).

Thus taking the limit as n→∞ on both sides and using the continuity of T ,

x∗ = limn→∞

T (xn) = T (x∗).

Hence x∗ is a fixed point of T .

(ii) Note that e−x2

and e−x4

are even functions and decreasing on [0,∞). Thus for any x ∈ [0, 1],

e−x2

− e−x4

≤ e−x2

− e−14

= e−x2

− 1

e≤ e−02

− 1

e= 1− 1/e.

Likewise, for any x ∈ [0, 1],

−(e−x2

− e−x4

) = e−x4

− e−x2

≤ 1− 1/e.

Thus for any x ∈ [0, 1],

|e−x2

− e−x4

| ≤ 1− 1/e.

For x > 1, x4 > x2, hence −x4 < −x2, so e−x2 − e−x4

> 0.

|e−x2

− e−x4

| = e−x2

− e−x4

≤ e−12

= 1/e.

Thus for all x ∈ R,

|f ′(x)| = |e−x2

− e−x4

| ≤ 1− 1/e.

Let x, y ∈ R. By the mean value theorem, there exists z ∈ (x, y) such that

f(x)− f(y) = f ′(z)(x− y).

Thus|f(x)− f(y)| ≤ |f ′(z)||x− y| ≤ (1− 1/e)|x− y|.

Thus f is a contraction, so by part (a), there exists α ∈ R with f(α) = α.

Fall 2011 #1. Let (X, d) be a compact metric space and let f : X → X be a map satisfying

d(f(x), f(y)) < d(x, y), for all x, y ∈ X with x 6= y.

Prove that there is a unique point x ∈ X so that f(x) = x.

Define δ : X → R by δ(x) = d(x, f(x)). For any x, y ∈ X,

δ(y) = d(y, f(y)) ≤ d(y, x) + d(x, f(x)) + d(f(x), f(y)) ≤ 2d(x, y) + δ(x),

thusδ(y)− δ(x) ≤ 2d(x, y).

28


By symmetry, this implies|δ(y)− δ(x)| ≤ 2d(x, y).

Let ε > 0. Then for any x, y ∈ X with d(x, y) < ε/2.

|δ(x)− δ(y)| ≤ ε,

so δ is a continuous function.Since X is compact and δ is continuous, there exists x0 ∈ X such that δ(x0) ≤ δ(x) for all x ∈ X.

Suppose for the sake of contradiction that δ(x0) 6= 0. Then x0 6= f(x0), so

δ(f(x0)) = d(f(x0), f(f(x0))) < d(x0, f(x0)) = δ(x0),

a contradiction. Hence δ(x0) = 0, so d(x0, f(x0)) = 0, and f(x0) = x0. Thus f has a fixed point.Suppose there exist x, y ∈ X with f(x) = x and f(y) = y. If x 6= y, then

d(x, y) = d(f(x), f(y)) < d(x, y),

a contradiction. Hence x = y, so there is a unique point x ∈ X so that f(x) = x.

Spring 2011 #12. Fall 2004 #5; Fall 2003 #7; Fall 2001 #3. (see text). Given a metric spaceM , and a constant 0 < r < 1, a continuous function T : M → M is said to be an r-contraction if it is acontinuous map and d(T (x), T (y)) < rd(x, y) for all x 6= y. A well-known fixed point theorem states thatif M is complete and T an r-contraction, then it must have a unique fixed point (don’t prove this). Thisresult is often used to prove the existence of solutions of differential equations with initial conditions.

1. Illustrate this technique for the (trivial) case

f ′(t) = f(t), f(0) = 1

by letting M be the space of continuous functions C([0, c]) for 0 < c < 1 with the uniform distance

d(f, g) = sup|f(t)− g(t)|,

and defining (Tf)(x) = 1 +∫ x

0f(t)dt. Carefully explain your steps.

2. What approximations do you obtain from the sequence

T (0), T 2(0), T 3(0), . . .?

1. Let 0 < c < 1 and M be the space of continuous functions C([0, c]) with the distance

d(f, g) = supt∈[0,c]

|f(t)− g(t)|.

Define T : M → M by (Tf)(x) = 1 +∫ x

0f(t)dt. For some f ∈ M , since f is continuous on a compact

set there exists B > 0 such that |f | ≤ B on [0, c]. Let ε > 0. Then for any x, y ∈ [0, c] with x ≤ y and|x− y| ≤ ε/B,

|(Tf)(y)− (Tf)(x)| = |∫ y

x

f(t)dt| ≤∫ y

x

Bdt = B(y − x) ≤ B(ε/B) = ε.

Hence Tf is continuous, so Tf ∈M , thus T is well-defined.It is well known that M is complete.Let (fn)∞n=1 be a Cauchy sequence in (M,d). Let ε > 0. Then there exists an integer N ≥ 1 such that

for all n,m ≥ N ,d(fn, fm) ≤ ε.

29


In particular, for any t ∈ [0, c],|fn(t)− fm(t)| ≤ ε.

Hence (fn(t))∞n=1 is a Cauchy sequence for each t ∈ [0, c]. Since R is complete, we can define f : [0, c] → Rby

f(t) = limn→∞

fn(t).

We now show that (fn)∞n=1 converge to f uniformly with respect to d. Then since the limit of a sequenceof continuous functions that converges uniformly is a continuous function, f is continuous, so f ∈M . Thuswe will have shown that M is complete.

Let ε > 0. Choose N so that for all n,m ≥ N , d(fm, fn) ≤ ε. Then for any t ∈ [0, c],

|f(t), fn(t)| ≤ |f(t)− fm(t)|+ |fm(t)− fn(t)| ≤ |f(t)− fm(t)|+ d(fm, fn) ≤ |f(t)− fm(t)|+ ε.

Letting m→∞,|f(t), fn(t)| ≤ ε.

Taking the sup of both sides over all t ∈ [0, c],

d(f, fn) ≤ ε.

Hence (fn)∞n=1 converges to f uniformly with respect to d.Let ε > 0. Select N so that d(fN , f) ≤ ε/3. Since fN is continuous, there exists δ > 0 so that if

|x− y| ≤ δ, then |fN (x)− fN (y)| ≤ ε/3. For any x, y ∈ [0, c] with |x− y| ≤ δ,

|f(x)−f(y)| ≤ |f(x)−fn(x)|+ |fn(x)−fn(y)|+ |fn(y)−f(y)| ≤ 2d(f, fn)+ |fn(x)−fn(y)| ≤ 2ε/3+ε/3 = ε.

Hence f is continuous.We now show T is continuous. Let ε > 0 and suppose f, g ∈M with d(f, g) ≤ ε/c. Then

d(Tf, Tg) = supt∈[0,c]

|(Tf)(t)− (Tg)(t)| = supt∈[0,c]

|∫ t

0

f(t)− g(t)dt| ≤ c supt∈[0,c]

(f − g) = c(d(f, g)) ≤ c(ε/c) = ε.

Thus T is continuous.Let f, g ∈M with f 6= g. It follows that

d(Tf, Tg) = supt∈[0,c]

|(Tf)(t)− (Tg)(t)| = supt∈[0,c]

∣∣∣∣∫ t

0

f(z)− g(z)dz

∣∣∣∣ ≤ c supz∈[0,c]

|f(z)− g(z)| = c(d(f, g)).

Thus T is a (c + ε)-contraction mapping. (Choose ε so that c + ε < 1.) By the given fixed point theorem,there exists a fixed point F of T .

We now show F satisfies the differential equation and initial condition. Clearly

F (0) = (TF )(0) = 1 +

∫ 0

0

F (t)dt = 1.

By the fundamental theorem of calculus, for any x ∈ [0, c],

F ′(x) = (TF )′(x) =d

dx(1 +

∫ x

0

F (t)dt) = F (x).

Thus F ′(x) = F (x) for all x ∈ [0, c]. Hence F is a solution to the given differential equation on [0, c].

2. Here we start with the constant 0 function, which is in M , and iterate T . The proof of the givenfixed point theorem shows that iterating T on any function in M will converge to the fixed point. Hence thesequence T (0), T 2(0), T 3(0), . . . converges to the solution F found in part 1.

As part of the proof, we obtain that for any natural numbers n,m,

d(Tn(0), Tn+m(0)) ≤ cn 1− cm

1− cd(T (0), 0).

30


Thus

d(F, Tn(0)) ≤ d(F, Tn+m(0)) + d(Tn+m(0), Tn(0)) ≤ d(F, Tn+m(0)) + cn1− cm

1− cd(T (0), 0).

Note that d(T (0), 0) = d(1, 0) = 1. Letting m→∞,

d(F, Tn(0)) ≤ cn

1− cd(T (0), 0) =

cn

1− c.

Thus for any t ∈ T ,

|F (t)− (Tn(0))(t)| ≤ cn

1− c.

Spring 2007 #7. Let f : R→ R be a twice continuously differentiable function with f ′′ uniformly bounded,and with a simple root at x∗ (i.e., f(x∗) = 0, f ′(x∗) 6= 0). Consider the fixed point iteration

xn = F (xn−1) where F (x) = x− f(x)

f ′(x).

Show that if x0 is sufficiently close to x∗, then there exists a constant C so that for all n,

|xn − x∗| ≤ C|xn−1 − x∗|2.

Note f ′ is continuous, so it is bounded away from zero in some open neighborhood U . Say 0 < C <f ′(x) < D there, without loss of generality. Now suppose x0 ∈ U and by induction that x1, . . . , xn−1 ∈ U .

By the mean value theorem, there exists yn−1 between x∗ and xn−1 such that

|xn − x∗| = |F (xn−1)− x∗| = |F (xn−1 − F (x∗)| = |xn−1 − x∗||F ′(yn−1)|.

Now

F ′(x) =f(x)f ′′(x)

(f ′(x))2.

Thus

|xn − x∗| = |xn−1 − x∗||f(yn−1)||f ′′(yn−1)||f ′(yn−1)|2

.

Apply the mean value theorem again to obtain zn−1 between x∗ and yn−1 such that f(yn−1) − f(x∗) =(y − x∗)f ′(zn−1). Since f(x∗) = 0, we obtain

|xn − x∗| = |xn−1 − x∗||yn−1 − x∗||f ′(zn−1)||f ′′(yn−1)|

|f ′(yn−1)|2.

Note yn−1, zn−1 ∈ U . Thus|f ′(zn−1)||f ′(yn−1)|2

≤ D

C2.

Say the second derivative is uniformly bounded by M . It follows that

|xn − x∗| ≤DM

C2|xn−1 − x∗|2.

If we forced |xn−1 − x∗| ≤ C2

MD , then |xn − x∗| ≤ |xn−1 − x∗|, hence xn ∈ U , completing the induction.

31


Inverse Function Theorem, Implicit Function Theorem

Inverse Function Theorem. Let f : U ⊂ Rn → Rn be a continuously differentiable function which hasinvertible derivative at a point p (the Jacobian determinant of f at p is non-zero). Then the inverse functionf−1 exists and is continuously differentiable in some neighborhood of f(p). Also,

Jf−1(f(p)) = [Jf (p)]−1.

Implicit Function Theorem. Let f : Rn+m → Rm be a continuously differentiable function, and let Rn+m

have coordinates (x,y). Fix a point (a,b) = (a1, . . . , an, b1, . . . , bm) with f(a,b) = c, where c ∈ Rm. If thematrix [(∂fi/∂yj)(a,b)] is invertible, then there exists an open set U containing a, an open set V containingb, and a unique continuously differentiable function g : U → V such that

(x, g(x)) : x ∈ U = (x,y) ∈ U × V : f(x,y) = c.

Fall 2001 #6. Suppose that F : R2 → R2 is a continuously differentiable function with F ((0, 0)) = (0, 0)and with the Jacobian of F at (0, 0) equal to the identity matrix (i.e., if F = (f1, f2) then ∂fi

∂xj|(0,0) = 1 if

i = j and = 0 if i 6= j). Outline a proof that there exists δ > 0 such that if a2 + b2 < δ, then there is apoint (x, y) in R2 with F (x, y) = (a, b). (Your argument will be part of the proof of the Inverse FunctionTheorem. You may use any basic estimation you need about the change in F being approximated by thedifferential of F without proof.)

See standard proof of the Inverse Function Theorem.

Fall 2004 #7. Observe that the point P = (1, 1, 1) belongs to the set S of points in R3 satisfying theequations

x4y2 + x2z + yz2 = 3.

Explain carefully how, in this case, the Implicit Function Theorem allows us to conclude that there existsa differentiable function g(x, y) such that (x, y, g(x, y)) lie in S for all (x, y) in a small open set containing(1, 1).

Define f : R3 → R by f(x, y, z) = x2y3 + x3z + 2yz4. Here n = 2 and m = 1. Here a = (1, 1) andb = 1. Note that f(a, b) = (1, 1, 1) = 3 and f is continuously differentiable. The relevant matrix of partialderivatives is

[(∂f/∂z)(a, b)] = [a31 + 8a2b

3] = [9]

which is clearly invertible. Thus by the Implicit Function Theorem, there exists an open set U containinga = (1, 1), an open set V containing b = 1 and a unique continuously differentiable function g : U → V suchthat

(x, y, g(x, y)) : (x, y) ∈ U = (x, y, z) ∈ U × V : f(x, y, z) = 3 ⊂ (x, y, z) : f(x, y, z) = 3 =: S.

Thus (x, y, g(x, y)) lie in S for all (x, y) ∈ U , where U is a small open set containing (1, 1).

Spring 2006 #4. Instead usex2y3 + x3z + 2yz4 = 4.

Prove that there exists a differentiable function g(x, y) defined in an open neighborhood N of (1, 1) in R2

such that g(1, 1) = 1 and (x, y, g(x, y)) lies in S for all (x, y) ∈ N .

Define f(x, y, z) = x2y3 + x3z + 2yz4. Note f(1, 1) = 4 and f is continuously differentiable. The matrixof partial derivatives under consideration is

[(∂f/∂z)(1, 1, 1)] = [[(x, y, z) 7→ (x3 + 6yz3)](1, 1, 1)] = [7],

32


which is clearly invertible. Hence the Implicit Function Theorem guarantees the existence of an open setN containing (1, 1), an open set V containing 1, and a unique continuously differentiable function g(x, y)defined in U such that

(x, y, g(x, y)) : (x, y) ∈ N = (x, y, z) ∈ N × V : f(x, y, z) = 4 ⊂ (x, y, z) : f(x, y, z) = 4 =: S.

Since f(1, 1, 1) = 4, then g(1, 1) = 1.

Spring 2007 #11. (a) Consider the equations

u3 + xv − y = 0, v3 + yu− x = 0.

Can these equations be solved uniquely for u, v in terms of x, y in a neighborhood of x = 0, y = 1, u = 1,v = −1? Explain your answer.

(b) Give an example in which the conclusion of the implicit function theorem is true, but the hypothesis isnot.

(a) Define f : R4 → R2 by f1(x, y, u, v) = u3 + xv − y and f2(x, y, u, v) = v3 + yu − x. Note thatf(0, 1, 1,−1) = (0, 0) and f is continuously differentiable. The matrix of partial derivatives under consider-ation is

[(∂fi/∂uj)(0, 1, 1,−1)] =

[∂f1∂u (0, 1, 1,−1) ∂f1

∂v (0, 1, 1,−1)∂f2∂u (0, 1, 1,−1) ∂f2

∂v (0, 1, 1,−1)

]=

[3 01 3

],

which has determinant 9 and is thus invertible. Hence the implicit function theorem guarantees the existenceof an open set U containing (0, 1), an open set V containing (1,−1), and a uniquely continuously differentiablefunction g(x, y) such that

(x, y, g1(x, y), g2(x, y)) : (x, y) ∈ U = (x, y, u, v) ∈ U × V : f(x, y, u, v) = 0.

Thus the equations can be solved uniquely in a neighborhood of x = 0, y = 1, u = 1, v = −1.

(b) Consider solving F (x, y) = y3 − x for y near (0, 0). The known solution is y = x1/3, but the implicitfunction theorem fails since the generated matrix is singular.

Spring 2010 #9. Assume that f(x, y, z) is a real-valued, continuously differentiable function such thatf(x0, y0, z0) = 0. If ∇f(x0, y0, z0) 6= 0, show that there is a differentiable surface, given parametrically by(x(s, t), y(s, t), z(s, t)) with (x(0, 0), y(0, 0), z(0, 0)) = (x0, y0, z0), on which f = 0.

Suppose without loss of generality that ∂f/∂z(x0, y0, z0) 6= 0. Then the Implicit Function Theoremguarantees the existence of an open set U ⊂ R2 containing (x0, y0), an open set V ⊂ R containing z0, and acontinuously differentiable function g : U → V such that

(x, y, g(x, y)) : (x, y) ∈ U = (x, y, z) : f(x, y, z) = 0.

Set s = x and t = y. Thus taking x(s, t) = s, y(s, t) = t, and z(s, t) = g(s, t) = g(x, y), it follows that

(x(s, t), y(s, t), z(s, t)) : (s, t) ∈ U

is a differentiable surface with (x(0, 0), y(0, 0), z(0, 0)) = (x0, y0, z0) on which f = 0.

Fall 2002 #6. Suppose F : R3 → R2 is continuously differentiable. Suppose for some v0 ∈ R3 and x0 ∈ R2

that F (v0) = x0 and F ′(v0) : R3 → R2 is onto. Show that there is a continuously differentiable functionγ : (−ε, ε)→ R3 for some ε > 0, such that

(i) γ′(0) 6= 0 ∈ R3, and

33


(ii) F (γ(t)) = x0 for all t ∈ (−ε, ε).

This problem is an immediate consequence of the implicit function theorem. Without loss of generality,assume x0 = (0, 0). Let v0 = (v1, v2, v3). Let M be the matrix representation of F ′(v0). Since F ′(v0) issurjective, two columns of M are linearly independent. Assume without loss of generality that these are thelast two columns of M . Then the matrix consisting of the last two columns of M is invertible, so we can applythe inverse function theorem. Thus there are continuously differentiable functions f, g : (v1 − ε, v1 + ε)→ Rsuch that

F (t, f(t), g(t)) = 0

for all t ∈ (v1 − ε, v1 + ε). Defining γ(t) = (v1 + t, f(v1 + t), g(v1 + t)) for all t ∈ (−ε, ε). Then F (γ(t)) =(0, 0) = x0 for all t ∈ (−ε, ε). Also, the final entry of γ′(0) is 1, so it is non-zero.

Spring 2002 #6. Suppose f : R3 → R is a continuously differentiable function with grad f 6= 0 at 0 ∈ R3.Show that there are two other continuously differentiable functions g : R3 → R, h : R3 → R such that thefunction

(x, y, z)→ (f(x, y, z), g(x, y, z), h(x, y, z))

from R3 to R3 is one-to-one on some neighborhood of 0.

Since ∇f(0) is nonzero, ∇f(0) is linearly independent and may be extended with two vectors v =(v1, v2, v3) and w = (w1, w2, w3) to a basis for R3. Define g(x, y, z) = (x, y, z) · (v1, v2, v3) and h(x, y, z) =(x, y, z) · (w1, w2, w3) so that ∇g = v and ∇h = w. In particular, the derivative of (f, g, h) at 0 has linearlyindependent columns ∇f(0), v, and w, so it is invertible. The Inverse Function Theorem then guaranteesthat (f, g, h) is invertible in a neighborhood of 0 (hence one-to-one).

Fall 2005 #4. Suppose F : [0, 1] → [0, 1] is a C2 function with F (0) = 0, F (1) = 0, and F ′′(x) < 0 forall x ∈ [0, 1]. Prove that the arc length of the curve (x, F (x)) : x ∈ [0, 1] is less than 3. (Suggestion:Remember that

√a2 + b2 < |a| + |b| when you are looking at the arc length formula - and at a picture of

what (x, F (x) could look like.)

The arc length of the curve (x, F (x)) : x ∈ [0, 1] is given by∫ 1

0

√1 + (F ′(x))2dx <

∫ 1

0

1 + |F ′(x)|dx = 1 +

∫ 1

0

|F ′(x)|dx.

Since F ′′(x) < 0 for all x ∈ [0, 1], F ′ is decreasing on [0, 1]. From F (0) = F (1) = 0, we conclude that Fincreases until it reaches its maximum, then decreases back to 0. Thus splitting the integral into two parts,to the left and right of the maximum occuring at c ∈ (0, 1),∫ 1

0

|F ′(x)|dx =

∫ c

0

|F ′(x)|dx+

∫ 1

c

|F ′(x)|dx =

∫ c

0

F ′(x)dx−∫ 1

c

F ′(x)dx = F (c)−F (0)−(F (1)−F (c)) = 2F (c).

By hypothesis, F (c) ≤ 1. Thus the arc length is less than 3.

34


Infinite sequences and series

Spring 2009 #1. Set a1 = 0 and define a sequence an via the recurrence

an+1 =√

6 + an for all n ≥ 1.

Show that this sequence converges and determine the limiting value.

We first show by induction that 0 ≤ an ≤ an+1 ≤ 3 for all n ≥ 0. For the base case n = 0, a0 = 0and a1 =

√6, so 0 ≤ a0 ≤ an+1 ≤ 3. Now assume inductively that 0 ≤ an ≤ an+1 ≤ 3. It follows that

an+2 =√

6 + an+1 ≤√

6 + 3 = 3. Also, an+2 ≥√

6 + 0 =√

6 ≥ 0. Finally,

an+2 =√

6 + an+1 ≥√

6 + an = an+1.

Thus 0 ≤ an+1 ≤ an+2 ≤ 3, completing the induction. Hence the sequence (an)∞n=0 is increasing and boundedabove by 3, so L := limn→∞ an exists.

Sending n→∞ inan+1 =

√6 + an

and using that the square root function is continuous,

L =√

6 + L,

hence L2 − L− 6 = 0, and L = 3 or L = −2. Clearly L ≥ 0, so we conclude L = 3.

Fall 2011 #4. Compute∞∑n=1

(−1)n

n2.

We start from the well-known sum∞∑n=1

1

n2=π2

6.

This implies that the given series is absolutely convergent. We can then derive its sum.

∞∑n=1

1

(2n)2=

1

4

∞∑n=1

1

n2=π2

24.

Thus∞∑n=1

1

(2n− 1)2=

∞∑n=1

1

n2−∞∑n=1

1

(2n)2=π2

6− π2

24=

3π2

24.

Therefore,∞∑n=1

(−1)n

n2=

∞∑n=1

1

(2n)2−∞∑n=1

1

(2n− 1)2=π2

24− 3π2

24= −π

2

12.

Spring 2013 #11. Define the Fibonacci sequence Fn by F0 = 0, F1 = 1, and recursively, Fn = Fn−1 +Fn−2

for n = 2, 3, 4, . . ..

(a) Show that the limit as n goes to infinity of Fn/Fn−1 exists and find its value.

(b) Prove that F2n+1F2n−1 − F 22n = 1 for all n ≥ 1.

(a) The Fibonacci recurrence is Fn+2 − Fn+1 − Fn = 0 for all n ≥ 0, F0 = 0, F1 = 1.

35


We guess that a solution takes the form Fn = λn. This implies λ2 − λ− 1 = 0, so we obtain solutions

λ1 =1

2(1 +

√5) and λ2 =

1

2(1−

√5).

The general solution is a linear combination of these. Using the initial conditions, we derive

Fn =1√5

(λn1 − λn2 ).

This implies

Fn/Fn−1 =λn1 − λn2

λn−11 − λn−1

2

.

As n→∞, since λ2 < 1, λn2 → 0, thus

Fn/Fn−1 → λn1/λn−11 = λ1 =

1

21 +√

5.

(b) This can be shown by a simple induction (easiest in the form Fn+1Fn−1 − F 2n = (−1)n).

Winter 2006 #1. Show that for each ε > 0 there exists a sequence of intervals (In) with the properties

Q ⊂∞⋃n=1

In and

∞∑n=1

|In| < ε.

Enumerate the rationals with the sequence (rn)∞n=1. Define In = [rn, rn + ε/2n+1]. Then Q ⊂⋃∞n=1 In

and∞∑n=1

|In| =∞∑n=1

ε/2n+1 = ε/2 < ε,

as desired.

Spring 2003 #2. Prove: If a1, a2, a3, . . . is a sequence of real numbers with

+∞∑j=1

|aj | < +∞,

then limN→+∞∑Nj=1 aj exists.

For each j ≥ 1, define pj = 12 (|aj | − aj) and qj = 1

2 (|aj | − aj). Then pj + qj = |aj | and pj − qj = ajfor each j. Note also that 0 ≤ pj ≤ |aj | and 0 ≤ qj ≤ |aj | for each j. Thus by the squeeze theorem, since∑∞j=1 |aj | converges,

∑∞j=1 pj and

∑∞j=1 qj converge. It follows that

∞∑j=1

an =

∞∑j=1

(pj − qj)

converges to∞∑j=1

pj −∞∑j=1

qj ,

so limN→+∞∑Nj=1 aj =

∑∞j=1 aj exists.

Spring 2010 #11. Suppose∑∞n=1 |an| < ∞. Let σ be a one-to-one mapping of N onto N. The series∑∞

n=1 aσ(n) is called a “rearrangement” of∑∞n=1 an. Prove that all rearrangements of

∑∞n=1 an are convergent

and have the same sum.

36


By the above exercise,∑∞n=1 an is convergent. Define S :=

∑∞n=1 an. Let ε > 0. It follows that

(∑Nn=1 an)∞N=1 is a Cauchy sequence, thus there exists N such that for all M ≥ N ,

|M∑

n=N+1

an| = |M∑n=1

an −N∑n=1

an| ≤ ε/2.

In addition, increase N as necessary such that

|N∑n=1

an − S| ≤ ε/2.

Now choose N ′ sufficiently large that 1, . . . , N ⊂ aσ(1), . . . , aσ(N ′). Fix m ≥ N ′. Let

M = maxσ(k) : 1 ≤ k ≤ m.

It follows that

|N∑n=1

an −m∑k=1

aσ(k)| ≤ |M∑

n=N+1

|an|| ≤ ε/2.

Then

|m∑k=1

aσ(k) − S| ≤ |m∑k=1

aσ(k) −N∑n=1

an|+ |N∑n=1

an − S| ≤ ε/2 + ε/2 = ε.

Hence∑∞k=1 aσ(k) exists and equals S.

Fall 2001 #2; Fall 2008 #5. Let N denote the positive integers, let an = (−1)n 1n , and let α be any real

number. Prove there is a one-to-one and onto mapping σ : N→ N such that

∞∑n=1

aσ(n) = α.

We proceed for a general series which is conditionally convergent but not absolutely convergent. Thisseries is conditionally convergent by the alternating series test, but not absolutely convergent. This is clearfrom the comparsion

∞∑n=1

|an| =1

1+

1

2+

1

3+ · · · ≤ 1

1+

1

2+

1

4+

1

4+ · · · ,

since the sum on the right is 1 + 12 + 1

2 + · · · , which does not converge.Define pj = 1

2 (|aj | + aj) and qj = 12 (|aj | − aj). Then pj = aj if aj is non-negative, and qj = −aj if aj

is negative. If both∑∞j=1 pj and

∑∞j=1 converge, then

∑∞j=1 aj =

∑∞j=1(pj − qj) converges, a contradiction.

Thus either∑∞j=1 pj or

∑∞j=1 qj diverges. Suppose the former. If

∑∞j=1 qj converges, then

∑∞j=1 pj =∑∞

j=1(aj + qj) converges, a contradiction. Hence both∑∞j=1 pj and

∑∞j=1 qj converge.

Reindex pj and qj to eliminate the 0 terms which do not correspond with some aj and preceed a termcorresponding to some aj . Clearly

∑∞j=1 pj and

∑∞j=1 qj still diverge.

Suppose α > 0 (the other cases are similar). Let Pj and Qj , j ≥ 1 be the partial sums of∑∞j=1 pj

and∑∞j=1 qj respectively. Select the smallest N1 such that PN1

> α. (Such an N1 exists since∑∞j=1 pj is

divergent and positive.) Then select the smallest N2 such that PN1−QN2

< α. Next, select N3 > N1 suchthat PN3

−QN2> α. By the zero test for sequences, pj and qj approach 0 as j → ∞. Thus it follows that

when continuing this procedure,PNk−QNk+1

approaches α. Define the bijection σ : N → N such that aσ(1) = p1, . . . , aσ(N1) = pN1, aσ(N1+1) =

q1, . . . , aσ(N2) = qN2, . . .. Then by construction,

∑Nj=1 aσ(j) converges to α as N →∞, so

∑∞n=1 aσ(n) = α.

37


Fall 2003 #3. Prove that the sequence a1, a2, . . . with

an =

(1 +

1

n

)nconverges as n→∞.

By comparing∑∞k=1

1k! to

∑∞k=1 2−k+1, which converges, we see that the former series converges. By the

binomial theorem,

an =

(1 +

1

n

)n=

n∑k=1

(n

k

)(1

n

)k

=

n∑k=1

1

k!

n(n− 1) · · · (n− k + 1)

nk

=

n∑k=1

1

k!

(1− 1

n

)(1− k − 1

n

)≤

n∑k=1

1

k!.

Taking the limit as n→∞, we see that limn→∞ an exists and does not exceed∑∞k=1

1k! .

Fall 2004 #1; Spring 2005 #9. Consider the following two statements:(A) The sequence (an) converges.(B) The sequence ((a1 + a2 + · · ·+ an)/n) converges.Does (A) imply (B)? Does (B) imply (A)? Prove your answers. (Summer 2005 #9 indicates that (A) implies(B) but (B) does not imply (A) in general).

The sequence an = (−1)n converges in the sense of (B) but not in the sense of (A). Thus (B) does notimply (A).

We show that (A) implies (B). Suppose the sequence (an)∞n=1 converges to s. Let ε > 0. Select N suchthat for all n ≥ N , |an − s| ≤ ε/2. Select N ′ > N such that

|a1 − s|+ · · ·+ |aN − s|N ′

≤ ε/2.

Then for any n ≥ N ′,

|a1 + · · ·+ ann

− s| = |a1 − s|+ · · ·+ |aN − s|n

+|aN+1 − s|+ · · ·+ |an − s|

n

≤ |a1 − s|+ · · ·+ |aN − s|N ′

+(n−N)ε

2n≤ ε/2 + ε/2 = ε.

Thus a1+···+ann converges to s.

Spring 2012 #4. For a sequence an of non-negative numbers, let sn :=∑Nn=1 an and suppose that sn

tends to a number s ∈ R in the Cesaro sense:

s = limn→∞

s1 + · · ·+ snn

.

Show that∑∞k=1 ak exists and equals s.

Suppose for the sake of contradiction that∑∞k=1 ak diverges. Then since all ak are positive, it must

diverge to +∞. Thus there exists N such that for all n ≥ N , sn =∑nk=1 an ≥ 10s. Then for any positive

integer M ,s1 + · · ·+ sN+M

N +M=s1 + · · ·+ sNN +M

+sN+1 + · · ·+ sN+M

N +M≥ (10s)M

N.

38


Letting M →∞, this means s is as large as desired, a contradiction.Thus

∑∞k=1 ak converges. By the previous problem,

s = limn→∞

s1 + · · ·+ snn

=

∞∑k=1

ak.

Spring 2004 #1. Let S denote the set of sequences a = (a1, a2, . . .), with ak = 0 or 1. Show that themapping Θ : S → R defined by

Θ((a1, a2, . . .)) =a1

10+

a2

102+ · · ·

is an injection. Include an explanation of why the infinite series converges. Hint: if a 6= b, you may assumethat

a = (a1, . . . , an−1, 0, an+1, . . .),

b = (b1, . . . , bn−1, 1, bn+1, . . .).

For any a,

Θ(a) =a1

10+

a2

102+ · · · ≤ 1

10+

1

10

2

=1

10/

9

10=

1

9,

thus Θ(a) is finite.Suppose a 6= b. Assume without loss of generality that

a = (a1, . . . , an−1, 0, an+1, . . .),

b = (b1, . . . , bn−1, 1, bn+1, . . .).

Then

|Θ(b)−Θ(a)| ≥ 10−n +

∞∑i=n+1

bi − ai10i

≥ 10−n −∞∑

i=n+1

1

10i= 10−n − 1

10n+1/

1

9=

1

10n+1.

Hence Θ(b) 6= Θ(a), so Θ is injective.

Spring 2006 #2. Let F (x) =∑∞n=0 anx

n be a power series with an ∈ R. Show that there exists a uniquenumber ρ ≥ 0 such that F (x) converges if |x| < ρ and F (x) diverges if |x| > ρ.

Suppose ρ ∈ R such that F (ρ) converges. We first show that F (x) converges for all |x| < |ρ|. Let x ∈ Rwith |x| < ρ. Since a convergent sequence is bounded, there exists B such that |anρn| ≤ B for all n. Defineα = |x|/|ρ|. Then

∞∑n=0

an|x|n ≤∞∑n=0

anαn|ρ|n ≤ B

∞∑n=0

αn,

which converges, thus F (x) converges.Let ρ = supx ≥ 0 : F (x) converges. Clearly 0 is in this set, so the supremum is non-negative. By the

above result, F (x) converges if |x| < ρ. If ρ = +∞, then F converges everywhere. Otherwise, ρ is somefinite non-negative real number. Suppose for the sake of contradiction that F (x) converges for some x with|x| > ρ. Then |x| is a member of x ≥ 0 : F (x) converges , but |x| > ρ, a contradiction. Hence F (x)diverges if |x| > ρ.

Winter 2006 #2. Let (an)n≥1 be a decreasing sequence of positive numbers such that∑∞n=1 an = ∞.

Under what condition(s) is the function

f(x) =

∞∑n=1

(−1)nanxn

39


well-defined and left-continuous at x = 1? Carefully prove your assertion.

Suppose limn→∞ an = 0. Then by the alternating series test, f(1) converges. Since f(1) =∑∞n=1(−1)nan

converges, this series is Abel-summable, so f is left-continuous at x = 1.

Fall 2007 #8. Suppose an ≥ 0 and∑∞n=1 an =∞. Does it follow that

∞∑n=1

an1 + an

=∞?

Prove your answer.

Yes. Suppose first that L := lim supn→∞ an 6= 0. Since the an are non-negative, L > 0. Let N be apositive integer. Then there exists n > N such that an > L/2. It follows that

an1 + an

≥ an > L/2.

Hence the limit of an1+an

as n → ∞ either does not exist, or does not equal 0, so the sum in question doesnot converge. Since the summands are positive, the series must diverge to +∞.

Otherwise, L := lim supn→∞ an = 0. Let ε > 0. Then there exists N such that for all n ≥ N , an < ε.Thus for all n ≥ N ,

an1 + an

>an

1 + ε.

Note that∞∑n=N

an1 + ε

=1

1 + ε

∞∑n=N

an = +∞,

thus∑∞n=1

an1+an

must also diverge to +∞, as desired.

Fall 2010 #4. (a) Show that given a real-valued continuous function f on [0, 1]× [0, 1] and an ε > 0, thereexist real-valued continuous functions g1, . . . , gn and h1, . . . , hn on [0, 1] for some finite n ≥ 1 so that∣∣∣∣∣f(x, y)−

n∑i=1

gi(x)hi(y)

∣∣∣∣∣ ≤ ε, 0 ≤ x, y ≤ 1.

(b) If f(x, y) = f(y, x) for all 0 ≤ x, y ≤ 1, can this be done with hi = gi for each i? Explain.

(a) Let ε > 0. Since f is continuous on a compact set, it is uniformly continuous. Thus there exists apositive integer N such that if |(x, y)− (x′, y′)| ≤ 1

N , then |f(x, y)− f(x′, y′)| ≤ ε/2.Let φ : R→ R be given by

φ(x) =

1 + x x ∈ [−1, 0],1− x x ∈ [0, 1],0 otherwise.

Define the real-valued continuous function gi(x) = φ(Nx+ i) for 0 ≤ i ≤ N . Consider the function

f(x, y) =

N∑i=0

gi(x)f(i/N, y).

For any (x, y) ∈ [0, 1], it is straightforward to show from continuity of f that

|f(x, y)− f(x, y)| ≤ ε/2.

40


By the Weierstrass Approximation Theorem in 1 dimension, there exists polynomials (real-valued con-tinuous functions) h0, . . . , hN such that for each 1 ≤ i ≤ N and any y ∈ [0, 1],

|hi(y)− f(i/N, y)| ≤ ε/4N.

It follows that

|f(x, y)−N∑i=0

gi(x)hi(y)| ≤ |f(x, y)− f(x, y)|+ |f(x, y)−N∑i=0

gi(x)hi(y)|

= |f(x, y)− f(x, y)|+ |N∑i=0

gi(x)(f(i/N, y)− hi(y))|

≤ ε/2 +N(ε/2N) = ε.

Adjust indices as necessary to obtain the desired expression.

(b) No. Take some f which is negative on (1, 1). Then if gi = hi,∑ni=1 gi(1)hi(1) =

∑ni=1 g

2i (1) is always

non-negative, so it cannot come arbitrarily close to f(1, 1). Hence such an approximation is not possible.

Fall 2012 #1. Let bn∞n=1 be a sequence of real numbers with bounded partial sums, i.e., there is M <∞such that for all N , |

∑Nn=1 bn| ≤ M , and let an∞n=1 be a sequence of positive numbers decreasing to 0.

Prove the series∑anbn converges.

We use summation by parts. Let sn denote the partial sums of∑nj=1 bj . For any positive integers N,M ,

M∑j=N

ajbj =

M∑j=N

aj(sj − sj−1) =

M∑j=N

ajsj −M∑j=N

ajsj−1

=

M∑j=N

ajsj −M−1∑j=N−1

aj+1sj

= aMsM − aNsN−1 +

M−1∑j=N

(aj − aj+1)sj .

Thus

|M∑j=N

ajbj | ≤M(aM + aN ) +M

M−1∑j=N

(aj − aj+1)

= M(aM + aN + aN − aM ) ≤ 2MaN .

Letting N →∞, since aN decreases to 0,M∑j=N

ajbj → 0

as N,M →∞. Hence the partial sums of∑anbn form a Cauchy sequence, so the series

∑∞n=1 anbn converges.

Spring 2013 #4. Denote by hn the n-th harmonic number:

hn = 1 +1

2+

1

3+ · · ·+ 1

n.

Prove that there is a limitγ = lim

n→∞(hn − lnn).

41


We show xn := hn − lnn is decreasing and bounded below, thus the desired limit exists.Recall lnx =

∫ x1

1ydy for all x ≥ 1. Thus hn is an upper Riemann sum for lnn, so hn − lnn ≥ 0. Thus

xn is bounded below by 0.For all n ≥ 1, let xn = hn − lnn. Fix n ≥ 1. Applying the mean value theorem to ln, there exists

c ∈ ( 1n+1 ,

1n ) such that

ln(n+ 1)− ln(n) =1

c((n+ 1)− n) =

1

c.

Thus

xn+1 − xn =1

n+ 1− (ln(n+ 1)− ln(n)) =

1

n+ 1− 1

c< 0.

Hence xn is decreasing.

Spring 2006 #3. Prove that the series

f(x) =

∞∑n=1

sin(nx)

n5/2

converges for all x ∈ R and that f(x) is a continuous function on R with a continuous derivative. Stateclearly any facts you assume.

We first show that∑∞j=1

sin(jx)j5/2

converges uniformly. Note that the terms are bounded by 1j5/2

, and∑∞j=1

1j5/2

< ∞ by the integral test. Thus the series converges by the Weierstrass M-test to a continuous

function. Also, the derivatives are continuous and the sum of the derivatives converge uniformly by theWeierstrass M-test (again using the integral test).

42


Partial Derivatives

Fall 2001 #5; Spring 2002 #5; Winter 2002 # 6; Spring 2003 #6; Fall 2005 #2. Suppose thatf : R2 → R is a continuous function such that the partial derivatives ∂f

∂x and ∂f∂y exist everywhere and are

continuous everywhere, and ∂∂x

(∂f∂y

)and ∂

∂y

(∂f∂x

)also exist, and are continuous everywhere. Prove that

∂

∂x

(∂f

∂y

)=

∂

∂y

(∂f

∂x

)at every point of R2.

Here I only assume ∂∂y

(∂f∂x

)exists. Define

δ(h, k) := f(h, k)− f(h, 0)− f(0, k) + f(0, 0).

Also let g(x) := f(x, k) − f(x, 0). Applying the mean value theorem twice, there exist ah ∈ (0, h) andbk ∈ (0, k) such that

δ(h, k) = g(h)− g(0) = g′(ah)h = (∂f

∂x(ah, k)− ∂f

∂x(ah, 0))h =

∂2f

∂y∂x(ah, bk)hk.

Thus since ∂2f∂y∂x is continuous,

limh→0

limk→0

δ(h, k)

hk= limh→0

limk→0

∂2f

∂y∂x(ah, bk) =

∂2f

∂y∂x(0, 0).

Hence ∂∂x

(∂f∂y

)exists and equals ∂2f

∂y∂x (0, 0).

Spring 2008 #5. Fall 2012 #6. (a) Let F (x, y) be a continuous function on the plane such that for everysquare S having its sides parallel to the axes,∫ ∫

S

F (x, y)dxdy = 0.

Prove F (x, y) = 0 for all (x, y).

(b) Assume f(x, y), ∂f(x,y)∂x , ∂∂y

(∂f(x,y)∂x

), and ∂

∂x

(∂f(x,y)∂y

)are all continuous in the plane. Use part (a) to

prove that∂

∂y

(∂f(x, y)

∂x

)=

∂

∂x

(∂f(x, y)

∂y

).

Hint: you may assume the double integral in (a) equals the iterated integral∫

(∫F (x, y)dx)dy and equals

the iterated integral∫

(∫F (x, y)dy)dx.

Suppose for the sake of contradiction that f(x∗, y∗) > 0. There exists δ > 0 such that if |(x, y)−(x′, y′)| ≤δ, then |f(x, y) − f(x′, y′)| ≤ |f(x∗, y∗)|/2. Take S with opposite corners [x − delta/

√2, y − delta/

√2]

and [x + δ/√

2, y + δ/√

2]. Then within S, |(x, y) − (x′, y′)| =√|x− x′|2 + |y − y′|2 ≤

√δ2 = δ, hence

|f(x, y)− f(x∗, y∗)| ≤ |f(x∗, y∗)|/2, so

|f(x, y)| ≥ f(x∗, y∗)− f(x∗, y∗)

2= f(x∗, y∗)/2.

It follows that ∫ ∫S

f(x, y) ≥∫ ∫

S

f(x∗, y∗)/2 > 0,

43


a contradiction. Thus f(x, y) = 0 for all (x, y).

(b) Let S be a square with its sides parallel to the axes. We compute∫ ∫S

∂2f

∂x∂y− ∂2f

∂y∂xdxdy =

∫ ∫S

∂2f

∂x∂ydxdy −

∫ ∫S

∂2f

∂y∂xdxdy = 0

where the final step follows from the fundamental theorem of calculus. Thus applying part (a) to the

continuous function ∂2f∂x∂y −

∂2f∂y∂x , we find ∂2f

∂x∂y −∂2f∂y∂x = 0.

Fall 2002 #5. Suppose f : R2 → R has partial derivatives at every point bounded by A > 0.

(a) Show that there is an M > 0 such that

|f((x, y))− f((x1, y1))| ≤M((x− x1)2 + (y − y1)2)1/2.

(b) What is the smallest value of M (in terms of A) for which this always works?

(c) Give an example where that value of M makes the inequality an equality.

(a) Using the mean value theorem,

|f(x, y)− f(x1, y1)| ≤ |f(x, y)− f(x1, y)|+ |f(x1, y)− f(x1, y1)| ≤ A|(x, y)− (x1, y)|+A|(x1, y)− (x1, y1)|

≤ A√

2√

(x− x1)2 + (y − y1)2.

Here we use the inequality

a+ b ≤√

2√a2 + b2

for the last inequality. So setting M = A√

2, we have the desired inequality.

(b) & (c) Use the example f(x, y) = x + y. Here A = 1 and taking (x, y) = (1, 1), (x′, y′) = (0, 0), weobtain the bound A

√2. Thus this is strict.

Spring 2003 #4. Consider the following equation for a function F (x, y) on R2:

∂2F

∂x2=∂2F

∂y2.

(a) Show that if a function F has the form F (x, y) = f(x+ y) + g(x− y) where f : R → R and g : R → Rare twice differentiable, then F satisfies this equation.

(b) Show that if F (x, y) = ax2+bxy+cy2, a, b, c ∈ R, satisfies this equation, then F (x, y) = f(x+y)+g(x−y)for some polynomials f and g in one variable.

(a) Straightforward with chain rule.

(b) Then 2a = 2c, so a = c. Write F (x, y) = a(x2 + y2) + bxy = A(x+ y)2 +B(x− y)2. Then

a = A+B, b = 2(A−B)

can be inverted. Hence we can take f(x+ y) = A(x+ y)2 and g(x− y) = B(x− y)2.

44


Differentiation

Spring 2004 #5. Suppose that G is an open set in Rn, f : G→ Rm is a function, and that x0 ∈ G.

(a) Carefully define what is meant by f ′(x0) : Rn → Rm.

(b) Suppose that I is a line segment in G connecting points p and q such that f ′(x) is defined for allx ∈ I. Show that if f is differentiable at all the points of I, then for some point c ∈ I,

||f(q)− f(p)||2 ≤ ||f ′(c)||||q − p||2.

Hint: let w be a unit vector with ||f(q)− f(p)||2 = (f(q)− f(p)) · w.

(a) f ′(x0) is the unique linear map from Rn to Rm such that

limx→x0;x∈G\x0

||f(x)− (f(x0) + (f ′(x0))(x− x0))||||x− x0||

= 0.

(b) If f(q) = f(p), the desired inequality is obvious. Otherwise, let

w =f(q)− f(p)

||f(q)− f(p)||2.

Then||f(q)− f(p)||2 = (f(q)− f(p)) · w.

Define g : G→ R byg(x) := f(x) · w.

Theng(q)− g(p) = (f(q)− f(p)) · w = ||f(q)− f(p)||2

andg′(x)(u) = (f ′(x)(u)) · w

for all x ∈ I, u ∈ Rn. By the mean value theorem from Rn to R, there exists c ∈ I such that

g(q)− g(p) = g′(c)(q − p) = (f ′(c)(q − p)) · w.

Thus by the Cauchy-Schwarz inequality,

||f(q)− f(p)||2 = g(q)− g(p) = (f ′(c)(q − p)) ≤ ||f ′(c)||2||q − p||2.

Winter 2002 #5; Spring 2009 #7. (a) Let f : U → Rk be a function on an open set U ⊂ Rn. Definewhat it means for f to be differentiable at a point x ∈ U .

(b) State carefully the Chain Rule for the composition of differentiable functions of several variables.

(c) Prove the Chain Rule you stated in (b).

(a) f is differentiable at x ∈ U if there exists a linear map f ′(x) : Rk → Rn such that

limy→x;y∈U\x

||f(y)− (f(x) + f ′(x)(y − x))||2||y − x||2

= 0.

45


(b) Let U ∈ Rk, V ∈ Rm, and consider f : U → V and g : V → Rn. Let x0 ∈ U and suppose that f isdifferentiable at x0 and g is differentiable at g(x0). Then g f is differentiable at x0 and as a compositionof linear maps,

(g f)′(x0) = g′(f(x0)) f ′(x0).

(c) Let Lf : Rk → Rm and Lg : Rm → Rn be the functions

Lf (x) = f(x) + f ′(x0)(x− x0), Lg(x) = g(x) + g′(x0)(x− x0).

We must show that||(g f)(x)− (Lg Lf )(x)|| ≤ ε||x− x0||

for all x in some neighborhood of x0. By the triangle inequality,

||(g f)(x)− (Lg Lf )(x)|| ≤ ||(g f)(x)− (Lg f)(x)||+ ||(Lg f)(x)− (Lg Lf )(x)||.

We shall handle the two terms on the right separately.For the first term, fix some M > ||f ′(x0)||. Since g is differentiable at f(x0), there exists r such that if

||y − f(x0)|| ≤ r, then

||g(y)− Lg(y)|| ≤ ε

2M||y − f(x0)||.

Since f is differentiable at x0, it is continuous at x0, so there exists δ1 > 0 such that if ||x − x0|| ≤ δ1,then ||f(x)− f(x0)|| ≤ r. Finally, since f is differentiable at x0, there exist M > 0 and δ2 > 0 such that if||x− x0|| ≤ δ2, then

||f(x)− f(x0)|| ≤M ||x− c||.

Thus for any x with ||x− x0|| ≤ min(δ1, δ2), taking y = f(x) above,

||(g f)(x)− (Lg f)(x)|| ≤ ε

2M||f(x)− f(x0)|| ≤ ε

2||x− x0||.

For the second term, since f is differentiable at x0, there exists δ3 > 0 such that if ||x− x0|| < δ3, then

||f(x)− Lf (x)|| ≤ ε

2||Lg||||x− x0||.

Thus||(Lg f)(x)− (Lg f)(x0)|| ≤ ||Lg||||f(x)− Lf (x)|| ≤ ε

2||x− x0||.

Combining the bounds yields the desired inequality to show that g f is differentiable at x0 with derivativeLg Lf .

Spring 2010 #10. Let f(x, y) be the function defined by

f(x, y) =xy√x2 + y2

when (x, y) 6= (0, 0) with f(0, 0) = 0.

(a) Compute the directional derivatives of f(x, y) at (0, 0) in all directions where they exist.

(b) Is f(x, y) differentiable at (0, 0)? Prove your answer.

(a) Let u = (u1, u2) be a unit vector. Then the directional derivative of f at (0, 0) in the direction of u,if it exists, is

limh→0

f(hu)− f(0, 0)

h= limh→0

h2u1u2

h|h|= limh→0

h

|h|u1u2.

46


In particular, the directional derivatives of f along the x and y axes exist and equal 0. But since thelimit for negative h does not equal the limit for positive h otherwise, all other directional derivatives of f donot exist.

(b) No. Suppose for the sake of contradiction that f is differentiable at (0, 0). Then since the x and ydirectional derivatives of f are 0, the derivative of f must be the 0 linear map. For any x, y > 0,

|f(x, y)− (f(0, 0) + 0)|||(x, y)− (0, 0)||2

=|xy|

x2 + y2.

Thus approaching (0, 0) by (x, x) as x → 0, this quotient is 12 , which does not approach 0. Hence f is not

differentiable at (0, 0).

Spring 2011 #10. Suppose that f is a function defined on an open subset G of R2 and that (x0, y0) ∈ G.

1. Define what it means for f to be differentiable at (x0, y0).

2. Show that if ∂f∂x and ∂f∂y exist and are continuous on an open set containing (x0, y0), then f is differentiable

at (x0, y0) ∈ G.

1. f is differentiable at (x0, y0) if there exists a linear map Df such that

lim(x,y)→(x0,y0)

||f(x, y)− (f(x0, y0) + (Df)(x− x0, y − y0))||2||(x, y)− (x0, y0)||2

= 0.

2. We focus on the case where f maps into the real numbers. Let ε > 0. Using continuity of the firstpartials, select δ > 0 such that if ||(x, y)− (x0, y0)|| ≤ δ, then∣∣∣∣∂f∂x (x, y)− ∂f

∂x(x0, y0)

∣∣∣∣ ≤ ε and

∣∣∣∣∂f∂y (x, y)− ∂f

∂y(x0, y0)

∣∣∣∣ ≤ ε.Fix x, y such that ||(x, y)− (x0, y0)|| ≤ δ. By the Mean Value Theorem, there exists some x∗ lying betweenx and x0 and y∗ lying between y and y0 such that

f(x, y)− f(x, y0) =∂f

∂y(x, y∗)(y − y0)

and

f(x, y0)− f(x0, y0) =∂f

∂x(x∗, y0)(x− x0).

Thus

f(x, y)− (f(x0, y0) + (x− x0)∂f

∂x(x0, y0) + (y − y0)

∂f

∂y(x0, y0))

= (x− x0)

[∂f

∂x(x∗, y0)− ∂f

∂x(x0, y0)

]+ (y − y0)

[∂f

∂y(x, y∗)− ∂f

∂y(x0, y0)

].

The expressions in braces have norm less than ε, hence by the Cauchy-Schwarz inequality, the norm of thisexpression is less than ε

√2||(x, y)− (x0, y0)||. Thus f is differentiable at (x0, y0).

Fall 2003 #2. Let f : R → R be an infinitely often differentiable function. Assume that for each elementx ∈ [0, 1] there is a positive integer m such that the m-th derivative of f at x is not zero. Prove that thereexists an integer M such that the following stronger statement holds: For each element x ∈ [0, 1], there is apositive integer m with m ≤M such that the m-th derivative of f at x is not zero.

For each positive integer n, let

Sn = x ∈ [0, 1] : there exists 0 < m ≤ n with f (m)(x) 6= 0.

47


Since f is infinitely differentiable, its derivatives are continuous, so the sets Sn are a union of open sets andhence open. By assumption,

⋃n≥1 Sn = [0, 1]. But [0, 1] is compact, so there exists some n1 < · · · < nk such

that [0, 1] =⋃ki=1 Sni

. Set M = nk. Note the Sn are increasing with n, hence

[0, 1] = SM ,

as desired.

Fall 2007 #2. Let f : (a, b) → R be continuous and differentiable in (a, b) \ c. If limx→c f′(x) = d ∈ R,

show that f is differentiable at c, and f ′(c) = d.

I assume c ∈ (a, b). Let ε > 0. Select δ > 0 such that if x ∈ (a, b) with |x− c| ≤ δ, then

|f ′(x)− d| ≤ ε.

Let 0 < δ′ < δ/2. Choose z ∈ (a, b) with |z − c| ≤ δ′ < δ/2. Since f is differentiable at z, there exists δ1 > 0such that if |x− z| ≤ δ1,

|f(x)− (f(z) + f ′(z)(x− z)| ≤ ε.

For any x ∈ (a, b) with |x− c| ≤ min(δ′/2, δ1/2), we have |x− z| ≤ δ1 and

|f(x)− (f(c) +d(x− c))| ≤ |f(x)− (f(z) + f ′(z)(x− z))|+ |f(z)− f(c)|+ |f ′(z)(x− z)−d(x− z)|+ |d(z− c)|

≤ ε+ ε+ dδ′.

Choosing ε and δ′ small enough, we see that f is differentiable at c with f ′(c) = d.

Fall 2007 #9. Suppose un : R→ R is differentiable and solves

u′n(x) = F (un(x), x),

where F is continuous and bounded.

(a) Suppose un → u uniformly. Show that u is differentiable and solves

u′(x) = F (u(x), x).

(b) Supposeu′(x) = F (u(x), x), u(x0) = y0

has a unique solution u : R→ R and un(x0) converges to y0 as n→∞. Show that un uniformly convergesto u.

(a) The given ODE is equivalent, via the fundamental theorem of calculus to the following integralequation:

un(x) = un(x0) +

∫ x

x0

F (un(t), t)dt.

If we show that u(x) also satisfies this integral equation, then by the fundamental theorem of calculus, it isdifferentiable with derivative satisfying the original ODE. Let ε > 0 and fix x 6= x0. Since F is continuous,there exists δ < ε/3 such that if |u − v|sup < δ, then |F (u, x) − F (v, x)| ≤ ε/(3(x − x0)). From uniformconvergence, there exists N such that if n ≥ N , |un − u|sup ≤ δ. Thus for all n ≥ N ,

|F (un, x)− F (u, x)| ≤ ε/3.

48


Also, for any x, |u(x)− un(x)| ≤ δ < ε. Therefore,

|u(x)− u(x0)−∫ x

x0

F (u(t), t)dt| = |u(x)− un(x) + un(x0)− u(x0) +

∫ x

x0

(F (un(t), t)− F (u(t), t))dt|

≤ |u(x)− un(x)|+ |u(x0)− un(x0)|+∫ x

x0

|F (un(t), t)− F (u(t), t)|dt

≤ ε/3 + ε/3 + (x− x0)ε

3(x− x0)= ε.

Thus u(x) satisfies the integral equation and hence the given ODE.

(b) It suffices to show the un converge uniformly, since part (a) then implies the un converge to somesolution u′ which satisfies the given system. By the assumption of uniqueness, u = u′, so the un converge tou.

We want to use the Arzela-Ascoli Theorem on the un, so we need them to be equicontinuous and uniformlybounded. Consider the compact set [−M,M ]. Since F is bounded (say by K),

|un(y)− un(x)| ≤∫ y

x

|F (un(t), t)|dt ≤ K|y − x|.

Thus the un are equicontinuous.Let ε > 0. Since the un(x0) converge to y0, there exists N such that for all n ≥ N , |un(x0) − y0| ≤ ε.

Then the above integral inequality is enough to guarantee that the un are uniformly bounded.So the un, along with any subsequence unk

, satisfy the hypothesis of the Arzela-Ascoli theorem on[−M,M ], so every subsequence of the un has a uniformly convergent sub-subsequence. Therefore, the unare uniformly convergent on [−M,M ].

NOTE: This does not obviously imply they converge to u on [−M,M ]. The statement to prove could befalse. We would need the guarantee of a unique solution on every closed interval with x0 as an endpoint.

Spring 2007 #8. Suppose the functions fn are twice continuously differentiable on [0, 1] and satisfy

limn→∞

fn(x) = f(x) for all x ∈ [0, 1], and

|f ′n(x)| ≤ 1, |f ′′n (x)| ≤ 1 for all x ∈ [0, 1], n ≥ 1.

Prove that f(x) is continuously differentiable on [0, 1].

Mean Value Theorem + Arzela-Ascoli on fn and f ′n implies uniform convergence of both of these se-quences. Uniform convergence of continuous functions to a continuous function then implies f and limn→∞ f ′nare continuous. Tao’s Theorem 14.7.1 implies f ′ = limn→∞ f ′n, so f ′ is continuous and f is continuouslydifferentiable.

49


Riemann integration

Fall 2003 #4; Fall 2010 #2; Spring 2007 #9. Let f : R → R be a continuous function. State thedefinition of the Riemann integral ∫ 1

0

f(x)dx

and prove that it exists.

First, we require that f is bounded on [0, 1] to be Riemann integrable. We define∫ 1

0

f = sup∫ 1

0

g : g a piecewise constant function on [0, 1] which minorizes f

and ∫ 1

0

f = inf∫ 1

0

g : g a piecewise constant function on [0, 1] which majorizes f,

integrate piecewise constant functions in the obvious way, and say that f is Riemann integrable if∫ 1

0

f =

∫ 1

0

f.

Suppose f : R→ R is continuous. Let ε > 0. Since f is continuous on a compact set, f is bounded anduniformly continuous. Choose δ > 0 such that if |x − y| ≤ δ, then |f(x) − f(y)| ≤ ε/2. Choose N suchthat 1

N ≤ δ. Form the equally spaced partition x0 = 0 < x1 < · · · < xN = 1 with spacing 1/N . Define

f : [0, 1]→ R so that f(x) = f(xn) + ε/2 for x ∈ [xn, xn+1) and f(1) = 1. Then f is piecewise constant andmajorizes f .

Likewise, define f . We compute

∫ 1

0

f −∫ 1

0

f ≤N−1∑n=0

1

Nε = ε.

Since ε was arbitrary, f is integrable.

Fall 2012 #2. Spring 2013 #1. Let f be a bounded, non-decreasing function on the closed interval [0, 1].

Prove that∫ 1

0f(x)dx exists.

Define the evenly-spaced partition 0 = x0 < · · · < xN = 1. Note that f : [0, 1] → R given by f(x) =f(xn+1) on (xn, xn+1] and f(0) = f(0) is piecewise constant and majorizes f . Likewise, define f by the left

endpoint. Then the integral of f − f is a telescoping series which tends to 0 as N →∞.

Fall 2007, #11. Let f be a bounded real function on [0, 1]. Show that f is Riemann integrable if and onlyif f3 is Riemann integrable.

Note f3 is bounded on [0, 1]. Use majoring piecewise constant functions of f to construct majorizingpiecewise constants of f3, and the identity

f3(x)− f3(y) = (f(x)− f(y))(f2(x) + f(x)f(y) + f2(y)).

Spring 2011 #9. Prove that if f(x) is a continuous function on [a, b] and f(x) ≥ 0, then∫ baf(x) = 0

implies that f = 0.

50


Standard.

Fall 2011 #5. Give an example of a function f(x) on [0, 1] with infinitely many discontinuities, but whichis Riemann integrable. Include proof (don’t just quote some theorem).

Just make something monotone increasing. Use 1/2n - sized intervals.

Fall 2010 #11. Find the function g(x) which minimizes∫ 1

0

|f ′(x)|2dx

among smooth functions f : [0, 1]→ R with f(0) = 0 and f(1) = 1. Is the optimal solution g(x) unique?

In general, the Euler-Lagrange equation provides such optimizing functions.Note that ∫ 1

0

(f ′(x)− 1)2 =

∫ 1

0

(f ′(x))2dx− 2

∫ 1

0

f ′(x)dx+ 1 =

∫ 1

0

(f ′(x))2dx− 1,

and the left hand side is minimized only by f(x) = x, thus only f(x) = x minimizes∫ 1

0

(f ′(x))2dx.

Spring 2011 #8; Fall 2008 #3; Spring 2008 #2. Give examples:

1. A function f(x) on [0, 1] which is not Riemann integrable, for which |f(x)| is Riemann integrable.

2. Continuous functions fn and f on [0, 1] such that fn(t) → f(t) for all t ∈ [0, 1] but∫ 1

0fn(t)dt does not

converge to∫ 1

0f(t)dt.

(a) -1 on rationals, 1 on irrationals.

(b) Small triangle near 0 with integral 1, length 1 / n, height n, converges to 0.

Fall 2002 #4; Spring 2009 #10; Spring 2013 #12. (a) Rigorously justify the following:∫ 1

0

dx

1 + x2= limN→∞

N∑n=0

(−1)n

2n+ 1.

(b) Deduce the value of 1− 13 + 1

5 −17 + · · · .

(a) For any |x| < 1, the series∞∑n=0

(−1)nx2n

is absolutely convergent (as a geometric series). Fix a < 1. The partial sums of this series converge on thecompact set [0, a], thus they converge uniformly on [0, a]. Thus for any a ∈ [0, 1),∫ a

0

dx

1 + x2=

∫ 1

0

∞∑n=0

(−1)na2n =

∫ 1

0

limN→∞

N∑n=0

(−1)na2n

51


= limN→∞

∫ 1

0

N∑n=0

(−1)na2n = limN→∞

N∑n=0

(−1)n

2n+ 1a2n+1.

By the alternating series test,∞∑n=0

(−1)n

2n+ 1

converges, hence by applying Abel’s Theorem in the final equality below,∫ 1

0

dx

1 + x2= lima→1−

∫ a

0

dx

1 + x2= lima→1−

limN→∞

N∑n=0

(−1)n

2n+ 1a2n+1 = lim

N→∞

N∑n=0

(−1)n

2n+ 1.

(b) We know∫ 1

0dx

1+x2 = arctan x|10 = arctan 1− arctan 0 = π4 , thus

1− 1

3+

1

5− 1

7+ · · · = π

4.

Winter 2002 #1. (a) State some reasonably general conditions under which this “differentiation underthe integral sign” formula is valid:

d

dx

∫ d

c

f(x, y)dy =

∫ d

c

∂f

∂xdy.

(b) Prove that the formula is valid under the conditions you gave in part (a).

(a) Let f : [a, b]× [c, d]→ R. Suppose ∂f∂x exists on (a, b)× [c, d] and extends to a continuous function on

[a, b]× [c, d]. Let

F (x) =

∫ b

a

f(x, y)dy.

Thend

dxF (x) =

∫ b

a

∂f

∂x(x, y)dy.

(b) For h with x+ h ∈ [a, b], we estimate∣∣∣∣∣F (x+ h)− F (x)

h−∫ b

a

∂f

∂x(x, y)dy

∣∣∣∣∣ =

∣∣∣∣∣∫ b

a

(f(x+ h, y)− f(x, y)

h− ∂f

∂x(x, y)

)dy

∣∣∣∣∣≤∫ b

a

∣∣∣∣f(x+ h, y)− f(x, y)

h− ∂f

∂x(x, y)

∣∣∣∣ dyBy the mean value theorem, there exists c ∈ (0, 1) such that

f(x+ h, y)− f(x, y)

h=∂f

∂x(x+ ch, y).

Since ∂f∂x is continuous on the compact set [a, b]× [c, d], it is uniformly continuous on this set. Choose δ such

that ||(x, y)− (x′, y′)|| ≤ δ implies ∣∣∣∣∂f∂x (x′, y′)− ∂f

∂x(x, y)

∣∣∣∣ ≤ ε

b− a.

52


Then using the above estimate, for h < δ,∣∣∣∣∣F (x+ h)− F (x)

h−∫ b

a

∂f

∂x(x, y)dy

∣∣∣∣∣ ≤∫ b

a

∣∣∣∣f(x+ h, y)− f(x, y)

h− ∂f

∂x(x, y)

∣∣∣∣ dy=

∫ b

a

∣∣∣∣∂f∂x (x+ ch, y)− ∂f

∂x(x, y)

∣∣∣∣ dy ≤ ∫ b

a

ε

b− a= ε.

Since ε > 0 was arbitrary, we have the desired equality.

Spring 2004 #3; Spring 2006 #1. Show that if fn are Riemann integrable functions on [0, 1] and fnconverges to f uniformly, then f is Riemann integrable.Fall 2004 #3. Show that if fn → f uniformly on the bounded closed interval [a, b], then∫ b

a

fn(x) dx→∫ b

a

f(x) dx.

Let ε > 0. There exists N such that for all n ≥ N and all x ∈ [a, b], |fn(x)−f(x)| ≤ ε/2. By definition of

the Riemann integral, there exists some fn which is piecewise constant, majorizes fn and∫ bafn −

∫ bafn ≤ ε.

It follows that fn + ε is piecewise constant and majorizes f . Thus for any n ≥ N ,∫ b

a

f ≤∫ b

a

(fn + ε) = (

∫ b

a

fn) + (b− a)ε ≤∫ b

a

fn + (b− a+ 1)ε.

Likewise, ∫ b

a

f ≥∫ b

a

fn − (b− a+ 1)ε.

Letting ε→ 0, we see that f is Riemann integrable and∫ b

a

fn →∫ b

a

f.

Spring 2010 #12. Assume that fn is a sequence of nonnegative continuous functions on [0, 1] such that

limn→∞∫ 1

0fn(x)dx = 0. Is it necessarily true that

(a) There is a B such that fn(x) ≤ B for x ∈ [0, 1] for all n?

(b) There are points x0 in [0, 1] such that limn→∞ fn(x0) = 0?Prove your answers.

(a) No, triangles to height n, width 1/n2.

(b) No, move triangles in (a) around.

Fall 2005 #3. (a) Prove that if fj : [0, 1] → R is a sequence of continuous functions which convergesuniformly on [0, 1] to a (necessarily continuous) function F : [0, 1]→ R then∫ 1

0

F 2(x)dx = limj→∞

∫ 1

0

f2j (x)dx.

53


(b) Give an example of a sequence fj : [0, 1] → R of continuous functions which converges to a continuousfunction F : [0, 1]→ R pointwise and for which

limj→∞

∫ 1

0

fj(x)2dx exists but

limj→∞

∫ 1

0

f2j (x)dx 6=

∫ 1

0

F 2(x)dx.

(fj converges to F “pointwise” means that for each x ∈ [0, 1], F (x) = limj→∞ fj(x)).

(a) Since F is continuous on a compact interval, it is bounded. Since the fj converge to F uniformly, itfollows that there exists J such that all fj with j ≥ J and f are uniformly bounded (say in [−M,M ]). Letε > 0. Choose J ′ > J such that |fj − F |sup ≤ ε/(2M) for all j ≥ J ′. It follows that

|f2j − F 2|sup = |fj − F |sup|fj + F |sup ≤ (ε/(2M))(2M) = ε.

Hence (f2j )∞j=1 converges uniformly to F 2. The result follows by previous exercises.

(b) Let fj(x) = j3/2(1/j − x) for x ∈ [0, 1/j] and 0 elsewhere. Then fj is continuous for each j and∫ 1

0

f2j (x) = j3

∫ 1/j

0

(1/j2 − 2x/j + x2)dx = j3(1/j3 − 1/j3 + 1/(3j3)) = 1/3.

Thus

limj→∞

∫ 1

0

f2j (x)dx = lim

j→∞1/3 = 1/3,

but F = 0, so

limj→∞

∫ 1

0

f2j (x)dx = 1/3 6= 0 =

∫ 1

0

F 2(x)dx.

Spring 2005 #8. Suppose f : R→ R is C1 (i.e., continuously differentiable). Show that

limn→∞

n∑j=1

∣∣∣∣f (j − 1

n

)− f

(j

n

)∣∣∣∣is equal to ∫ 1

0

|f ′(t)|dt.

Note f ′ is uniformly compact on [0, 1]. Let ε > 0. Let N be large enough that for all x, y ∈ [0, 1]with |x − y| < 1

N , we have ||f ′(x)| − |f ′(y)|| ≤ ε. Fix n ≥ N . By the mean value theorem, there exists

xn ∈ (j−1/n, j/n) such that f( j−1n )−f( jn ) = 1

nf′(xn). Also let |f ′| achieve its maximum at yn ∈ [j−1/n, j/n]

for each n. Then ∣∣∣∣∣∣∫ 1

0

|f ′(t)|dt−n∑j=1

∣∣∣∣f(j − 1

n)− f(

j

n)

∣∣∣∣∣∣∣∣∣∣ ≤

∣∣∣∣∣∣ 1nn∑j=1

|f ′(yn)| − 1

n

n∑j=1

|f ′(xn)|

∣∣∣∣∣∣≤ 1

n

n∑j=1

||f ′(yn)| − |f ′(xn)|| ≤ 1

n(nε) ≤ ε.

Hence

limn→∞

n∑j=1

∣∣∣∣f (j − 1

n

)− f

(j

n

)∣∣∣∣ =

∫ 1

0

|f ′(t)|dt.

54


Fall 2007 #5. (a) Show that, given a continuous function f : [0, 1]→ R, which vanishes at x = 1, there isa sequence of polynomials vanishing at x = 1 which converges uniformly to f on [0, 1].

(b) If f is continuous on [0, 1], and∫ 1

0

f(x)(x− 1)kdx = 0 for each k = 1, 2, . . . ,

show that f(x) is identically 0.

(a) Fix ε > 0. By the Weierstrass Approximation Theorem, there exists a polynomial P (x) such that|P (x)− f(x)| ≤ ε for all x ∈ [0, 1]. Consider the sequence of polynomials (P (x)(1− xn))∞n=1. Clearly thesepolynomials vanish at x = 1. We show they converge uniformly to f on [0, 1]. Since f(1) = 0, |P (1)| < ε/2.By continuity of P , there exists a δ > 0 such that |P (x)| < ε when |1 − x| < δ. Let M = sup[0,1] P (x)and let n be large enough that xn < ε/M on [0, 1 − δ]. Then if x ∈ [0, 1], either x > 1 − δ, in which case|xnP (x)| ≤ |P (x)| < ε, or x ≤ 1− δ, in which case |xnP (x)| < (ε/M)|P (x)| ≤ ε.

For any x ∈ [0, 1],|P (x)(1− xn)− f(x)| ≤ |P (x)− f(x)|+ |P (x)xn| ≤ 2ε.

Thus we have the desired uniform convergence.

(b) Let ε > 0. Select a sequence of polynomials (Pn)∞n=1 vanishing at x = 1 which converges uniformlyto f on [0, 1]. Choose N such that |Pn − f |sup ≤ ε for all n ≥ N . Now each Pn can be written as a finitesum of polynomials of the form (x− 1)k for k ≥ 1, hence∫ 1

0

f(x)Pn(x)dx = 0.

Since f is bounded on [0, 1], fPn also converges uniformly to f2. Thus by uniform convergence,∫ 1

0

f2(x)dx =

∫ 1

0

f(x)( limn→∞

Pn)dx = limn→∞

∫ 1

0

f(x)Pn(x)dx = limn→∞

0 = 0.

From here it follows easily that f is identically 0.

Spring 2007 #6. Consider the integral equation

y(t) = y0 +

∫ t

0

f(s, y(s))ds

where f(t, y) is continuous on [0, T ]× R and is Lipschitz in y with Lipschitz constant K. Assume that youhave shown that the iterates defined by

yn(t) = y0 +

∫ t

0

f(s, yn−1(s))ds, y0(t) identically y0

converge uniformly to a solution y(t) of the given integral equation. Show that if Y (t) is a solution of thegiven integral equation and satisfies |Y (t)− y0| ≤ C for some constant C and all t ∈ [0, T ], then Y (t) agreeswith y(t) on [0, T ].

Let ε > 0. We estimate

|Y (t)− y(t)| ≤ |Y (t)− yn(t)|+ |y(t)− yn(t)|.

The second term on the right can be made arbitrarily small by uniform convergence, so we focus on the first.It follows that

|Y (t)− yn(t)| ≤ Kn

∫ t

0

· · ·∫ t

0

|Y (sn)− y0|ds1 . . . dsn ≤ C(Kt)n.

55


Choosing n arbitrarily large, this implies Y (t) = y(t) for all t ∈ [0, 1/K). Now repeating the previousargument with the initial condition at t = i/K for each i necessary, we obtain Y (t) = y(t) on [0, T ].

Fall 2010 #10. Suppose f is bounded and Lipschitz continuous. For k ∈ N, define xk(t) : [0, 1] → R byxk(0) = 0 and

xk(t) = xk(n2−k) + (t− n2−k)f(xk(n2−k))

forn2−k < t ≤ (n+ 1)2−k, n ∈ N.

Explain why xk(t) uniformly converges to a solution x(t) : [0, 1]→ R of the ODE

x′(t) = f(x(t)), x(0) = 0,

as k →∞.

This is Euler’s Method of solving ODEs. Petersen’s solution in class is a less direct (and probably simpler)approach than this:

Let B be a bound for f and suppose |f(x1)− f(x2)| ≤ L|x1 − x2| for all x1, x2 ∈ R.We now verify that xk(t) converges to some x(t) as k → ∞. Let t ∈ [0, 1]. Fix ε > 0. Select K

such that 2−K ≤ ε. If t = 0, then clearly |xk(t) − xk′(t)| = 0 ≤ ε. Now for m ≥ 0 assume inductivelythat (xk(t))∞k=1 converges uniformly to some x(t) for all t ≤ m2−K . Increase K as necessary so that|xk(t)− xk′(t)| ≤ ε for all k, k′ ≥ K. Let k, k′ ≥ K. Now there exists n, n′ such that n2−k < t ≤ (n+ 1)2−k

and n′2−k′< t ≤ (n′ + 1)2−k. It follows that

|xk(t)− xk′(t)| = |xk(n2−k)− xk′(n′2−k′) + (t− n2−k)f(xk(n2−k))− (t− n′2−k

′)f(xk′(n

′2−k′)|

≤ |xk(n2−k)− xk′(n′2−k′)|+B|n2−k − n′2−k

′|+ (t− n′2−k

′)|f(xk(n2−k))− f(xk′(n

′2−k))|

≤ ε+B2−K + 2−KL|xk(n2−k)− xk′(n′2−k′)| ≤ (1 +B + Lε)ε.

After finitely many such iterations, we cover all of [0, 1]. Hence there exists x(t) such that xk → x pointwise.The above analysis also shows the sequence is uniformly Cauchy, hence its convergence to x is uniform.

Then since the xk are continuous, x is continuous. We now verify that x(t) is a solution to the ODE. Letε > 0 and fix t ∈ (0, 1]. Choose δ for continuity of f at x(t). Clearly x(0) = limk→∞ xk(0) = limk→∞ 0 = 0.Select K such that |xk − x|sup ≤ δ for all k ≥ K. By the continuity of f , we can adjust t by a tiny amountso that t does not take the form n2−k. Let k ≥ K and select n such that n2−k < t ≤ (n + 1)2−k. By thedefinition of xk,

x′k(t) = f(xk(n2−k)).

Let h < 2−k+1. Then the derivative quotient for xk is ε-close to f(xk(n2−k)). Then

|x(t+ h)− x(t)

h− f(x(t))| ≤ 1

2−k+1(|x(t+ h)− xk(t+ h)|+ (x(t)− xk(t))|)

+|xk(t+ h)− xk(t)

h− f(xk(t))|+ |f(xk(t))− f(x(t))|

≤ 2k−12ε+ ε+ ε.

Hence the limit as h→ 0 of the left hand side is 0, so x(t) is differentiable with x′(t) = f(x(t)), as desired.

Fall 2008 #10. Given v = (v1, . . . , vn) ∈ Rn, we let ||v|| = (∑|vj |2)1/2. If f = (f1, . . . , fn) : [a, b]→ Rn is

a continuous function, we define∫ b

a

f(t)dt =

(∫ b

a

f1(t)dt, . . . ,

∫ b

a

fn(t)dt

).

56


Prove that

||∫ b

a

f(t)dt|| ≤∫ b

a

||f(t)||dt.

Let y = (y1, . . . , yk), with yj =∫ bafj(t)dt. Then by definition y =

∫ baf(t)dt. Note that

||y||2 =

n∑j=1

y2j =

n∑j=1

yj

∫ b

a

fj(t)dt =

∫ b

a

n∑j=1

yjfj(t)dt

.

By the Cauchy-Schwarz inequality,n∑j=1

yjfj(t) ≤ ||y||||f(t)||,

thus since the integral is monotonic,

||y||2 ≤∫ b

a

||y||||f(t)||,

so

||y|| ≤∫ b

a

||f(t)||.

Fall 2009 #10. (i) Let I = [0, 2]. If f : I → R is a continuous function such that∫If(x)dx = 36, prove

that there is an x ∈ I such that f(x) = 18.

(ii) Consider I2 ⊂ R2, and let g : I2 → R be a continuous function such that∫I2g(x, y)dxdy = 36. Prove

that there is (x, y) ∈ I2 such that g(x, y) = 9.

(a) Let h(t) =∫ t

0f(x)dx. Then by the fundamental theorem of calculus, h is differentiable with h′(x) =

f(x). The mean value theorem then ensures there exists x ∈ [0, 2] = I with

36 =

∫I

f(x)dx = h(2)− h(0) = h′(x)(2− 0) = 2f(x).

Thus f(x) = 18.

(b) Suppose for the sake of contradiction that there is not (x, y) ∈ I2 such that g(x, y) = 9. Then theintermediate value theorem implies that either g(x, y) > 9 for all (x, y) ∈ I2 or g(x, y) < 9 for all (x, y) ∈ I2.Either leads to a contradiction with

∫I2g(x, y)dxdy = 36.

Spring 2009 #12. Let F : R3 → R3 and ρ : R3 → R be smooth functions. Show that

div(F ) = ρ

for all points (x, y, z) ∈ R3 if and only if∫ ∫∂Ω

F · dS =

∫ ∫ ∫Ω

ρdxdydz

for all balls Ω (with all radii r > 0 and all possible centers). [You may use the various standard theorems ofvector calculus without proof.]

Write F (x, y, z) = (F1(x, y, z), F2(x, y, z), F3(x, y, z)). Since F is smooth, Fi is smooth. We verify thedivergence theorem for one of the components, and the others follow similarly. Then by adding up the results

57


on both sides, we get the desired equality. We can express Ω = (x, y, z) : f1(x, y) ≤ z ≤ f2(x, y), (x, y) ∈ D,where D is the unit disc. We can split ∂Ω into S1 = (x, y, z) : z = f1(x, y), (x, y) ∈ D and S2 = (x, y, z) :z = f2(x, y), (x, y) ∈ D. Then by the fundamental theorem of calculus and the definition of surface integrals,∫ ∫ ∫

Ω

∂F3

∂zdzdxdy =

∫ ∫D

∫ f2(x,y)

f1(x,y)

∂F3

∂zdzdxdy

=

∫ ∫D

F3((x, y, f2(x, y)))− F3((x, y, f1(x, y)))dxdy

=

∫ ∫D

F3((x, y, f2(x, y)))−∫ ∫

D

F3((x, y, f1(x, y))dxdy

=

∫ ∫S2

(0, 0, F3) · dS +

∫ ∫S1

(0, 0, F3) · dS =

∫ ∫∂Ω

(0, 0, F3) · dS.

Fall 2010 #12. Let us define D(t) = x2 + y2 ≤ r2(t) ⊂ R2, where r(t) : R → R is continuouslydifferentiable. For a given smooth, nonnegative function u(x, t) : R2×R→ R, express the following quantityin terms of a surface integral:

d

dt

(∫D(t)

u(x, t)dx

)−∫D(t)

ut(x, t)dx.

[You may use various theorems in Calculus without proof.]

We use Liebniz’s Rule for differentiation under the integral sign, which requires the appropriate derivativeto exist and be continuous.

First, switch to polar coordinates:∫D(t)

u(x, t)dx =

∫ 2π

0

∫ r(t)

0

u(r, θ, t)rdrdθ.

Then by Liebniz’s Rule,

d

dt

(∫ r(t)

0

u(r, θ, t)rdr

)= u(r(t), θ, t)

dr(t)

dtr +

∫ r(t)

0

∂u

∂t(r, θ, t)dr.

Applying Liebniz’s Rule again on the outer integral, we obtain

d

dt

∫D(t)

u(x, t)dx = rdr(t)

dθ

∫ 2π

0

u(r(t), θ, t)dθ +

∫D(t)

∂u

∂t(x, t)dx.

Thus the desired difference is

rdr(t)

dθ

∫ 2π

0

u(r(t), θ, t)dθ.

58


Taylor Series

Fall 2003 #5. Assume f : R2 → R is a function such that all partial derivatives of order 3 exist and arecontinuous. Write down (explicitly in terms of partial derivatives of f) a quadratic polynomial P (x, y) in xand y such that

|f(x, y)− P (x, y)| ≤ C(x2 + y2)3/2

for all (x, y) in some small neighborhood of (0, 0), where C is a number that may depend on f but not on xand y. Then prove the above estimate.

Let

P (x, y) = f(0) + fx(0)x+ fy(0)y +1

2[fxx(0)x2 + fxy(0)xy + fyx(0)xy + fyy(0)y2].

Since f is continuous, fxy = fyx and

P (x, y) = f(0) + fx(0)x+ fy(0)y +1

2[fxx(0)x2 + 2fxy(0)xy + fyy(0)y2].

By Taylor’s Theorem with Remainder, for any v = (x, y) ∈ R2, defining g(t) = f((1 − t)0 + tv) = f(tv),there exists t∗ ∈ (0, 1) such that

R(v) := f(v)− P (v) =g(3)(t∗)

3!.

It follows thatg(3)(t∗)

3!=∑|α|=3

∂αf(t∗v)vα

α!.

Since the third partials of f are continuous, they are bounded on the compact set B(0, 1). Let M be thisbound. Then for any (x, y) ∈ B(0, 1),

|g(3)(t∗)

3!| ≤M |x3 + x2y + y2x+ y3| ≤ 4Mmax|x|, |y|3 ≤ 4M(x2 + y2)3/2.

Note M depends solely on f , not x and y.

Winter 2006 #5. Consider a function f(x, y) which is twice continuously differentiable. Suppose that fhas its unique minimum at (x, y) = (0, 0). Carefully prove that then at (0, 0),

∂2f

∂x2

∂2f

∂y2≥(∂2f

∂x∂y

)2

.

[You may use without proof that the mixed partials are equal for C2 functions.]

Consider the Hessian matrix,

H = H(0, 0) =

(∂2f∂x2 (0, 0) ∂2f

∂x∂y (0, 0)∂2f∂y∂x (0, 0) ∂2f

∂y2 (0, 0)

).

By Taylor’s Theorem,

f(x, y) = f(0, 0) +∇f(0, 0) · (x, y) +1

2(x, y) ·H(x, y) +R(x, y).

Since (0, 0) is the unique minimum of f , it is a critical point, so ∇f(0, 0) = 0. Suppose for the sake ofcontradiction that det(H) < 0. Then H has a positive eigenvalue λ1 and a negative eigenvalue λ2. Choosev2 with Hv2 = λ2v2. Then f(tv2) = f(0, 0) + 1

2 t2λ2|v2|2 + R(tv2), so f(tv2) has a local maximum when

t = 0. Thus, (0, 0) cannot be the unique minimum point of f , a contradiction.

59


Fall 2010 #3. Suppose f : R→ R and g : R2 → R have continuous derivatives up to order three.

(a) State Taylor’s Theorem with remainder for each of f and g.

(b) Using the statement for f , prove the statement for g.

(a) For any x0, x ∈ R, there exists λ between x and x0 such that

f(x) = f(x0) + (x− x0)f ′(x0) +1

2f ′′(x0)(x− x0)2 +

1

3!f (3)(λ).

Likewise, for any (x0, y0), (x, y) ∈ R2, there exists a function h(x0,y0) : R2 → R such that

g((x, y)) = g((x0, y0)) +∂g

∂x(x−x0) +

∂g

∂y(y− y0) +

1

2(∂2g

∂x2(x−x0)2 + 2

∂2g

∂x∂y(x−x0)(y− y0) +

∂2g

∂y2(y− y0)2)

+∑|α|=3

h(x0,y0)((x, y))((x, y)− (x0, y0))α,

and lim(x,y)→(x0,y0)

h(x0,y0)((x, y)) = 0.

(b) It suffices to verify the statement for g when (x0, y0) = 0. Let (x, y) ∈ R2 and define f(t) := g(tx, ty).Then f has continuous derivatives up to order three. By the chain rule,

f ′(t) = ∇f(tx, ty) · (x, y),

andf ′′(t) = (x, y) ·H(tx,ty)(x, y)t.

By Taylor’s Theorem in 1 dimension, there exists a function h0 : R→ R such that limt→0 h(t) = 0 and

f(t) = f(0) + f ′(t)t+1

2f ′′(t)t2 +

1

3!h0(t)t3.

Evaluating at t = 1, we reproduce Taylor’s Theorem in 2 dimensions for g.

Spring 2008 #3. Assuming that f ∈ C4[a, b] is real, derive a formula for the error of approximation E(h)when the second derivative is replaced by the finite-difference formula

f ′′(x) ≈ f(x+ h)− 2f(x) + f(x− h)

h2,

and h is mesh size. (Assume that x, x+ h, x− h ∈ (a, b)).

By Taylor’s Theorem, for some λ ∈ (x, x+ h),

f(x+ h) = f(x) + hf ′(x) +h2

2f ′′(x) +

h3

3!f (3)(x) +

h4

4!f (4)(λ).

Likewise, for some λ′ ∈ (x− h, x),

f(x+ h) = f(x)− hf ′(x) +h2

2f ′′(x)− h3

3!f (3)(x) +

h4

4!f (4)(λ′).

Thus

f(x+ h)− 2f(x) + f(x− h) = h2f ′′(x) +h4

4!(f (4)(λ) + f (4)(λ′)).

60


Since f (4) is continuous, it follows that the error in approximation is about h2

12 |f4(x)| for small h.

Fall 2007 #4. Suppose that f : R → R is twice differentiable and its second derivative, f ′′ satisfies|f ′′(x)| ≤ B.

(a) Prove that

|2Af(0)−∫ A

−Af(x)dx| ≤ A3

3B.

(b) Use the result of part (a) to justify the following estimate:∣∣∣∣∣∫ b

a

f(x)dx− b− an

n∑k=1

f(a+2k − 1

2n(b− a))

∣∣∣∣∣ ≤ Cn−2,

where C is a constant that does not depend on n.

(a) Fix an A > 0. By Taylor’s Theorem, there exists, for any [−A,A], some c ∈ [−A,A] such that

f(x)− f(0)− f ′(0)x =f ′′(c)c2

2.

Thus for any x ∈ [−A,A], |f(0) + f ′(0)x− f(x)| ≤ A2B/2. We have

|2Af(0)−∫ A

−Af(x)dx| = |

∫ A

−Af(0) + f ′(0)x− f(x)− f ′(0)xdx|

≤ |∫ A

−Af(0) + f ′(0)x− f(x)dx|+ |

∫ A

−Af ′(0)xdx|

≤∫ A

−A|f(0) + f ′(0)x− f(x)|dx

≤∫ A

−A

A2B

2dx =

A3

3B.

(b) Fix n. For 1 ≤ k ≤ n, put fk(x) = f(x− a− 2k−12n (b− a)). Then∫ b

a

f(x)dx =

n∑k=1

∫ 12n (b−a)

− 12n (b−a)

fk(x)dx.

By applying part (a) with A = 12n (b− a) to the fk(x), we obtain

|∫ b

a

f(x)dx− b− an

n∑k=1

f(a+2k − 1

2n(b− a))| = |

n∑k=1

∫ 12n (b−a)

− 12n (b−a)

fk(x)dx− b− an

n∑k=1

fk(0)|

≤n∑k=1

|∫ 1

2n (b−a)

− 12n (b−a)

fk(x)dx− b− an

fk(0)|

≤n∑k=1

B

3(b− a2n

)3 =B(b− a)3

24n−2.

61


Winter 2006 #3. Consider a function f : [a, b] → R which is twice continuously differentiable (includingthe endpoints). Let a = x0 < x1 < · · · < xn = b be the uniform partition of [a, b], i.e., xi+1 − xi = (b− a)/nfor all 0 ≤ i < n. Show that there exists M such that for all n ≥ 1,∣∣∣∣∣ 1n

(1

2f(x0) + f(x1) + · · ·+ f(xn−1) +

1

2f(xn)

)−∫ b

a

f(x)dx

∣∣∣∣∣ ≤ M

n2.

[Recall that the sum is an approximation of the integral in the Trapezoid Rule. It may be instructive to firstsolve the problem for n = 1 and then address the general case.]

In the case n = 1, consider the line p(x) = f(a) + (x− a)(f(b))/(b− a). Note p(a) = f(a), p(b) = f(b),

and∫ bap(x) = ( f(a)

2 + f(b)2 )(b− a). See the problem below for the general theory.

The general case follows easily from this case.

Spring 2013 #2. The approximation from “Simpson’s Rule” for∫ baf(x)dx is

S[a,b]f =

[2

3f

(a+ b

2

)+

1

3

(f(a) + f(b)

2

)](b− a).

If f has continuous derivatives up to order three, prove that∣∣∣∣∣∫ b

a

f(x)dx− S[a,b]f

∣∣∣∣∣ ≤ C(b− a)4 max[a,b]|f (3)(x)|,

where C does not depend on f .

We first want to find a quadratic polynomial p(x) such that p(a) = f(a), p(b) = f(b), and p((a+ b)/2) =f((a+ b)/2). This is

p(x) = f(a)(x− (a+ b)/2)(x− b)(a− (a+ b)/2)(a− b)

+f((a+b)/2)(x− a)(x− b)

(a− (a+ b)/2)(b− (a+ b)/2)+f(b)

(x− a)(x− (a+ b)/2)

(b− a)(b− (a+ b)/2).

Integrating p from a to b apparently gives the left hand side of the desired inequality.Suppose p matches f in N + 1 places x0, . . . , xN . For some η ∈ [a, b],

f(x)− pN (x) =(x− x0) · · · (x− xN )

(N + 1)!f (N+1)(η).

Thus

|∫ b

a

fdx−∫ b

a

pNdx| ≤(b− a)N+2

(N + 1)!max[a,b](|f (N+1)(x)|).

62


Jacobian

Spring 2002 #7; Winter 2002 #7. Suppose F : R2 → R2 is continuously differentiable and that theJacobian matrix of F is everywhere nonsingular. Suppose also that F (0) = 0 and that ||F ((x, y))|| ≥ 1 forall (x, y) with ||(x, y)|| = 1. Prove that

(x, y) : ||(x, y)|| < 1 ⊂ F ((x, y) : ||(x, y)|| < 1).

(Hint: Show, with U = (x, y) : ||(x, y)|| < 1, that F (U) ∩ U is both open and closed in U).

Let V = F (U). Define O1 = V ∩ U . If O1 is closed in U , then O2 = (V ∩ U)c ∩ U is open. Thus if O1

is also open, we have that U = O1 ∪ O2 with O1, O2 disjoint open sets. But U is connected, and 0 ∈ O1,hence O2 = ∅. Thus U = O1 = V ∩ U , so F (U) ⊃ U .

Thus it suffices to show that O1 = F (U) ∩ U is both open and closed in U .Let (x0, y0) ∈ F (U) ∩ U . We are given that F ′((x0, y0)) is nonsingular. Thus by the inverse function

theorem, the range of F on B((x0, y0), r) contains an open neighborhood of F ((x0, y0)) for sufficiently smallr > 0. Thus F (U) is open, hence U ∩ F (U) is open in U .

Let pn be a sequence of points in F (U) ∩ U converging to some p∞ ∈ U . Then there exist qn ∈ Uwith pn = F (qn). By the Bolzano-Weierstrass Theorem, there exists a subsequence qnk

which convergesto some q∞ ∈ U . By continuity of F ,

p∞ = F (q∞).

Now if ||q∞|| = 1, then ||p∞|| = ||F (q∞)|| ≥ 1, so p∞ /∈ U , a contradiction. Hence ||q∞|| < 1, so q∞ ∈ U andp∞ ∈ F (U) ∩ U . Hence F (U) ∩ U is closed relative to U .

Fall 2003 #6. Let U = (x, y) : x2 + y2 < 1 be the standard unit ball in R2 and let ∂U denote itsboundary. Suppose F : R2 → R2 is continuously differentiable and that the Jacobian determinant of Fis everywhere non-zero. Suppose also that F (x, y) ∈ U for some (x, y) ∈ U and F (x, y) /∈ U ∪ ∂U for all(x, y) ∈ ∂U . Prove that U ⊂ F (U).

Same reasoning as the previous problem. These hypotheses are just a bit more general.

Lagrange multipliers

Spring 2003 #5. Consider the function F (x, y) = ax2 + 2bxy + cy2 on the set A = (x, y) : x2 + y2 = 1.

(a) Show that F has a maximum and minimum on A.(b) Use Lagrange multipliers to show that if the maximum of F on A occurs at a point (x0, y0), then the

vector (x0, y0) is an eigenvector of the matrix

(a bb c

).

(a) A is a compact set, since x2 +y2 is continuous. Thus the continuous function F achieves its maximumand minimum on A.

(b) If a maximum occurs at (x, y, z), the method of Lagrange multipliers implies there exists λ such that(∇F )(x, y, z) = λ(∇g)(x, y, z). In this case,

(2ax+ 2by, 2bx+ 2cy) = λ(2x, 2y).

Thus (a bb c

)(xy

)= λ

(xy

),

so (x, y) is an eigenvector of the given matrix.

63


Miscellaneous

Spring 2005 #5. For a subset X ⊂ R, we say that X is algebraic if there exists a family F of polynomialswith rational coefficients, so that x ∈ X if and only if p(x) = 0 for some p ∈ F .

(a) Show that the set Q of rational numbers is algebraic.

(b) Show that the set R \Q of irrational numbers is not algebraic.

(a) Let F = (x− q) : q ∈ Q. Then for every q ∈ Q, (x− q)(q) = 0, so Q is algebraic.

(b) The set of algebraic numbers is countable, but R \Q is uncountable.

Fall 2009 #3. The purpose of this problem is to give a multi-variable calculus proof of the geometric andarithmetic means inequality along the concrete steps below. The inequality has numerous other proof andnaturally you are not allowed to use it (or them) below.

Let Rn+ ⊂ Rn be the (open) subset of vectors all whose coordinates are positive, and f : Rn+ ⊂ R bedefined by:

f(x1, . . . , xn) = x1 + · · ·+ xn +1

x1 · x2 · · ·xn.

(i) Explain carefully why f attains a global (not necessarily unique) minimum at some p ∈ Rn+. (Hint: whathappens when xi → 0,∞?)

(ii) Find p.

(iii) Deduce that if all xi ∈ R are positive and∏xi = 1 then

∑xi ≥ n, with equality iff xi = 1 for all i.

(This is a special case of the geometric and arithmetic means inequality, from which the general statementcan be immediately deduced - no need to write down this part here.)

(i) As some xi → 0 or xi →∞, f(x1, . . . , xn)→∞. Thus we can restrict to a compact set, and f attainsits minimum on the compact set.

(ii) The gradient of f is 0, hence p = (1, 1, . . . , 1).

(iii) Obvious from (ii).

Spring 2012 #5. Prove that there is a unique continuous function y : [0, 1]→ R solving the equation

y(x) = ex +y(x2)

2, x ∈ [0, 1].

Define F (x, y(x)) = ex+ y(x2)2 . Then F is a contraction in y, so it has a unique fixed point. The contraction

condition forces Fn(x, y(x)) to converge uniformly to the fixed point, so the fixed point is continuous.

Spring 2012 #6. Let γ be a smooth curve from (1, 0) to (1, 0) in R2 \ (0, 0) winding once around theorigin in the clockwise direction. Compute the integral

I(γ) :=

∫γ

ydx− xdyx2 + y2

.

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The problem does not specify we should show the integral is path independent, so we must merely choosea curve and compute. Let γ be the unit circle with clockwise orientation, given by the parametrization

γ(t) = (cos(t), sin(t))

for t ∈ [0, 2π). Then

I(γ) :=

∫γ

ydx− xdyx2 + y2

=

∫ 2π

0

sin(t)(− sin(t))− cos(t) cos(t)dt

1= −2π.

Spring 2013 #5. Define the polynomials Un(x), n = 0, 1, 2, . . . as follows:

U1(x) = 1, U2(x) = 2x, Un+1(x) = 2xUn(x)− Un−1(x).

(a) Prove that

Un(cos(θ)) =sin(nθ)

sin(θ).

(b) Prove that the polynomials Un(x) satisfy:∫ 1

−1

Um(x)Un(x)√

1− x2dx =

0 when m 6= n,π/2 when m = n.

(a) We proceed by induction. For n = 1,

Un(cos(θ)) = 1 = sin(θ)/ sin(θ) = sin(nθ)/ sin(θ).

Now assume inductively that

Un(cos(θ)) =sin(nθ)

sin(θ)

for all non-negative integers n ≤ k. Then

Uk+1(cos(θ)) = 2 cos(θ)Uk(cos(θ))− Uk−1(cos(θ)) = 2 cos(θ)sin(nθ)

sin(θ)− sin((n− 1)θ)

sin(θ)

=2 cos(θ) sin(nθ)− (sin(nθ) cos(θ)− sin(θ) cos(nθ))

sin(θ)

=sin(nθ) cos(θ) + sin(θ) cos(nθ)

sin(θ)=

sin((n+ 1)θ)

sin(θ),

which completes the induction.

(b) Make the change of variables x = cos(θ), where θ ranges from π to 0. Then using part (a), theintegral under consideration becomes∫ 0

π

sin(mθ)

sin(θ)

sin(nθ)

sin(θ)sin(θ)(− sin(θ))dθ =

∫ π

0

sin(mθ) sin(nθ)dθ.

Use the identity

sin(α) sin(β) =1

2(cos(α− β)− cos(α+ β))

to finish the proof.

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Spring 2005 #11. Let us make Mn(C) into a metric space in the following fashion:

dist(A,B) =

∑i,j

|Ai,j −Bi,j |2

1/2

(which is just the usual metric on Rn2

.

(a) Suppose F : R→Mn(C) is continuous. Show that the set

x ∈ R : F (x) is invertible

is open (in the usual topology on R).

(b) Show that on the set given above, x 7→ |F (x)|−1 is continuous.

(a) The determinant function is continuous, thus x ∈ R : F (x) is invertible = x ∈ R : det(F (x)) 6= 0is open as a pullback of an open set.

(b) I assume |F (x)| = det(F (x)). Then this is the composition of the inverse function, the determinant,and F , which are all continuous on the given set, hence the composition is continuous.

Fall 2011 #2. A function f : Rn → R is called convex if f satisfies

f(αx+ (1− α)y) ≤ αf(x) + (1− α)f(y), for all x, y ∈ Rn, 0 ≤ α ≤ 1.

Assume that f is continuously differentiable and that, for some constant c > 0, the gradient ∇f satisfies

(∇f(x)−∇f(y)) · (x− y) ≥ c(x− y) · (x− y), for all x, y ∈ Rn,

where · denotes the dot product. Show that f is convex.

Fix x, y ∈ Rn. Let ϕ(t) = f(tx + (1 − t)y) and ψ(t) = tf(x) + (1 − t)f(y) for all t ∈ [0, 1]. We wantto show ϕ − ψ ≤ 0 on [0, 1]. Note (ϕ − ψ)(0) = (ϕ − ψ)(1) = 0. It is sufficient to show that d

dt (ϕ − ψ) ismonotone increasing in t. (The mean value theorem guarantees this by contradiction.) We compute

d

dt(ϕ(t)− ψ(t)) = ∇ftx+(1−t)y · (x− y)− (f(x)− f(y)).

Suppose t1 ≤ t2. Then by assumption,

(∇ft2x+(1−t2)y −∇ft1x+(1−t1)y) · (t2 − t1)(x− y) ≥ c(t2 − t1)2(x− y) · (x− y).

Thus(∇ft2x+(1−t2)y) · (x− y) > (∇ft1x+(1−t1)y) · (x− y),

so ddt (ϕ− ψ) is monotone increasing.

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Linear Algebra

Important definitions and facts:

A monic polynomial p ∈ F[t] is said to be irreducible if the only monic polynomials from F[t] that dividep and 1 and p.

We say v 6= 0 is an eigenvector of L : V → V if there exists λ ∈ F with Lv = λv. In this case λ is calledan eigenvalue of L. The eigenspace corresponding to λ is

Eλ := ker(L− λ1V ).

The sum of eigenspaces for distinct eigenvalues is their direct sum.

The dual space of V is V ′ = hom(V,F).

The annihilator of a subspace M ⊂ V is the subspace M ⊂ V ′ such that

M = f ∈ V ′ : f(x) = 0 for all x ∈M.

The geometric multiplicity of an eigenvalue λ is dim(Eλ). The algebraic multiplicity of an eigenvalue λis the number of times λ appears as a root of the characteristic polynomial. Geometric multiplicity of aneigenvalue is always less than or equal to its algebraic multiplicity.

The characteristic polynomial of L is χL(t) = det(L− t1V ). Its roots are eigenvalues of L. Over C, thereare always n eigenvalues with λ1 · · ·λn = (−1)na0 and tr(L) = λ1 + · · ·+λn = −an−1. Complex roots alwayscome in conjugate pairs.

An involution L is a linear operator such that L2 = 1V .

The minimal polynomial µL(t) of L is the monic polynomial of smallest rank such that µL(L) = 0. Alleigenvalues of L are roots of P . Note L is invertible if and only if a0 6= 0. Note µL divides any polynomialp with p(L) = 0. In particular, µL divides χL.

If two linear operators on an n-dimensional vector space have the same minimal polynomials of degreen, then they have the same Frobenius canonical form and are similar.

An operator L : V → V is said to be diagonalizable if we can find a basis for V that consists of eigenvectorsof L. In other words, the matrix representation for L is a diagonal matrix. Note this depends on V as wellas L.

A subspace M ⊂ V is said to be L-invariant if L(M) ⊂M .

The companion matrix of a monic polynomial p(t) ∈ F[t] is

Cp =

0 0 · · · 0 −α0

1 0 · · · 0 −α1

0 1 · · · 0 −α2

......

. . ....

...0 0 · · · 1 −αn−1

,

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when p(t) = tn + αn−1tn−1 + · · ·+ α1t+ α0.

The characteristic and minimal polynomials of Cp are both p(t) and all eigenspaces are one-dimensional.Thus Cp is diagonalizable if and only if all the roots of p(t) are distinct and lie in F.

The cyclic subspace corresponding to x is

Cx = spanx, L(x), L2(x), . . . , Lk−1(x),

where k is the smallest integer with

Lk(x) ∈ spanx, L(x), L2(x), . . . , Lk−1(x).

An inner product on a vector space over F (R or C) is an F valued pairing (x|y) for x, y ∈ V , i.e., a map(x|y) : V × V → F, that satisfies

(1) (x|x) ≥ 0 and vanishes only when x = 0.(2) (x|y) = (y|x).(3) For each y ∈ V , the map x→ (x|y) is linear.The associated norm is ||x|| =

√(x|x).

The adjoint is the transpose but with each entry conjugated, notation A∗ = At. This satisfies

(Ax|y) = (x|A∗y).

Note (L2L1)∗ = L∗1L∗2 and if L is invertible, (L−1)∗ = (L∗)−1. Note λ is an eigenvalue for L if and only if λ

is an eigenvalue for L∗. Moreover, these eigenvalue pairs have the same geometric multiplicity.

An orthogonal projection is a projection (E2 = E) for which the range and the null space are orthogonalsubspaces. E is orthogonal if and only if E = E∗ (E is self-adjoint, or Hermitian). Skew-adjoint: E∗ = −E.When over a real space, this becomes symmetric and skew-symmetric.

A map is completely reducible or semi-simple if for each invariant subspace one can always find a comple-mentary invariant subspace. Both self-adjoint and skew-adjoint maps are completely reducible (if L(M) ⊂M ,then L(M⊥) ⊂M⊥.

Two inner product spaces V and W are isometric, if we can find an isometry L : V → W , i.e., anisomorphism such that (L(x)|L(y)) = (x|y).

Let L : V →W be an isomorphism. Then L is an isometry if and only if L∗ = L−1. When V = W = Rn,isometries are called orthogonal matrices. When V = W = Cn, isometries are called unitary matrices.

Gram-Schmidt procedure: Given a linearly independent set x1, . . . , xm, set e1 = x1/||x1|| and induc-tively,

zk+1 = xk+1 − (xk+1|e1)e1 − · · · − (xk+1|ek)ek

andek+1 = zk+1/||zk+1||.

The operator norm is||L|| = sup||Lx|| : ||x|| = 1.

This is finite in a finite dimensional inner product space. Note this implies ||L(x)|| ≤ ||L|| ||x|| for all x ∈ V .

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The orthogonal complement to M in V is

M⊥ = x ∈ V : (x|z) = 0 for all z ∈M.

An operator L : V → V on an inner product space is normal if LL∗ = L∗L. All self-adjoint, skew-adjoint,and isometric operators are clearly normal.

Two n × n matrices A and B are said to be unitarily equivalent if A = UBU∗, where U ∈ Un (U isunitary). If U is orthogonal, A and B are said to be orthogonally equivalent. If A and B are unitarilyequivalent, A is normal / self-adjoint / skew-adjoint / unitary if and only if B is normal / self-adjoint /skew-adjoint / unitary. Two normal operators are unitarily equivalent if and only if they have the sameeigenvalues (counted with multiplicities).

A unitary matrix U satisfies U∗U = UU∗ = I. An orthogonal matrix O satisfies OOt = OtO = I.

Tr(AB) = Tr(BA).

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Important Theorems:

The Fundamental Theorem of Algebra. Any complex polynomial of degree ≥ 1 has a root.

Characterizations of Diagonalizability. 1. V is n-dimensional, and the sum of the geometric mul-tiplicities of all the eigenvalues is n. In particular, if all the eigenvalues are distinct, then the operator isdiagonalizable.

2. There exists p ∈ F[t] such that p(L) = 0 and

p(t) = (t− λ1) · · · (t− λk)

where λ1, . . . , λk are distinct.Corollary: L is diagonalizable if and only if µL(t) factors as

µL(t) = (t− λ1) · · · (t− λk),

with λi distinct. Since µL divides χL, all the λi must be eigenvalues of L.Note a polynomial p has a multiple root if and only if p and Dp have a common root.

Cayley-Hamilton Theorem. Let L be a linear operator on a finite dimensional vector space. Then Lis a root of its own characteristic polynomial:

χL(L) = 0.

In particular, µL(t) divides χL(t).

Cyclic Subspace Decomposition. Let L : V → V be a linear operator on a finite dimensional vectorspace. Then V has a cyclic subspace decomposition

V = Cx1⊕ · · · ⊕ Cxk

where each Cx is a cyclic subspace. In particular, L has a block diagonal matrix representation where eachblock is a companion matrix:

[L] = Cp1 ⊕ Cp2 ⊕ · · · ⊕ Cpkand χL(t) = p1(t) · · · pk(t). Moreover, the geometric multiplicity satisfies

dim(ker(L− λ1V )) = number of pis such that pi(λ) = 0.

In particular, we see that L is diagonalizable if and only if all of the companion matrices Cp individuallyhave distinct eigenvalues. (In general this decomposition is not unique.)

Frobenius Canonical Form (Rational Canonical Form). Let L : V → V be a linear operator on afinite dimensional vector space. Then V has a cyclic subspace decomposition such that the block diagonalform of L,

[L] = Cp1 ⊕ Cp2 ⊕ · · · ⊕ Cpkhas the property that pi divides pi−1 for each i = 2, . . . , k. Moreover, the monic polynomials p1, . . . , pk areunique. (The pi are called the similarity invariants or invariant factors for L. Similar matrices have the samesimilarity invariants.)

Jordan-Chevalley decomposition. Let L : V → V be a linear operator on an n-dimensional complexvector space. Then L = S +N , where S is diagonalizable, Nn = 0, and SN = NS.

70


For p(t) = (t− λ)n, Cp is similar to a Jordan block

[L] =

λ 1 0 · · · 0 00 λ 1 · · · 0 0

0 0 λ. . .

......

0 0 0. . . 1 0

......

... · · · λ 10 0 0 · · · 0 λ

.

Moreover the eigenspace for λ is 1-dimensional and is generated by the first basic vector.

Jordan Canonical Form Let L : V → V be a complex linear operator on a finite dimensional vectorspace. Then we can find L-invariant subspaces M1, . . . ,Ms such that

V = M1 ⊕ · · · ⊕Ms

and each L|Mi has a matrix representation of the form

[L] =

λi 1 0 · · · 0 00 λi 1 · · · 0 0

0 0 λi. . .

......

0 0 0. . . 1 0

......

... · · · λi 10 0 0 · · · 0 λi

where λi is an eigenvalue for L. Note each eigenvalue corresponds to as many blocks as the geometricmultiplicity of the eigenvalue.

Cauchy-Schwarz Inequality|(x|y)| ≤ ||x|| ||y||

In Rn, (x|y) is usually x · y.

Uniqueness of Inner Product Spaces. An n-dimensional inner product space over R, respectivelyC, is isometric to Rn, respectively Cn.

Orthogonal Sum Decomposition. Let V be an inner product space and M ⊂ V a finite dimensionalsubspace. Then V = M ⊕M⊥ and for any orthonormal basis e1, . . . , em for M , the projection onto M alongM⊥ is given by

projM (x) = (x|e1)e1 + · · ·+ (x|em)em.

Also projM (x) is the one and only point closest to x among all points in M .

Polarization. Let L : V → V be a linear operator on a complex inner product space. Then L = 0 ifand only if (L(x)|x) = 0 for all x ∈ V .

Existence of Eigenvalues for Self-adjoint Operators. Let L : V → V be self-adjoint and V finitedimensional. Then L has a real eigenvalue.

The Spectral Theorem. Let L : V → V be a self-adjoint operator on a finite dimensional inner productspace. Then there exists an orthonormal basis e1, . . . , en of eigenvectors of L. Moreover, all eigenvalues of Lare real.

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The Spectral Theorem for Normal Operators. Let L : V → V be a normal operator on a complexinner product space. Then there is an orthonormal basis e1, . . . , en of V consisting of eigenvectors of L.

Schur’s Theorem. Let L : V → V be a linear operator on a finite dimensional complex inner productspace. It is possible to find an orthonormal basis e1, . . . , en such that the matrix representation [L] is uppertriangular in this basis.

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Recurring Linear Algebra Problems

Fall 2001 #9; Fall 2002 #10; Spring 2006 #9. Let A ∈ Mn(C) (Mn(R)) be a normal (self-adjoint)matrix. Prove that there exists an orthonormal basis of Cn (Rn) such that the matrix of A is diagonal withrespect to this basis.

Real case: Let S be a real n-dimensional inner product space. We proceed by induction on n to showthat for each self-adjoint map L : S → S, there exists an orthonormal basis for S consisting of eigenvectors ofL. The case n = 1 is trivial. Let n ≥ 2 and assume inductively that for all real n− 1-dimensional real innerproduct spaces T and self-adjoint maps L′, there exists an orthonormal basis of T consisting of eigenvectorsof L′.

Lemma 1: If L = L∗ and λ ∈ C such that for some v ∈ Cn with v 6= 0, Lv = λv, then λ is necessarily real.

Proof: Using that L is self-adjoint,

λ(v|v) = (λv|v) = (T (v)|v) = (v|T ∗(v)) = (v|T (v)) = (v|λv) = λ(v, v).

Thus since v 6= 0, (v|v) 6= 0, so λ = λ, and λ is real.

Lemma 2: Let λ be an eigenvalue of L and V be the eigenspace corresponding to λ. Then V ⊥ is L-invariant.

Proof: For any v ∈ V and u ∈ V ⊥,

(L(u)|v) = (v|L∗(u)) = (v|L(u)) = λ(v|u) = 0.

Thus L(u) ∈ V ⊥, so L(V ⊥) ⊂ V ⊥ and V ⊥ is L-invariant.

Let A be a matrix representation of L. Let χA(t) = det(A− tIdn×n) be the characteristic polynomial ofL. The fundamental theorem of algebra guarantees there exists λ ∈ C such that χL(λ) = 0. Thus A−λIdn×ntaken as an operator on Cn is not invertible, so ker(A− λIdn×n) 6= 0. Hence for some v ∈ Cn with v 6= 0,L(v) := Av = λv. By Lemma 1, λ ∈ R.

Thus χA(λ) = det(A− λIdn×n) = 0, so there exists v ∈ Rn with v 6= 0 and L(v) = λv. Let V = span(v).Since V is 1-dimensional, it follows that V ⊥ is n − 1 dimensional. By Lemma 2, V ⊥ is L invariant, thusL|V ⊥ : V ⊥ → V ⊥. Since L is self-adjoint, L|V ⊥ is also self-adjoint. Applying the induction hypothesisto LV ⊥ , there exists an orthonormal basis u2, . . . , un of V ⊥ consisting of eigenvectors of L|V ⊥ . Settingu1 = v

||v|| ∈ V , u1 is a unit vector perpendicular to all ui for i ≥ 2 and an eigenvector of L. Hence u1, . . . , unis an orthonormal basis for S consisting of eigenvectors of L. This completes the induction.

Let LA : Rn → Rn be the linear map corresponding to A given by LA(v) := Av. Applying the aboveresult with S = Rn, we see there exists an orthonormal basis for Rn consisting of eigenvectors of LA. Thuswith respect to this basis, LA is diagonal, as desired.

Complex case: Decompose L = B + iC, where B and C are self adjoint and BC = CB. Since B isself-adjoint, the above proof implies there exists an eigenvalue α such that ker(B − α1V ) 6= 0. ThenC : ker(B −α1V )→ ker(B −α1V ). Then the restriction of C to ker(B −α1V ) is self-adjoint, so there existsx ∈ ker(B − α1V ) such that C(x) = βx. Then

L(x) = (α+ iβ)x,

so we have found an eigenvalue and eigenvector of L. We also have

L∗(x) = B(x)− iC(x) = (α− iβ)x.

Thus spanx is both L and L∗ invariant. It follows that M = (spanx)⊥ is also L and L∗ invariant. Hence(L|M )∗ = L∗|M , so L|M : M → M is also normal. We can use induction as in the real case above to finishthe proof.

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Spring 2002 #11; Spring 2006 #10; Spring 2008 #10; Spring 2010 #3. Let V be a complex (real)inner product space and let TλΛ be a collection (or two) normal (self-adjoint) operators. If TλΛ pairwisecommute, prove that there exists an orthonormal basis for V consisting of vectors that are simultaneouslyeigenvectors of each Tλ.

Real case: S, T are self-adjoint commuting operators. Let λ1, . . . , λk be the distinct eigenvalues of T andlet Ei = ker(T − λiI). Since T is self-adjoint,

V =

k⊕i=1

Ei

with Ei orthogonal to Ej for i 6= j. Suppose Ei has dimension di. Suppose x ∈ Ei. Then since S and Tcommute,

T (Sx) = S(Tx) = S(λix) = λi(Sx).

Hence Sx ∈ Ei. Thus Ei is S-invariant. Hence S|Ei is self-adjoint. Thus there exists vectors vi1, . . . , vidithat constitute an orthonormal basis for Ei and are simultaneously eigenvectors of S and T . Then

k⋃i=1

di⋃j=1

vij

is a basis of orthogonal vectors for V , since the Ei’s are orthogonal, and it consists of simultaneous eigen-vectors of S and T , as desired.

Complex Case: Induction on the number in the collection. Use the trick B = 12 (A + A∗) and C =

12i (A − A

∗), so A = B + iC, and B,C are self-adjoint. Since A is normal, A∗A = AA∗, thus BC = CB.Then proceed as above.

Fall 2005 #7. Let A be a real n×m matrix. Prove that the maximal number of linearly independent rowsof A is equal to the maximum number of linearly independent columns of A.

Suppose the maximal number of linearly independent rows of A is r. Let x1, . . . , xr be a basis of the rowspace of A. Suppose the ci are scalars such that

0 = c1(Ax1) + c2(Ax2) + · · ·+ cr(Axr) = A(c1x1 + · · ·+ crxr).

Then v = c1x1 + · · ·+ crxr is in the row space of A, but since Av = 0, v is orthogonal to every vector in therow space of A. Thus v is orthogonal to itself, so v = 0. Thus

c1x1 + · · ·+ crxr = 0.

But x1, . . . , xr is a basis of the row space, hence c1 = · · · = cr = 0. Thus Ax1, . . . , Axr are linearlyindependent. Now each Axi is a vector in the column space of A, hence the dimension of the column spaceof A is at least r, the dimension of the row space of A. Applying the same argument to the transpose of A,we see that the dimension of the row space of A is at least the dimension of the column space of A. Hencethese dimensions are equal.

Fall 2008 #12; Fall 2011 #12; Fall 2012 #9. Let A be an m× n real matrix and let b ∈ Rm. SupposeAx and Ay are both minimal distance to b (minimizing among members of Im(A)). Prove x− y ∈ ker(A).

Then ||Ax− b|| = ||Ay − b||. It follows that

||A((x+ y)/2)− b|| ≤ ||A(x)/2− b/2||+ ||A(y)/2− b/2|| = 1

2(||A(x)− b||+ ||A(y)− b||) = ||A(x)− b||.

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By the assumption, we conclude ||A((x+ y)/2)− b|| = ||Ax− b||. Suppose for the sake of contradiction thatx− y /∈ ker(A) so that Ax 6= Ay. It follows that Ax 6= b, Ay 6= b, and A((x+ y)/2) 6= b. Then the triangleswith points Ax,A((x + y)/2), b and Ay,A((x + y)/2), b are both isosceles and similar to one another. Itfollows that the two angles at A((x + y)/2) are equal. Since A is linear, these angles add to 180 degrees,so they are both right angles. But then one leg of a right triangle has equal length to its hypotenuse, acontradiction. Hence x− y ∈ ker(A).

Fall 2003 #6; Fall 2011 #11; Fall 2008 #6; Fall 2007 #6. State and prove the Rank-Nullity Theorem.

Let T : V →W be a linear mapping, where V is finite dimensional. Then

dim(V ) = dim(Ker(T )) + dim(Im(T )).

Proof: Note that the images of a basis of V will span Im(T ), hence Im(T ) is finite dimensional. Choosea basis w1, . . . , wn of Im(T ). There exist preimages v1, . . . , vn with wi = T (vi) for 1 ≤ i ≤ n. Select a basisu1, . . . , uk of Ker(T ). The result will follow once we show that u1, . . . , uk, v1, . . . , vn is a basis of V .

Let v ∈ V . Since T (v) ∈ Im(T ), there exist b1, . . . , bn such that

T (v) = b1w1 + · · ·+ bnwn.

ThenT (b1v1 + · · ·+ bnwn − v) = 0,

so there exist scalars a1, . . . , ak such that

b1v1 + · · ·+ bnvn − v = a1u1 + · · ·+ akuk.

Thus u1, . . . , uk, v1, . . . , vn span V .Now let a1, . . . , ak, b1, . . . , bn be scalars such that

a1u1 + · · ·+ akuk + b1v1 + · · ·+ bnvn = 0.

Applying T ,b1w1 + · · ·+ bnwn = 0.

Since w1, . . . , wn are linearly independent, wi = 0 for 1 ≤ i ≤ n. Then

a1u1 + · · ·+ akuk = 0.

Since u1, . . . , un are linearly independent, ai = 0 for 1 ≤ i ≤ k. Thus u1, . . . , uk, v1, . . . , vn are linearlyindependent and thus a basis for V .

Fall 2001 #7; Fall 2002 #7; Fall 2012 #12. Let T : V → W be a linear transformation of finitedimensional vector spaces. Define the transpose of T and then prove the following:

1. (Im(T )) = ker(T t)2. Rank(T ) = Rank(T t), where the rank of a linear transformation is the dimension of its image.

The transpose of T is the linear map T t : W ∗ → V ∗ defined by T t(f) = f T . In Petersen’s book, thisis called the dual map of T

1. Direction from definitions.

2. Use the Dimension Theorem and the fact that dim W + dim W = dim V for any subspace W ⊂ V .

75


Fall 2001 #10; Winter 2002 #9; Fall 2008 #7. Let V be a complex vector space and let T : V → V bea linear map. Let v1, . . . , vn be non-zero vectors in V , each an eigenvector for a different eigenvalue. Provethat v1, . . . , vn is linearly independent.

We proceed by induction. The statement is clear for n = 1. Now suppose n ≥ 2 and the desired statementholds for n− 1. Suppose a1v1 + · · ·+ anvn = 0. Applying T ,

a1λ1v1 + · · ·+ anλnvn = T (0) = 0 = λn(a1v1 + · · ·+ anvn).

It follows thata1(λ1 − λn)v1 + · · ·+ an−1(λn−1 − λn)vn−1 = 0.

Since v1, . . . , vn−1 are linearly independent, and λ1 − λi 6= 0 for all 1 ≤ i ≤ n− 1, ai = 0 for 1 ≤ i ≤ n− 1.Thus

anvn = 0,

so since vn 6= 0, an = 0. Hence v1, . . . , vn are linearly independent, which completes the induction.

Winter 2002 #11; Winter 2006 #10; Spring 2004 #10; Spring 2003 #8. Let V be a finitedimensional complex inner product space and let T : V → V be a linear map. Prove that there exists anorthonormal ordered basis for V such that the matrix representation of T with respect to this basis is uppertriangular.

Suppose V has dimension n. We show by induction that there exists a flag of invariant subspaces

0 ⊂ V1 ⊂ V2 ⊂ · · · ⊂ Vn−1 ⊂ V,

where dim(Vk) = k and each Vk is T -invariant with Vk = spane1, . . . , ek. Clearly if n = 1, then V1 = Vand we are finished.

Now assume inductively we have such a flag for all n−1 dimensional spaces V . Select an eigenvalue/vectorpair T ∗(v) = λv (using the fundamental theorem of algebra) and define Vn−1 = v⊥ = x ∈ V : (x|v) = 0.Since V is n-dimensional, v⊥ is n− 1-dimensional and

(T (x)|y) = (x|T ∗(y)) = (x|λy) = λ(x|y) = 0.

Thus Vn−1 is T -invariant. Applying the induction hypothesis, there exists a flag 0 ⊂ V1 ⊂ · · · ⊂ Vn−1,where dim(Vk) = k and T (Vk) ⊂ Vk. Since Vn−1 ⊂ V and T : V → V , we have the desired flag to Vimmediately, completing the induction.

Now choose unit vectors ek ∈ Vk ∩ V ⊥k−1 for 1 ≤ k ≤ n. Then these vectors must form an orthonormalbasis of V . Then since T (ek) ∈ Vk, we can express T (ek) as a linear combination of e1, . . . , ek for each k.Hence we obtain an upper triangular matrix representation for T with respect to the basis e1, . . . , ek.

Fall 2002 #9; Winter 2006 #7. Let V be a complex inner product space. State and prove the Cauchy-Schwarz inequality.

The Cauchy-Schwarz inequality says for any x, y ∈ V ,

|(x|y)| ≤ ||x|| ||y||.

Proof: If y = 0, the inequality is obvious. Otherwise, define

z = x− projy(x) = x− (x|y)

(y|y)y.

Then(z|y) = (x|y)− (x|y) = 0,

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so z is orthogonal to y. Applying the pythagorean theorem to

x =(x|y)

(y|y)y + z,

||x||2 =|(x|y)|2

||y||2+ ||z||2 ≥ |(x|y)|2

||y||2.

Thus||x|| ||y|| ≥ |(x|y)|.

Spring 2003 #7; Spring 2004 #7. Let V be a finite dimensional real vector space. Let W1 and W2 besubspaces of V . Prove the following:

1. W 1 ∩W 2 = (W1 +W2).2. (W1 ∩W2) = W 1 +W 2 .

1. Let f ∈ W 1 ∩W 2 . Then for any x ∈ W1, f(x) = 0, and for any y ∈ W2, f(y) = 0. Thus for anyx+ y ∈W1 +W2, with x ∈W1 and y ∈W2,

f(x+ y) = f(x) + f(y) = 0 + 0 = 0.

Hence f ∈ (W1 + W2). Conversely, suppose f ∈ (W1 + W2). Then for any x ∈ W1, x ∈ W1 + W2, sof(x) = 0, hence f ∈W 1 . Likewise, f ∈W 2 , so f ∈W 1 ∩W 2 . Thus we have shown part 1.

2. Let f ∈ (W1 ∩W2). Then f |W1∩W2= 0. Note that

f = f |V \W1+ f |W1 .

Now (f |V \W1)|W1

= 0 and (f |W1)|W2

= 0, thus extending these functions by 0 to the rest of the space,f |V \W1

∈W 1 and f |W1∈W 2 . Hence f ∈W 1 +W 2 . Conversely, suppose f + g ∈W 1 +W 2 , where f ∈W 1

and g ∈W 2 . Then for any x ∈W1 ∩W2,

(f + g)(x) = f(x) + g(x) = 0 + 0 = 0,

hence f + g ∈ (W1 ∩W2). Thus we have shown part 2.

Fall 2003 #8b. Let T be a linear transformation from a finite dimensional vector space V with an innerproduct to a finite dimensional vector space W also with an inner product. Show that the kernel (null space)of T ∗ is the orthogonal complement of the range of T .

By definition,ker(T ) = x ∈ V : Lx = 0,

andim(T ∗)⊥ = x ∈ V : (x|T ∗z) = 0 for all z ∈W.

Fix x ∈ V and use that (Tx|z) = (x|T ∗z) for all z ∈W . Then if x ∈ ker(T ), then x ∈ im(T ∗)⊥. Conversely,if x ∈ im(T ∗)⊥, then 0 = (x|T ∗z) = (Tx|z) for all z ∈W . Picking z = Tx, we see that Tx = 0, so x ∈ ker(T ).Thus

ker(T ) = im(T ∗)⊥.

Replacing T in the above argument by T ∗ and using that T ∗∗ = T , we obtain

ker(T ∗) = im(T )⊥.

Spring 2006 #8; Fall 2009 #9. If A ∈ M2n+1(R) is such that AAt = Id2n+1, then prove that one of 1or −1 is an eigenvalue of A.

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Note that det(T − λI) is a real polynomial with degree 2n + 1. By the intermediate value theorem, ithas a real root λ. Thus there exists v ∈ R2n+1 with v 6= 0 such that Av = λv.

Now(v|v) = (AA∗v|v) = (v|(AA∗)∗v) = (v|A∗Av) = (Av|Av) = (λv|λv) = |λ|2(v|v).

Since v 6= 0, (v|v) 6= 0, hence |λ|2 = 1. Since λ is real, λ = ±1.

Spring 2007 #1. Spring 2011 #6. Let V and W be finite dimensional real inner product spaces, andlet A : V → W be a linear transformation. Let w ∈ W . Show that the elements v ∈ V for which the norm||Av − w|| is minimal are exactly the solutions to the equations A∗Ax = A∗w.

Define S = Im(A) and let e1, . . . , en be an orthonormal basis for S. The vector in S closest to w is

s := projS(w) := (w|e1)e1 + · · ·+ (w|en)en.

A straightforward calculation shows that w − s ∈ S⊥.Using that w − s ∈ S⊥, for any z ∈ S,

(z, w − s) = 0.

Let x ∈ V such that Ax = s. Then for any y ∈ V , Ay ∈ S, thus

0 = (Ay,Ax− w) = (y,A∗(Ax− w)),

hence A∗(Ax− w) = 0. This implies A∗Ax = A∗w.Conversely, suppose A∗Ax = A∗w. Let s = Ax. Then by the above calculation in reverse, w − s ∈ S⊥.

Thus for any t ∈ S, t− s and w − s are perpendicular, so

||s− w||2 ≤ ||s− w||2 + ||s− t||2 = ||t− w||2,

hence ||s− w|| ≤ ||t− w||. Thus v = x is a minimizer of ||Av − w||.

Spring 2008 #11. Spring 2011 #2. Show that a positive power of an invertible matrix with complexentries is diagonalizable if and only if the matrix itself is diagonalizable.

Suppose A is an invertible n × n matrix and k be a positive integer. There exists an invertible matrixV such that V AV −1 is diagonal. It follows that (V AV −1)k = V AkV −1 is also diagonal. Hence Ak isdiagonalizable.

Conversely, suppose Ak is diagonalizable. Then µAk(t) = (t − λ1) · · · (t − λk) with λi distinct. SinceµAk(Ak) = 0,

(Ak − λ1) · · · (Ak − λk).

The λi are eigenvalues of Ak, thus since A is invertible, λi = 6= 0. It follows that the minimal polynomial ofA divides

p(t) := (tk − λ1) · · · (tk − λk).

Suppose p(r) = 0, so there exists j with rk = λj . Since λi 6= 0, r 6= 0. Note that

Dp(r) = (krk−1)

k∑i=1

(rk − λ1) · · · ˆ(rk − λi) · · · (rk − λk)

= (krk−1)

k∑i=1

(λj − λ1) · · · ˆ(λj − λi) · · · (λj − λk)

= (krk−1)(λj − λ1) · · · ˆ(λj − λj) · · · (λj − λk) 6= 0.

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Hence p has no multiple roots, thus the minimal polynomial of A has no multiple roots. Since we are workingover C, the minimal polynomial of A necessarily factors as

µA(t) = (t− η1) · · · (t− ηk),

with ηi distinct, so A is diagonalizable.

Winter 2011 #4. Fall 2012 #11. Show that an n by n matrix can be factored as A = LU where L isa lower triangular matrix with ones along the diagonal and U is an upper triangular matrix provided eachdeterminant det(Aj) (for j ∈ 1, . . . , n − 1) is non-zero (where Aj is the submatrix of A consisting of thefirst j rows and first j columns of A).

Assume det(Aj) 6= 0 for each j. We just need to show that Gaussian elimination does not need pivoting.We prove by induction on k that the kth step does not need pivoting. This holds for k = 1, since A1 = (a11),so, a11 6= 0. Assume that no pivoting was necessary for the first k steps (1 ≤ k ≤ n − 1). In this case, wehave

Ek · · ·E1A = Uk,

where L = Ek · · ·E1 is a unit lower-triangular matrix and Uk[1..k, 1..k] is upper-triangular. Since det(Ak+1) 6=0, (Uk)k+1,k+1 6= 0. Thus we can multiply Uk by unit lower-triangular matrices on the left to eliminate en-tries (Uk)i,k+1 for k + 1 ≤ i ≤ n. Thus the (k + 1)st step of Gaussian elimination does not need pivoting,completing the induction.

Fall 2011 #10. Spring 2013 #9. Let A be a real orthogonal matrix. Show that A is similar to a blockdiagonal matrix, where each block is a scalar (which is a real eigenvalue of A) or of the form

Tα,θ = α

[cos(θ) sin(θ)− sin(θ) cos(θ)

],

where α, θ ∈ R.

Any complex eigenvalues of A come in conjugate pairs. And as an orthogonal matrix, all eigenvalues ofA take the form eiθ for some θ. Since A is diagonalizable, there exists an invertible V ∈ Mn(C) such thatV AV −1 is a diagonal matrix consisting of eigenvalues of A.

Suppose without loss of generality that λ, λ are the first eigenvalues of A in the diagonal form. Define

U =1√2

(1 i1 −i

)and note that

U−1

(λ 0

0 λ

)U

has the desired form.Also let V be a matrix with first column v and second column v. It follows that UV is real, so A is

similar to a block diagonal matrix of the desired form.

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Other Linear Algebra Qualifying Exam Problems

Fall 2001 #8. Spring 2004 #8 (similar) Let T : R3 → R3 be the rotation by 60 degrees counterclockwiseabout the plane perpendicular to the vector (1, 1, 1) and S : R3 → R3 be the reflection about the planeperpendicular to the vector (1, 0, 1). Determine the matrix representation of S T in the standard basise1, e2, e3. You do not have to multiply the resulting matrices but you must determine any inverses thatarise.

Let v1 = 1√3(1, 1, 1), v2 = 1√

2(1, 0,−1), and v3 = v1 × v2 = 1√

6(−1, 2,−1). Because v1 is fixed under

T and v2 and v3 are rotated counterclockwise by 60 degrees, the matrix representation of T in the basisβ = v1, v2, v3 is

[T ]ββ =

1 0 00 cos(60) − sin(60)0 sin(60) cos(60)

.

Let γ be the standard basis. The change of basis matrix is [Id]γβ = (v1, v2, v3). Since the columns are

orthonormal, [Id]γβ is orthogonal, so [Id]βγ = ([Id]γβ)−1 = ([Id]γβ)t.

Do the same with w1 = 1√2[1, 0, 1], and two other perpendicular vectors and call this basis α. Then the

reflection just takes w1 → −w1 and leaves the others unchanged. Basis for S w.r.t. α is easy. The desiredmatrix representation is

[S T ]γγ = [S]γγ = [Id]γα[S]αα[Id]γα[Id]γβ [T ]ββ [Id]βγ .

Fall 2002 #8. Let T be the rotation of an angle 60 degrees counterclockwise about the origin in the planeperpendicular to (1, 1, 2) in R3.

i. Find the matrix representation of T in the standard basis. Find all eigenvalues and eigenspaces of T .ii. What are the eigenvalues and eigenspaces of T if R3 is replaced by C3?[You do not have to multiply any matrices out but must compute any inverses.]

(i) Let v1 = 12 (1, 1, 2). Select v2 = 1√

3(1, 1,−1). v3 = v1 × v2. Call this orthonormal basis α and the

standard basis β.[T ]αα = ....[Id]βα = [v1v2v3].

Then [T ]ββ = [Id]βα[T ]αα[Id]αβ = [Id]βα[T ]αα([Id]βα)t.(ii) Find the eigenvalues and eigenspaces corresponding to [T ]αα:

[T ]αα =

1 0 0

0 12 −

√3

2

0√

32

12

.

Eigenvalues : 1, eπi/3, e2πi/3.

Spring 2002 #8. Let V be a finite dimensional real vector space. Let W ⊂ V be a subspace and

W := f : V → F linear |f = 0 on W.

Prove thatdim(V ) = dim(W ) + dim(W ).

Let w1, . . . , wk be a basis for W , and extend it to a basis w1, . . . , wk, vk+1, . . . , vn for V . Define fi(v) = 1for v = vk+i and 0 otherwise. Then fi ∈ W⊥ and they are linearly independent. Also, any function in W⊥

can be expressed as a linear combination of these. Hence fin−ki=1 is a basis for W⊥, so we get the desireddimension equation.

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Spring 2002 #9. Find the matrix representation in the standard basis for either rotation by an angle θ inthe plane perpendicular to the subspace spanned by the vectors (1, 1, 1, 1) and (1, 1, 1, 0) in R4. [You do nothave to multiply the matrices out but must compute any inverses.]

Additional perpendicular vectors are (1,−1, 0, 0) and (1, 1,−2, 0). Then follow previous problem.

Spring 2002 #10. Let V be a complex inner product space and W a finite dimensional subspace. Letv ∈ V . Prove that there exists a unique vector vW ∈W such that

||v − vW || ≤ ||v − w||

for all w ∈W . Deduce that equality holds if and only if w = vW .

Let w1, . . . , wk be an orthogonal basis for W . Define

vW = projW v :=(v|w1)

(w1|w1)w1 − · · · −

(v|wk)

(wk|wk)wk.

Let s = v − vW . Then (s|wi) = 0 for all 1 ≤ i ≤ k. Hence s ∈W⊥. Thus for any w ∈W , s and vW − w areperpendicular. By the pythagorean theorem,

||v − w||2 = ||v − vW ||2 + ||w − vW ||2 ≥ ||v − vW ||2,

hence||v − w|| ≥ ||v − vW ||

with equality if and only if w = vW . This equality case forces uniqueness.

Winter 2002 #8. Let T : V → W and S : W → X be linear transformations of finite dimensional realvector spaces. Prove that

rank(T ) + rank(S)− dim(W ) ≤ rank(S T ) ≤ maxrank(T ), rank(S).

From the Dimension Theorem,

dim(V ) = rank(T ) + nullity(T ),

dim(W ) = rank(S) + nullity(S),

anddim(V ) = rank(S T ) + nullity(S T ).

Thusrank(T ) + rank(S)− dim(W ) = rank(S T ) + nullity(S T )− nullity(T )− nullity(S).

Show thatnullity(S T ) ≤ nullity(S) + nullity(T )

by appealing to a basis. This gives the first inequality.Also, by appealing to a basis, we can check rank(S T ) ≤ minrank(T ), rank(S).

Winter 2002 #10. Let V be a finite dimensional complex inner product space and f : V → V a linearfunctional. Show that there exists a vector w ∈ V such that f(v) = (v|w) for all v ∈ V .

Let v1, . . . , vn be an orthonormal basis for V . Let f1, . . . , fn be the associated dual basis. Inparticular, this means fi(v) = (v, vi) for all v ∈ V . There exists αi with

f = α1f1 + · · ·+ αnfn.

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Thus

f(v) =

n∑i=1

αifi(v) =

n∑i=1

αi(v, vi) = (v,

n∑i=1

αivi).

Thus if we takew = α1v1 + · · ·+ αnvn,

then for any v ∈ V , f(v) = (v, w) for all v ∈ V .

Fall 2003 #9. Consider a 3 by 3 real symmetric matrix with determinant 6. Assume (1, 2, 3) and (0, 3,−2)are eigenvectors with eigenvalues 1 and 2.

(a) Give an eigenvector of the form (1, x, y) for some real x, y which is linearly independent of the twovectors above.

(b) What is the eigenvalue of this eigenvector?

(a) By the spectral theorem, the eigenspaces of the eigenvalues are orthogonal. Thus the cross productof the given vectors, scaled appropriately, is the answer.

(b) The eigenvalue must be 3, since the determinant is the product of the eigenvalues.

Fall 2003 #10. (a) Let t ∈ R such that t is not an integer multiple of π. For the matrix

A =

(cos(t) sin(t)− sin(t) cos(t)

)prove that there does not exist a real valued matrix B such that BAB−1 is a diagonal matrix.

(b) Do the same for the matrix

A =

(1 λ0 1

)where λ ∈ R \ 0.

(a) If there were such a matrix, the diagonal entries would be the eigenvalues. Thus the eigenvalueswould be real. However, the eigenvalues satisfy

(cos(t)− λ)2 + sin2(t) = 0,

soλ2 − 2 cos(t)λ+ 1 = 0,

andλ = cos(t)± i sin(t) = e±it,

which are not real.

(b) Just check directly. Or note that the geometric multiplicity of the only eigenvalue 1 is 1, so A can’tbe diagonalizable.

Spring 2003 #9. Let A ∈ M3(R) satisfy det(A) = 1 and AtA = I = AAt where I is the 3 × 3 identitymatrix. Prove that the characteristic polynomial of A has 1 as a root.

Show complex eigenvalues come in conjugate pairs. Then since det(A) = 1 is the product of the eigenval-ues, and the product of conjugates is positive, there exists a real positive root. Then for a positive eigenvaluewith eigenvector v 6= 0,

(v, v) = (AtAv, v) = (Av,Av) = |λ|2(v, v).

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Since (v, v) 6= 0, and λ is a positive real number, λ = 1.

Spring 2003 #10. Let V be a finite dimensional real inner product space and T : V → V a hermitianlinear operator. Suppose the matrix representation of T 2 in the standard basis has trace zero. Prove that Tis the zero operator.

Let A be the matrix representation of T . Since T is hermitian, A is diagonalizable. Thus there existsan invertible matrix P such that A = PDP−1 with D a diagonal matrix consisting of eigenvalues of T . Itfollows that

A2 = PD2P−1,

thus the trace of A2 is

tr(PD2P−1) = tr(PP−1D2) = tr(D2) =

n∑i=1

λ2i .

Since this is assumed to be 0, λi = 0 for all i. It follows that D = 0, hence A = 0, so T = 0.

Fall 2004 #8. Let A = (aij) be a real, n×n symmetric matrix and let Q(v) = v ·Av (ordinary dot product)be the associated quadratic form defined for v = (v1, . . . , vn) ∈ Rn.

1. Show that ∇Qv = 2Av where ∇Qv is the gradient at v of the function Q.2. Let M be the minimum value of Qv on the unit sphere Sn = v ∈ R : ||v|| = 1 and let u ∈ Sn be the

vector such that Q(u) = M . Prove, using Lagrange multipliers, that u is an eigenvector of A with eigenvalueM .

Let Ai be the ith column of A. By the product rule,

∂Q

∂vi(v) = v · ai + ei ·Av.

Since A is symmetric,ei ·Av = etiAv = etiA

tv = (Aei)tv = ativ = ai · v,

thus∂Q

∂vi(v) = 2ai · v = 2ativ,

and∇Qv = 2Av.

(b) Let g(u) = ||u||2−1. Then we are trying to minimize Qu subject to the constraint g(u) = 0. Lagrangemultipliers yields the equation

∇Qu = λ∇g(u).

From part (a), this becomes2Au = λ(2u).

Thus Au = λu, so u is an eigenvector of A. Then by definition of M ,

M = Qu = u ·Au = u · λu = λ|u|2 = λ.

Fall 2004 #9. Let T : Cn → Cn be a linear transformation and P (X) a polynomial such that P (T ) = 0.Prove that every eigenvalue of T is a root of P (X).

Let λ be an eigenvalue of T with an eigenvector v. A simple induction shows (P (T ))(v) = P (λ)v. Sincethe expression on the left is zero and v 6= 0, P (λ) = 0.

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Fall 2004 #10. Let V = Rn and let T : V → V be a linear transformation. For λ ∈ C, the subspace

V (λ) = v ∈ V : (T − λI)Nv = 0 for some N ≥ 1

is called a generalized eigenspace.1. Prove that there exists a fixed number M such that V (λ) = ker((T − λI)M ).2. Prove that if λ 6= µ, then V (λ) ∩ V (µ) = 0. Hint: use the following equation by raising both sides

to a high power:T − λIµ− λ

+T − µIλ− µ

= I.

1. Let v1, . . . , vn be a basis for V (λ). Let N be the maximum over i of the N such that (T−λI)Nvi = 0for each i. Then for any v ∈ V (λ), (T − λI)Nv = 0. Hence V (λ) = ker((T − λI)M ).

2. Suppose λ 6= µ. Suppose v ∈ V (λ) ∩ V (µ). Note

T − λIµ− λ

+T − µIλ− µ

= I.

Raising both sides to the power 2M , note that every expression on the left has a factor of (T − λI)M or(T − µI)M . Thus when evaluating at v, the left hand side is 0, but the right hand side is v.

Spring 2004 #9. Let V be a finite dimensional real inner product space under ( , ) and T : V → V a linearoperator. Show the following are equivalent:

a) (Tx, Ty) = (x, y) for all x, y ∈ V ,b) ||T (x)|| = ||x|| for all x ∈ V ,c) T ∗T = IdV , where T ∗ is the adjoint of T .d) TT ∗ = IdV .[T is an orthogonal map.]

(a) =⇒ (b). Then for any x ∈ V .

||T (x)||2 = (Tx, Tx) = (x, x) = ||x||2.

Since the norm is non-negative, this yields (b).(b) =⇒ (a). Then for any x, y ∈ V ,

(Tx+Ty, Tx+Ty) = (Tx, Tx+Ty)+(Ty, Tx+Ty) = (Tx, Tx)+(Tx, Ty)+(Ty, Tx)+(Ty, Ty) = (Tx, Tx)+(Ty, Ty)+2(Tx, Ty),

thus

2(Tx, Ty) = (T (x+ y), T (x+ y))− (Tx, Tx)− (Ty, Ty) = (x+ y, x+ y)− (x, x)− (y, y) = 2(x, y).

Hence(Tx, Ty) = (x, y).

(b) =⇒ (c). For any x ∈ V ,

((T ∗T − IdV )x, (T ∗T − IdV )x) = (T ∗Tx, T ∗Tx)− 2(T ∗Tx, IdV x) + (IdV x, IdV x)

= (Tx, TT ∗Tx)− 2(Tx, Tx) + (x, x) = (x, T ∗Tx)− 2(x, x) + (x, x) = 0.

Thus (T ∗T − IdV )x = 0 for all x, so T ∗T = IdV .(c) =⇒ (b). For any x ∈ V ,

(T (x), T (x)) = (x, T ∗Tx) = (x, x).

The equivalence of (b) and (d) follows analogously to the equivalence of (b) and (c).

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Fall 2005 #6. (a) Prove that if P is a real-coefficient polynomial and if A is a real symmetric matrix, thenthe eigenvalues of P (A) are exactly the numbers P (λ), where λ is an eigenvalue of A.

(b) Use part (a) to prove that if A is a real symmetric matrix, then A2 is non-negative definite.(c) Check part (b) by verifying directly that det(A2) and trace(A2) are non-negative when A is real

symmetric.

(a) Send A to diagonal matrix. P (A) is similar to P (D).

(b) A2 is non-negative definite if for all v, vtA2v is non-negative.All eigenvalues of A are real, so all eigenvalues of A2 are non-negative.Suppose P−1AP = D. For any v ∈ V , there exists w with v = Pw. Then

vtA2v = wt(P−1A2P )w = wtD2w =

n∑i=1

λ2iw

2i ≥ 0.

(c) det(A2) = (det(A))2 ≥ 0.

trace(A2) = trace((

n∑k=1

aikakj)) =

n∑i=1

n∑k=1

a2ik ≥ 0.

Fall 2005 #9. Suppose V1 and V2 are subspaces of a finite-dimensional vector space V .(a) Show that

dim(V1 ∩ V2) = dim(V1) + dim(V2)− dim(span(V1, V2))

where span(V1, V2) is by definition the smallest subspace that contains both V1 and V2.(b) Let n = dim(V ). Use part (a) to show that, if k < n, then an intersection of k subspaces of dimension

n− 1 always has dimension at least n− k. (Suggestion: Do induction on k.)

(a) Let v1, . . . , vk be a basis for V1 ∩ V2. Then there exists a basis v1, . . . , vk, w1, . . . , wr for V1 andv1, . . . , vk, w

′1, . . . , w

′s for V2. It follows that wi /∈ V2 and w′i /∈ V1. Hence v1, . . . , vk, w1, . . . , wr, w

′1, . . . , w

′s is

a basis for span(V1, V2). This gives the desired equality.

(b) k = 1 obvious. Assume true for k. Let k + 1 < n. Let V1, . . . , Vk+1 be subspaces of V of dimensionn− 1. Then since span(

⋂i≤k Vi, Vk+1) ⊂ V ,

dim(⋂

i≤k+1

Vi) = dim(⋂i≤k

Vi) + dim(Vk+1)− dim(span(⋂i≤k

Vi, Vk+1))

≥ (n− k) + (n− 1)− n ≥ n− (k + 1),

which completes the induction.

Fall 2005 #10. (a) For each n = 2, 3, 4, . . ., is there an n× n matrix A with An−1 6= 0 but An = 0?(b) Is there an n× n upper triangular matrix A with An 6= 0 but An+1 = 0?

(a). Yes, take the linear map which sends en to 0 and ei to ei+1 for each i < n.(b). I assume the field has characteristic 0.No. Because A is upper triangular, An+1 is diagonal and (An)ii = (Aii)

n. Thus if An+1 = 0, then Aii = 0for all i, in which case A2 = 0. If n ≥ 2, then An = 0. Otherwise, n = 1, and the statement is obvious.

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Spring 2005 #1. Given n ≥ 1, let tr : Mn(C)→ C denote the trace of a matrix:

tr(A) =

n∑k=1

Ak,k.

(a) Determine a basis for the kernel (or null-space) of tr.(b) For X ∈ Mn(C), show that tr(X) = 0 if and only if there exists an integer m and matrices

A1, . . . , Am, B1, . . . , Bm ∈Mn(C) so that

X =

m∑i=1

AjBj −BjAj .

(a) Eij , i 6= j and E11 − Eii for i ≥ 2.Dimension of image of trace is 1. Thus the dimension of the kernel of the trace is dim(Mn(C)) −

dim(range(trace)) = n2 − 1. Just need to show linear independence.

(b) ⇐= is obvious since tr(AB) = tr(BA).=⇒: Suppose A ∈Mn(C) such that tr(A) = 0. Then we can write A in terms of the basis from part (a).

Note Ei,j = [Ei,i, Ei,j ] for i 6= j and E1,1 − Ei,i = [E1,j , Ej,1] for all i.

Spring 2005 #2. Let V be a finite-dimensional vector space, and let V ∗ denote the dual space; that is,the space of linear maps φ : V → C. For a set W ⊂ V , let

W⊥ = φ ∈ V ∗ : φ(w) = 0 for all w ∈W.

For a subset U ⊂ V ∗, let⊥U = v ∈ V : φ(v) = 0 for all φ ∈ U.

(a) Show that for any subset W ⊂ V , ⊥(W⊥) = span(W ). Recall that the span of a set of vectors is thesmallest vector sub-space that contains these vectors.

(b) Let W ⊂ V be a linear subspace. Give an explicit isomorphism between (V/W )∗ and W⊥. Showthat it is an isomorphism.

(a) Suppose w ∈ W . Then for all φ ∈ W⊥, φ(w) = 0, hence w ∈⊥ (W⊥). Since ⊥(W⊥) is a subspaceand Span(W ) is the smallest subspace containing W we have ⊥(W⊥) ⊃ Span(W ).

Conversely, select a basis v1, . . . , vk for Span(W ) and then extend it to a basis v1, . . . , vn for W . Let φibe the corresponding dual basis vectors. Suppose w ∈⊥ (W⊥). Then there exist ai with w =

∑ni=1 aivi,

with ai ∈ C. Since φj ∈W⊥ for k + 1 ≤ j ≤ n,

aj = φj(w) = 0.

Hence w ∈ Span(W ), so Span(W ) =⊥ (W⊥).

(b) Define φ : (V/W )∗ → W⊥ by φ(T ) = [v 7→ T (v + W )]. If w ∈ W , then (φ(T ))(w) = T (0) = 0, soφ(T ) ∈W⊥. Thus this map is well defined. If φ(T ) = 0, then T (v +W ) = 0 for all v ∈ V , so T = 0. Henceφ is injective. Also,

dim((V/W )∗) = dim(V/W ) = dim(V )− dim(W ) = dim(W⊥).

Hence φ must be surjective as well. Hence φ is an isomorphism.

Spring 2005 #3. Let A be a Hermitian-symmetric n× n complex matrix. Show that if (Av|v) ≥ 0 for allv ∈ Cn, then there exists an n× n matrix T so that A = T ∗T .

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By the spectral theorem for Hermitian operators, we may select an orthonormal basis v1, . . . , vn andnumbers λ1, . . . , λn ∈ R such that Avi = λivi. We claim λi ≥ 0 for all i. Note

λi(vi|vi) = (Avi|vi) ≥ 0.

Define T by T (vi) =√λivi. Then T ∗(vi) =

√λivi, hence A(vi) = TT ∗(vi), so A = TT ∗.

Spring 2005 #4. Let A = Mn(C) denote the set of all n× n matrices with complex entries. We say thatI ⊂ A is a two-sided ideal in A if

(i) For all A,B ∈ I, A+B ∈ I.(ii) For all A ∈ I and B ∈ A, AB and BA belong to I.Show that the only two-sided ideals in A are 0 and A itself.

Let I ⊂ A be a two-sided ideal. Suppose there is a nonzero A = (aij) ∈ I. By multiplying A by a suitablepermutation matrix and diagonal matrix we see by (ii) that the matrix Eij ∈ I. Thus by (i), Mn(C) ∈ I.

Spring 2006 #7. A matrix T (with entries, say, in the field C of complex numbers) is diagonalizable ifthere exists a non-singular matrix S such that STS−1 is diagonalizable. Prove that if a, λ ∈ C with a 6= 0,then the following matrix is not diagonalizable:

T =

1 a 00 1 a0 0 λ

.

Note the eigenvalues are 1 and λ and both have eigenspaces of dimension 1. Thus the geometric multi-plicities of the eigenvalues do not add to 3, the dimension of the matrix, hence it is not diagonalizable.

Spring 2006 #10. Let Y be an arbitrary set of commuting matrices in Mn(C) (i.e., AB = BA for allA,B ∈ Y ). Prove that there exists a non-zero vector v ∈ Cn which is a common eigenvector of all elementsof Y .

We first work out the case of two matrices with AB = BA. Let λ be an eigenvalue of A and let Vλ bethe eigenspace of λ. If v ∈ Vλ, then

A(Bv) = B(Av) = λ(Bv).

Hence Bv ∈ Vλ, so Vλ is invariant under B. Consider the restriction of B to Vλ. This restricted operator hasat least one eigenvalue and eigenvector (say w ∈ Vλ is an eigenvector). Then w is an eigenvector for both Aand B.

Inductively, suppose v is a common eigenvector for A1, . . . , An−1 with eigenvalues λ1, . . . , λn−1. Thenthe commutative property implies the eigenspaces corresponding to each eigenvalue and matrix are invariantunder An. Take the intersection of all such eigenspaces. This is a non-empty set since it contains v. Then therestriction of An to this intersection has an eigenvalue w, which is then a common eigenvalue for A1, . . . , An.

Winter 2006 #8. Let T : V → W be a linear transformation of finite dimensional real inner productspaces. Show that there exists a unique linear transformation T t : W → V such that

(T (v)|w)W = (v, T t(w))V

for all v ∈ V and w ∈W .

Let v1, . . . , vn and w1, . . . , wm be orthonormal bases for V and W respectively. Let A be the matrixrepresentation of T relative to these bases. Define the map T t : W → V to have matrix representation At

with respect to these bases. It follows that

(v, T tw)V = (Tv,w)W .

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Suppose S and S′ both satisfy

(Tv,w)W = (v, Sw)V = (v, S′w)V

for all v ∈ V , w ∈W .Then

(v, Sw − S′w)V = 0

for all v ∈ V , hence Sw − S′w = 0, so S = S′.

Fall 2007 #3. Let T be a linear transformation of the vector space V into itself. If Tv and v are linearlydependent for each v ∈ V , show that T must be a scalar multiple of the identity.

Take v ∈ V and λ such that Tv = λv. Then for any w ∈ V , Tw = cw and T (v + w) = c′(v + w). Thisimplies c′(v+w) = T (v+w) = Tv+Tw = λv+ cw. Hence (c′−λ)v = (c− c′)w. This is only generally trueif V is 1-dimensional or λ = c = c′. In either case, T (v) = λv for all v ∈ V .

Fall 2007 #7. Let A(x) be a function on R whose values are n× n matrices. Starting from the definitionthat the derivative A′(x) is the matrix you get by differentiating the entries in A(x), show that when A(x)is invertible and differentiable for all x, A−1(x) is differentiable, and

(A−1)′(x) = −A−1(x)A′(x)A−1(x).

Suppose A(x) is invertible and differentiable for all x. Note

A−1 = (det(A))−1adj(A),

and since the adjugate is the transpose of the cofactor matrix, which consists of sums of products of elementsof A, the adjugate is differentiable, hence A−1 is differentiable.

By the definition of A−1(x),In = A−1(x)A(x).

Supposing A−1(x) is differentiable, differentiating the i, j entry of each side and combining results,

0 = (A−1)′(x)A(x) +A−1(x)A′(x).

Thus(A−1)′(x) = −A−1(x)A′(x)A−1(x).

Fall 2007 #10. Suppose that vjnj=1 is a basis for the complex vector space Cn.(a) Show that there is a basis wjnj=1 such that (wj |vk) = δjk. Here (·, ·) is the standard inner product,

(w, v) = w1v1 + · · ·+ wnvn.(b) If the vj ’s are eigenvectors for a linear transformation T of Cn, show that the wj ’s are eigenvectors

for T ∗, the adjoint of T with respect to the inner product.

(a) The matrix A = (v1 . . . vn) is invertible since the vi are linearly independent. Let wj be the jth row

vector of A−1. Then(wj , vk) = (A−1A)jk = (In)jk = δjk.

(b) Suppose Tvj = λjvj . Then [T ∗] = [T ]t. Let B be the diagonal matrix with entries λj . Then[T ]A = AB, so A−1[T ] = BA−1. Taking the transpose and conjugate of both sides,

[T ∗](A−1)t = [T ]t(A−1)t = (A−1)tBt = (A−1)tB.

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Thus T ∗wj = λjwj .

Fall 2007 #12. (a) Suppose that x0 < x1 < · · · < xn are points in [a, b]. Define linear functions on Pn,the vector space of polynomials of degree less than or equal to n, by setting

lj(p) = p(xj)

for j = 0, . . . , n. Show that the set ljnj=0 is linearly independent.(b) Show that there are unique coefficients cj such that∫ b

a

p(x)dx =

n∑j=0

cj lj(p)

for all p ∈ Pn.

Set a0l0 + · · ·+ anln = 0. Then taking pi(x) =∏j 6=i(x− xj),

0 = (a0l0 + · · ·+ anln)(pi) = a0p(x0) + · · ·+ anp(xn) = αi∏j 6=i

(xi − xj).

Thus ai = 0. This works for any i, hence the lj are linearly independent.

(b) From part (a) and the fact that (Pn)∗ has dimension n+ 1, we see that ljnj=0 forms a basis for (Pn)∗.

Since the map which takes p to∫ bap(x)dx is in (Pn)∗, then there are unique coefficients cj with the desired

equality.

Spring 2007 #2. Let V,W,Z be n-dimensional vector spaces and T : V → W and U : W → Z be lineartransformations. Prove that if the composite transformation UT : V → Z is invertible, then both T and Uare invertible. (Do not use determinants in your proof!)

Since UT is invertible, it is one-to-one and onto. Clearly this makes U onto. By the Dimension Theorem,it follows that U is one-to-one. If T is not one-to-one, then UT is not one-to-one, a contradiction. Hence Tis one-to-one, and it follows by the Dimension Theorem that T is onto. Thus U and T are bijections, henceinvertible.

Spring 2007 #3. Consider the space of infinite sequences of real numbers

S = (a0, a1, . . .) : an ∈ R, n = 0, 1, 2, . . .

endowed with the standard operations of addition and scalar multiplication. For each pair of real numbersA and B, prove that the set of solutions (x0, x1, . . .) of the linear recursion

xn+2 = Axn+1 +Bxn

for n ≥ 0 is a linear subspace of S of dimension 2.

Clearly the set of solutions forms a subspace. Show that (1, 0, . . .) and (0, 1, . . .) is a basis for the setof solutions. Clearly any solution is a linear combination of these two. Also, these are necessarily linearlyindependent vectors because of the first two components.

Spring 2007 #4. Suppose that A is a symmetric n × n real matrix with distinct eigenvalues λ1, . . . , λl,(l ≤ n). Find the sets

X = x ∈ Rn : limk→∞

(xtA2kx)1/k exists

89


andL = lim

k→∞(xtA2kx)1/k : x ∈ X,

where Rn is identified with the set of real column vectors, and xt denotes the transpose of x.

The answers are X = Rn and L = λ1, . . . , λl.By the spectral theorem, A is diagonalizable, so there exists an orthonormal basis v1, . . . , vn such that

Avi = λivi (different λi than the problem statement). Then for any x ∈ X, there exist ai such that

x = a1v1 + · · ·+ anvn.

We computextA2kx = λ2

1a1v21 + · · ·+ λ2

nanv2n.

Suppose λi is the dominant eigenvalue. Then if ai 6= 0,

(xtA2kx)1/k = λ2.

(More detail here on exam.)Picking x = vi for each i, we can recover λ2

i in this way.

Spring 2007 #5. Let T be a normal linear operator on a finite dimensional complex inner product linearspace V . Prove that if v is an eigenvector of T , then v is also an eigenvector of its adjoint T ∗.

Since T is normal TT ∗ − T ∗T = 0.

(Tv, Tv) = (v, T ∗Tv) = (v, TT ∗v) = (v, T ∗∗T ∗v) = (T ∗v, T ∗v).

Note that (T − λI) is also normal. Thus

0 = ||(T − λI)v|| = ||(T − λI)∗v|| = ||(T ∗ − λ∗I)v||.

Hence T ∗v = λ∗v.

Fall 2008 #8. Must the eigenvectors of a linear transformation T : Cn → Cn span Cn? Prove yourassertion.

No, take the nilpotent transformation T (v1) = 0, T (v2) = v1. This only has 0 eigenvalues, and eigenvec-tors (x, 0). Clearly these do not span C2.

Fall 2008 #9. (a) Prove that any linear transformation T : Cn → Cn must have an eigenvector.(b) Is (a) true for any linear transformation T : Rn → Rn?

(a) Characteristic polynomial has a root.

(b) No, pick a non-trivial rotation.

Fall 2008 #11. Consider the Poisson equation with periodic boundary conditions on [0, 1]:

∂2u

∂x2= f, x ∈ (0, 1)

u(0) = u(1).

A second order accurate approximation to the problem is given by the solution to the following system ofequations

Au = ∆x2f

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where

A =

−2 1 0 · · · 0 11 −2 1 0 · · · 00 1 −2 1 0 · · ·

. . .. . .

. . .

0 · · · 0 1 −2 11 0 0 · · · 1 −2

,

u = [u0, u1, . . . , un−1], f = [f0, f1, . . . , fn−1] and ui ≈ u(xi) with xi = i∆x, ∆x = 1/n, and fi = f(xi) fori = 0, . . . , n− 1.

(a) Show that the matrix A is singlar.(b) What condition must f satisfy so that a solution exists?

(a) Apply it to (1, . . . , 1) to get 0. If the matrix had an inverse, applying the inverse to 0 would yield 0,not (1, . . . , 1), a contradiction.

(b) Based on part (a), f must satisfy f0 + · · · + fn−1 = 0. Then we can reduce the system to an n − 1by n− 1 matrix that is not singular, hence a solution exists.

Spring 2008 #8. Assume V is an n-dimensional vector space over the rationals Q, and T is a Q-lineartransformation T : V → V such that T 2 = T . Prove that every vector v ∈ V can be written uniquely asv = v1 + v2 such that T (v1) = v1 and T (v2) = 0.

Write v = T (v) + (v − T (v)). Then T (T (v)) = T (v) and T (v − T (v)) = T (v)− T (v) = 0.Note that if v = v1 + v2 with T (v1) = v1 and T (v2) = 0, then T (v) = T (v1) = v1 and (v − T (v)) =

v − T (v1) = v − v1 = v2. Hence v1 and v2 are unique.

Spring 2008 #9. Let V be a vector space over R.(a) Prove that if V is odd dimension, and if T is an R-linear transformation T : V → V of V , then T has

a non-zero eigenvector v ∈ V .(b) Show that for every even positive integer n, there is a vector space V over R of dimension n, and an

R-linear transformation T : V → V of V , such that there is no non-zero v ∈ V satisfying T (v) = λv for someλ ∈ R.

(a) See previous problems. (b) Take a block diagonal matrix consisting of 2 by 2 non-trivial rotations.

Spring 2008 #10. Suppose A is an n × n complex matrix such that A has n distinct eigenvalues. Provethat if B is an n× n complex matrix such that AB = BA, then B is diagonalizable.

Let vi, λi be eigenvector/eigenvalue pairs for A. Note v1, . . . , vn is linearly independent. It follows thatA(Bvi) = B(Avi) = λBvi. Thus Bvi is an eigenvector of A with eigenvalue λi. The eigenspace of λi mustbe one-dimensional, hence Bvi = civi for some ci ∈ C. It follows that B is diagonal with respect to the basisv1, . . . , vn.

Spring 2008 #11. Assume A is an n× n complex matrix such that for some positive integer m the powerAm = Im. Prove that A is diagonalizable.

Note that xm − 1 has no repeated roots over C. If mxm−1 = 0, then x = 0, but this is not a root ofxm − 1. Hence the minimal polynomial of A has no repeated roots. Thus the minimal polynomial factorsinto distinct monic linear factors with coefficients in C. Hence A is diagonalizable.

Spring 2008 #12. Let A be an n× n real symmetric (ai,j = aj,i) matrix, and let S be the unit sphere of

91


Rn. Let x ∈ S be such that(Ax, x) = sup

S(Ay, y)

where (z, y) =∑zjyj is the usual inner product on Rn. (By compactness, such x exists.)

(a) Prove that (x, y) = 0 implies (Ax, y) = 0. Hint: Expand

(A(x+ εy), x+ εy).

(b) Use (a) to prove x is an eigenvector for A.(c) Use induction to prove Rn has an orthonormal basis of eigenvectors for A.Note: If you use part (c) to prove part (a) or part (b), then your solution should include a proof of part

(c) that does not use part (a) or part (b).

(a) Assume (x, y) = 0. Then

(A(x+ εy), x+ εy) = (Ax, x+ εy) + (εy, x+ εy) = (Ax, x) + (Ax, εy) + (εy, x) + (εy, εy)

= (Ax, x) + ε(Ax, y) + 0 + ε2||y||2.

By the choice of x, we must have 1||x+εy||2 (A(x+ εy), x+ εy) ≤ (Ax, x). Hence

(Ax, x) + ε(Ax, y) + ε2||y||2 ≤ (||x||2 + ε2||y||2)(Ax, x) = (Ax, x) + ε2||y||2(Ax, x),

and(Ax, y) ≤ ε||y||2((Ax, x)− 1)

Replacing ε by −ε in the analysis above,

(Ax, y) ≥ −ε||y||2((Ax, x)− 1).

Thus taking ε→ 0, we find that (Ax, y) = 0.

(b) Form an orthonormal basis x, y2, . . . , yn for Rn. Write Ax = c1x + c2y2 + · · · + cnyn. Then for alli ≥ 2, (x, yi) = 0, hence by part (a), ci = (Ax, yi) = 0. Hence Ax = c1x, so x is an eigenvector of A.

(c) This is the usual argument in the proof of the Spectral Theorem.

Fall 2009 #4. Let V be a finite dimensional R-vector space equipped with an inner product. For a vectorsubspace U ⊂ V , denote by U⊥ its orthogonal complement, i.e., the set of v ∈ V such that (v|u) = 0 for allu ∈ U . Show that

dim(U) + dim(U⊥) = dim(V ).

Let u1, . . . , uk be an orthonormal basis for U , and extend it to an orthonormal basis u1, . . . , un for V . Itfollows that uk+1, . . . , un is an orthonormal basis for U⊥, hence we get the desired equality.

Fall 2009 #5. Show that if α1, . . . , αn ∈ R are all different, and some a1, . . . , an ∈ R satisfy∑aie

αit = 0

for all t ∈ (−1, 1), then necessarily ai = 0 for all 1 ≤ i ≤ n. (Hint: you may use the differentiation operatorand a theorem in Linear Algebra on distinct eigenvalues.)

Note eαit are eigenvectors of the differentiation operator on smooth functions with support on (−1, 1).Moreover these eigenvectors have distinct eigenvalues, thus they are linearly independent. (Same proof asin finite dimensional case for infinite dimensions and a finite number of eigenvectors.)

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Or, repeatedly differentiate to get the Vandermonde matrix, which has non-zero determinant.

Fall 2009 #7. Let V isomorphic to Rn be an n-dimensional vector space over R and denote by End(V )the vector space of R-linear transformations of V . (Note that dim(End(V )) = dim(V )2 = n2.) Then forT ∈ End(V ) show that the dimension of the subspace W of End(V ) spanned by T k for k running throughnon-negative integers, satisfies the inequality dim(W ) ≤ dim(V ) = n.

Let T ∈ End(V ). By the Cayley-Hamilton Theorem, χT (T ) = 0, and χT has degree n. This implies thatT 0, . . . , Tn−1 is a basis for W , hence we get the desired inequality.

Fall 2009 #8. For a matrix A ∈ Mn(R), define eA :=∑∞n=0

An

n! . Let v0 ∈ Rn. Prove that the functionv : R → Rn given by v(t) = eAtv0 solves the linear differential equation v′(t) = Av(t) with the initialcondition v(0) = v0. Explain precisely which theorems in calculus you are using in your proof and why theyare applicable.

Note thateA(t+h) − eAt

h=eAh − 1Fn

heAt.

By definition,

eAh − 1Fn

h=

∞∑n=1

1

h

Anhn

n!=

∞∑n=1

Anhn−1

n!= A+

∞∑n=2

Anhn−1n!.

Now we estimate

||∞∑n=2

Anhn−1

n!|| ≤

∞∑n=2

||A||n|h|n−1

n!= ||A||

∞∑n=2

||Ah||n−1n! ≤ ||A||∞∑n=1

||Ah||n = ||A||||Ah|| 1

1− ||Ah||,

which approaches 0 as |h| → 0. Thus

lim|h|→0

eA(t+h) − eAt

h=

(lim|h|→0

eAh − 1Fn

h

)eAt = AeAt.

Fall 2009 #11. Fall 2010 #6. (i) State the Cayley-Hamilton Theorem for matrices A ∈Mn(C).(ii) Prove it directly for diagonalizable matrices.

(iii) Identify Mn(C) isomorphic to Cn2

though some (say, the natural) linear isomorphism. Through thisidentification Mn(C) becomes a metric space with the Euclidean metric. Fact: The subset of diagonalizable

matrices in Mn(C) (isomorphic to Cn2

) is dense. Use this fact, together with part (ii), to prove the Cayley-Hamilton Theorem.

(i) χA(A) = 0.

(ii) A = P−1DP . χA(t) = χD(t) = (t− λ1) · · · (t− λn). Note

χA(A) = χD(A) = χD(D) = (D − λ1) · · · (D − λn) = 0,

since D − λi has the ith row all 0.

(iii) The isomorphism comes by taking each entry. Take A ∈Mn(C). Let An be a sequence of diagonal-izable matrices converging to A. It follows that Akn converges to Ak and thus p(An) converges to p(A) forany polynomial P . By (ii), χAn

(An) = 0. We estimate

||χA(A)|| ≤ ||χA(A)− χA(An)||+ ||χA(An)− χAn(An)||+ ||χAn

(An)||

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= ||χA(A)− χA(An)||+ ||χA(An)− χAn(An)||.The first term is small since χA is continuous. The second is small since det is continuous.

Fall 2009 #12. Let V be an n (≥ 2)-dimensional vector space over C with a set of basis vectors e1, . . . , en.Let T be a linear transformation of V satisfying T (e1) = e2, . . . , T (en−1) = en, T (en) = e1.

(i). Show that T has 1 as an eigenvalue and write down an eigenvector with eigenvalue 1. Show that upto scaling it is unique.

(ii) Is T diagonalizable? (Hint: calculate the characteristic polynomial.)

(i) e1 + · · ·+ en. Fairly straightforward to show unique up to scaling.

(ii) Yes. The characteristic polynomial is tn − 1. Clearly the minimal polynomial is also tn − 1. Both ofthese have n distinct complex eigenvalues, thus T is diagonalizable over C.

Spring 2009 #2. Compute the norm of the matrix

A =

[2 1

0√

3

].

That is, determine the maximum value of the length of Ax over all unit vectors x.

We wish to maximize f(x, y) = ||A[x, y]t|| =√

(2x+ y)2 + (√

3y)2 =√

4x2 + 4xy + 4y2 = 2√x2 + xy + y2 =

2√

1 + xy given x2 + y2 = 1. Now (x− y)2 ≥ 0, so 2xy ≤ x2 + y2 = 1. Note that x = y = 1√2

achieves this

maximum for xy, hence the maximum value of the length of Ax is

2√

3/2 =√

6.

Spring 2009 #3. We wish to find a quadratic polynomial P obeying

P (0) = α, P ′(0) = β, P (1) = γ, andP ′(1) = δ

where ′ denotes differentiation.(a) Find a minimal system of linear constraints on (α, β, γ, δ) such that this is possible.(b) When the constraints are met, what is P? Is it unique? Explain your answer.

(a) P (t) = at2 + bt+ c. The conditions imply c = α, b = β. Thus

P (t) = at2 + βt+ α.

From P (1) = γ,γ = a+ β + α,

so a = γ − β − α andP (t) = (γ − β − α)t2 + βt+ α.

Finally, the condition P ′(1) = δ implies

2(γ − β − α) + β = δ.

Thus we get the constraint2α+ β − 2γ + δ = 0.

(b) If this constraint is met, P is unique and given by

P (t) = (γ − β − α)t2 + βt+ α.

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The matrix equation for a, b, c in terms of α, β, γ involves an invertible matrix, so P is unique.

Spring 2009 #5. Compute eAt when

A =

2 1 50 1 31 0 1

.Recall that eAt is defined by the property that a smooth vector function x(t) obeys:

dx

dt(t) = Ax(t) if and only if x(t) = eAtx(0).

The characteristic polynomial of A is λ2(4− λ), so A3 = 4A2. Thus for all k ≥ 2,

Ak = 4k−2A2.

Then by the series expansion for eAt,

eAt = In + tA+t2A2

2+t3A3

3!+ · · · = In + tA+

t2A2

2+t34A2

3!+ · · ·

= In + tA+t2A2

2+A2

16(e4t − (1 + 4t+ 8t2)) = In + tA+

A2

16(e4t − 1− 4t).

We can use uniqueness of the solution to the differential equation with initial condition to verify theseries formula for etA.

Spring 2009 #8. (a) Show that(A|B) = tr(ABt)

defines an inner product on Mn×n(R). More precisely, show that it obeys the axioms of an inner product.(b) Given C ∈Mn×n(R), we define a linear transformation

ΦC : Mn×n(R)→Mn×n(R)

byΦC(A) = CA−AC.

Compute the adjoint of ΦC . Check that when C is symmetric, then ΦC is self-adjoint.(c) Show that whatever the choice of C, the map ΦC is not onto.

(a) Note

(A,A) = tr(AAt) = tr((

n∑k=1

aikajk)ij)

=

n∑k=1

a1ka1k + · · ·+n∑k=1

ankank =

n∑i,j=1

a2ij ≥ 0,

with equality if and only if aij = 0 for all i, j, in which case, A = 0.Clearly (A+B,C) = (A,C) + (B,C) and (kA,B) = k(A,B).Finally, (A,B) = tr(ABt) = tr((ABt)t) = tr(BAt) = (B,A). Thus (·, ·) is an inner product.

(b) We need that for all A,B,

tr((CA−AC)Bt) = tr(A(Φ∗CB)t).

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The left hand side is

tr(CABt)− tr(ACBt) = tr(ABtC)− tr(ACBt) = tr(A(BtC − CBt)) = tr(A(CtB −BCt)t).

Thus the adjoint isΦ∗C(B) := CtB −BCt = ΦCt(B).

Hence when C is symmetric, Φ∗C(B) = ΦCt(B) = ΦC(B), so ΦC is self-adjoint.

(c) If C = 0, ΦC = 0, so it is clearly not onto. Otherwise, note ΦC(C) = 0, so the kernel of ΦC is at leastone-dimensional. Hence by the Dimension Theorem, ΦC cannot be onto.

Spring 2009 #9. Let us say that a real symmetric n× n matrix A is a reflection if A2 = In and

rank(A− In) = 1.

Given distinct unit vectors x, y ∈ Rn show that there is a reflection with Ax = y and Ay = x. Moreover,show that the reflection A with these properties is unique.

For any orthogonal matrix P , A is a reflection if and only if PAP−1 is a reflection, thus we can shift toany orthonormal basis to solve the problem.

Let w1 = 12 (x+y) and w2 = 1

2 (x−y), then w1 = w1/||w1|| and w2 = w2/||w2||. It follows that w1 ·w2 = 0.Extend this to some orthonormal basis w1, . . . , wn. Note x = ||w1||w1 + ||w2||w2 and y = ||w1||w1−||w2||w2.With respect to this basis, we take

For uniqueness, note that if Rx = y and Ry = x, then R(x−y) = −(x−y). Also, since rank(R−In) = 1,ker(R− In) = (x− y)⊥. Thus for all v ∈ (x− y)⊥, Rv = v. Thus Rwi = Awi for all i, so R = A.

Spring 2009 #11. (a) Explain the following (overly informal) statement:Every matrix can be brought to Jordan normal form; moreover the normal form is essentially unique.No proofs are required; however, all statements must be clear and precise. All required hypotheses must

be included. The meaning of the phrases ’brought to’, ’Jordan normal form’, and ’essentially unique’ mustbe defined explicitly.

(b) Define the minimal polynomial of a matrix. How may it be determined for a matrix in Jordan normalform?

(a) More precisely, every n × n with entries in C is similar to an essentially unique matrix in Jordannormal form.

Two matrices A and B are said to be similar if there exists an invertible matrix P such that B = PAP−1.A Jordan block Jmλ is an m ×m matrix with diagonal entries λ, superdiagonal entries 1, and all other

entries 0. A matrix is in Jordan normal form if it has Jordan blocks along the diagonal, and is zero elsewhere.Given a matrix A, its Jordan normal form is unique up to reordering of the Jordan blocks. In other

words, two matrices in Jordan normal form are similar if and only if they are composed of the same Jordanblocks.

(b) The minimal polynomial of a matrix A is the unique monic polynomial p of minimal degree thatsatisfies p(A) = 0. Clearly the minimal polynomial of a Jordan block Jmλ is (t−λ)m. For a matrix in Jordannormal form, this implies the minimal polynomial is

p =

s∏i=1

(t− λi)Mi ,

where λi is the set of distinct eigenvalues of A, and Mi is the maximal size of a λi-Jordan block in A.

Fall 2010 #5. Prove or disprove: For any two subsets S and S′ of a vector space V ,

96


(a) span(S) ∩ span(S′) = span(S ∩ S′),(b) span(S) + span(S′) = span(S ∪ S′).

(a) False. Take V = R, S = 0, 1 and S′ = 0, 2. Then

span(S) ∩ span(S′) = R ∩ R = R 6= 0 = span(S ∩ S′).

(b) True. For any a ∈ span(S) and b ∈ span(S′), since span(S∪S′) is a vector space, a+b ∈ span(S∪S′).Thus span(S) + span(S′) ⊂ span(S ∪ S′). Conversely, note that span(S) + span(S′) is a vector space whichcontains S ∪ S′, since 0 ∈ span(S) and 0 ∈ span(S′). Thus as the smallest subspace containing S ∪ S′,span(S ∪ S′) ⊂ span(S) + span(S′).

Fall 2010 #7. Let V and W be inner product spaces over C such that dim(V ) ≤ dim(W ) < ∞. Provethat there is a linear transformation T : V →W satisfying

(T (v)|T (v′))W = (v|v′)V

for all v, v′ ∈ V .

We want to construct an isometry from V to a subspace of W . Let v1, . . . , vn be an orthogonal basis forV and w1, . . . , wm be a basis for W . By the assumptions, n ≤ m. Define T (vi) = wi for 1 ≤ i ≤ n. Then forany v, v′ ∈ V , we can write v = a1v1 + · · ·+ anvn and v′ = b1v1 + · · ·+ bnvn so that

(T (v)|T (v′))W = (a1w1 + · · ·+ anwn|b1w1 + · · ·+ bnwn)W = a1b1 + · · ·+ anbn

= (a1v1 + · · ·+ anvn|b1v1 + · · ·+ bn)V = (v|v′)V .

Fall 2010 #8. Let W1 and W2 be subspaces of a finite dimensional inner product space V . Prove that(W1 ∩W2)⊥ = (W1)⊥ + (W2)⊥.

Let v ∈ (W1 ∩W2)⊥. Write v = v1 + v2, where v1 = projW⊥1 v. Then v1 ∈W⊥1 and v− v1 ∈W1. For anyw ∈W2,

(v2|w) = (v − v1|w).

Write w = w1 + w2, where w1 ∈ W1 and w2 ∈ W⊥1 . Then it follows that the right hand side is zero. Hencev2 ∈W⊥2 , and v ∈ (W1)⊥ + (W2)⊥.

Conversely, suppose v1 + v2 ∈ (W1)⊥+ (W2)⊥. Then for any w1 ∈W1, (v1|w1) = 0 and for any w2 ∈W2,(v2|w2) = 0. Thus for any w ∈W1 ∩W2, w ∈W1 and w ∈W2, hence

(v1 + v2|w) = (v1|w) + (v2|w) = 0 + 0 = 0.

Hence v1 + v2 ∈ (W1 ∩W2)⊥.

Fall 2010 #9. Consider the following iterative method

xk+1 = A−1(Bxk + c)

where c is the vector (1, 1)t and A and B are the matrices

A =

(2 00 2

), B =

(2 11 2

).

(a) Assume the iteration converges; to what vector x does the iteration converge?(b) Does this iteration converge for arbitrary initial vectors x0?

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(a) We must solve the matrix equation

L =

(1 1

212 1

)L+

(1212

).

Thus L1 = L1 + 12L2 + 1

2 , so L2 = −1 and likewise, L1 = −1. Hence it converges to L = (−1,−1).

(b) No, note if x0 has positive entries, then xk has positive entries for all k. Thus it cannot converge to−1.

Spring 2010 #1. Let u1, . . . , un be an orthonormal basis of Rn and let y1, . . . , yn be a collection of vectorsin Rn satisfying

∑i ||yi||2 < 1. Prove that the vectors u1 + y1, . . . , un + yn are linearly independent.

Define T (uk) = −yk and extend it by linearity. Let x ∈ Cn and write x =∑nj=1 αjuj . Then

||Tx|| = ||n∑j=1

αjyj || ≤n∑j=1

|αj |||yj || ≤

n∑j=1

|αj |21/2 n∑

j=1

||yj ||2 =

n∑j=1

||yj ||2 ||x||.

Hence

||T || ≤n∑j=1

||yj ||2 < 1.

It follows that I−T is invertible with inverse∑∞i=0 T

i. Since the uk form a basis and (I−T )(uk) = uk + yk,and invertible linear maps take bases to bases, we have that u1 + y1, . . . , un + yn form a basis.

Spring 2010 #5. Let A,B be two n× n complex matrices which have the same minimal polynomial M(t)and the same characteristic polynomial P (t) = (t−λ1)a1 · · · (t−λk)ak , where the λi are distinct. Prove thatif P (t)/M(t) = (t− λ1) · · · (t− λk), then these matrices are similar.

Using Jordan canonical form, there must be a Jordan block Jai−1λi

and J1λi

for each i. Hence the twomatrices have the same Jordan canonical form, up to rearranging the Jordan blocks, hence they are similar.

Spring 2010 #6. Let A =

(4 −41 0

).

(i) Find Jordan form J of A and a matrix P such that P−1AP = J .(ii) Compute A100 and J100.(iii) Find a formula for an, when an+1 = 4an − 4an−1 and a0 = a, a1 = b.

(i) A has minimal polynomial and characteristic polynomial (t− 2)2. Thus its Jordan form is

J =

(2 10 2

).

Since J is the representation of A in the basis consisting of columns of P , we need A(v1) = 2v1 andA(v2) = v1 + 2v2. Picking v2 = (1 0)t, we obtain

P = (v1 v2) =

(2 11 0

).

(ii) Jn =

(2n n2n−1

0 2n

). Higher entries get (nCk)λn−k.

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(iii) Straightforward.

Fall 2011 #7. Let f : R → Mn×n be a continuous function. Show that the function g(t) = rank(f(t)) islower semi-continuous, meaning that if a sequence tn converges to t then g(t) ≤ lim infn g(tn). Is g alwayscontinuous?

Recall that if A ∈Mn(R) \ 0 then rank(A) can be computed to be the largest k such that there existsa k × k submatrix B of A such that det(B) 6= 0.

Let f : R→Mn(F) be a continuous function. Therefore, for each 1 ≤ k ≤ n and selection 1 ≤ i1 < i2 <· · · < ik ≤ n and 1 ≤ j1 < j2 < · · · < jk ≤ n, the function fk,i1,...,ik,j1,...,jk : R → Mk(R) defined as thek × k submatrix of f(t) with rows i1, . . . , ik and columns j1, . . . , jk is a continuous function. Therefore thecomposition hk,i1,...,ik,j1,...,jk of the determinant with this function is continuous.

Fix t ∈ R and let (tn)n≥1 ⊂ R be such that limn→∞ tn = t. If f(t) = 0 there is nothing to prove.Otherwise, f(t) 6= 0. Then taking k = rank(f(t)) there exists a selection 1 ≤ i1 < i2 < · · · < ik ≤ n and1 ≤ j1 < j2 < · · · < jk ≤ n such that hk,t1,...,tk,j1,...,jk(t) 6= 0. Since h... is continuous, there exists an N ∈ Nsuch that h···(tn) 6= 0 for all n ≥ N . Hence

g(tn) = rank(f(tn)) ≥ k = rank(f(t)) = g(t)

for all n ≥ N . Hence g(t) ≤ lim infn g(tn), as desired.To see that g is not always continuous, take f(t) = tIn. Then for t 6= 0, g(t) = n, but g(0) = 0.

Fall 2011 #8. Assume that a complex matrix A satisfies ker((A − λI)) = ker((A − λI)2) for all λ ∈ C.Show from first principles (i.e., without using the theory of canonical forms) that A must be diagonalizable.

Suppose for the sake of contradiction that µA(t) has a multiple root λ. Then

0 = µA(A) = (A− λI)2(q(A))(v)

for all v. But since ker((A− λI)2) = ker((A− λI)), this means

(A− λI)(q(A))(v) = 0

for all v. But this contradicts the definition of the minimal polynomial. Hence the minimal polynomial of Ahas no repeated roots. This implies that A is diagonalizable.

Fall 2011 #10. Let A be a 3× 3 real matrix with A3 = I. Show that A is similar to a matrix of the form 1 0 00 cos(θ) − sin(θ)0 sin(θ) cos(θ)

for some (real) θ. What values of θ are possible?

Note eigenvalues satisfy λ3 = 1. Thus λ = 1, e2π/3 or e4π/3. And if λ is an eigenvalue, then so is λ. Also,the product of the eigenvalues is det(A) = 1. Hence all the eigenvalues are 1 or λ1 = 1, λ2 = e2π/3, andλ3 = e4π/3. In the first case, the characteristic polynomial is t3 − 1, so by the Cayley-Hamilton Theorem,(A − I)3 = 0. Using A3 = I, this implies A2 = A, so A = 0 or A = 1. Clearly A = 1, and this is in thedesired form for θ = 0.

In the second case, defining θ = 2π/3, the second and third eigenvalues are eiθ and e−iθ. Let v1, v2, v3

be eigenvectors of A corresponding to λ1, λ2, λ3. We can write

A = [v1v2v3]

1 0 00 λ2 00 0 λ3

[v1v2v3]−1.

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Now conjugate the lower two rows by

U =1√2

(1 i1 −i

)and note UV is real.

Fall 2011 #11. Suppose V,W , and U are finite dimensional vector spaces over R and that T : V → Wand S : W → U are linear operators. Suppose further that T is one-to-one, S is onto, and S T = 0. Provethat ker(S) ⊃ image(T ) and that

−dim(V ) + dim(W )− dim(U) = dim(ker(S)/image(T )).

From S T = 0, we obtain ker(S) ⊃ image(T ). We know

dim(ker(S)/image(T )) = dim(ker(S))− dim(image(T )).

By the dimension theorem, since T is one-to-one

dim(V ) = dim(ker(T )) + dim(image(T )) = dim(image(T )).

By the dimension theorem, since S is onto,

dim(W ) = dim(ker(S)) + dim(image(S)) = dim(ker(S)) + dim(U).

Putting these together directly yields the desired equation.

Spring 2011 #1. Let A be a 3 by 3 matrix with complex entries. Consider the set of such A that satisfytr(A) = 4, tr(A2) = 6, and tr(A3) = 10. For each similarity (i.e. conjugacy) class of such matrices, give onemember in Jordan normal form. The following identity may be helpful:

If b1 = a1 + a2 + a3, b2 = a21 + a2

2 + a23, and b3 = a3

1 + a32 + a3

3, then

6a1a2a3 = b31 + 2b3 − 3b1b2.

We deduce from the algebra that the eigenvalues of A are 1, 1, and 2. Thus the Jordan normal form musthave a 1 × 1 block with eigenvalue 2 and a 2 × 2 block with diagonal entries 1 and either a 0 or 1 in theupper right.

Spring 2011 #3. Show that for any Hermitian (i.e. self-adjoint) operator H on a finite dimensional innerproduct space there exists a unitary operator U such that UHU∗ is diagonal. (You may use a basis if youneed to!)

The spectral theorem gives an orthonormal basis such that H is diagonal with respect to that basis; thechange of basis matrix consists of the orthogonal basis vectors and hence is unitary.

Spring 2011 #4. Let A be an n×n real matrix. Define an LU decomposition of A. State a necessary andsufficient condition on A for the existence of such a decomposition. Suppose we normalize the decompositionby requiring that the diagonal entries of L are 1. Show that in this case, if the LU decomposition exists,then it is unique. Give the LU decomposition of the matrix(

4 36 3

).

If the principal minors are invertible, the LU decomposition exists. Suppose A = L1U1 = L2U2. ThenL−1

2 L1 = U2U−11 , where the left hand side is unit lower triangular and the right hand side is upper triangular.

It follows that both sides are the identity, so L1 = L2 and U1 = U2.

100


(4 36 3

)=

(1 032 1

)(4 30 − 3

2

).

Fall 2012 #7. Let A be an invertible n × n matrix with entries in C. Suppose that the set of powers An

of A, for n ∈ Z, is bounded. Show that A is diagonalizable.

Since Ak = λkv for an eigenvector v, it is clear that if Ak is bounded for all integers k, then |λ| = 1.Suppose for the sake of contradiction that A is not diagonalizable. Then A has a Jordan normal form with ablock of size at least 2, so there exist unit vectors v1, v2 such that Av1 = λv1 and Av2 = v1 + λv2. It followsthat Akv2 = kλk−1v1 + λkv2, so ||Ak|| ≥ |Akv2| ≥ k− 1. Since k was arbitrary, the set of powers of A is notbounded, a contradiction.

Fall 2012 #8. Let H be an n × n Hermitian matrix with non-zero determinant. Use H to define anHermitian form [·, ·] by the formula: for x, y ∈ Cn (column vectors!), [x, y] = xtHy. Let W be a complexsubspace of Cn such that [w1, w2] = 0 for all w1, w2 ∈ W . Show that dim(W ) ≤ n/2. Give also for each nan example of an H for which dim(W ) = n/2 if n is even or dim(W ) = (n− 1)/2 if n is odd.

Select an orthonormal basis e1, . . . , en so that H is diagonal with respect to this basis. Then the diagonalentries must be non-zero real numbers.

With respect to the orthonormal basis,

[w,w] = λ1|w1|2 + λ2|w2|2 + · · ·+ λn|wn|2.

Pair up positive diagonal entries with negative to get the example H and W . We can always form such abasis, and then nothing can be added to it.

Fall 2012 #10. Let A be a linear operator on a four dimensional complex vector space that satisfies thepolynomial equation P (A) = A4 + 2A3 − 2A − I = 0. Let B = A + I, and suppose dim(range(B)) = 2.Finally, suppose that |tr(A)| = 2. Give a Jordan canonical form of A.

Factor P (A) = (A − I)(A + I)3 and note this must be the characteristic polynomial of A. Thus theeigenvalues of A are 1 and −1. Since dim(range(B)) = 2, the dimension theorem implies dim(ker(A −(−1)I)) = dim(ker(B)) = 2, thus the geometric multiplicity of −1 is 2. From the trace condition, −1 musthave multiplicity 3 and 1 must have multiplicity 1. Hence the Jordan form has a 2× 2 block with −1 on thediagonal, and 1 on the superdiagonal, as well as a 1× 1 block with −1, and a 1× 1 block with 1.

Spring 2012 #7. Let F be the finite field of p elements, let V be an n-dimensional vector space over Fand let 0 ≤ k ≤ n. Compute the number of invertible linear maps V → V . It is acceptable if your solutionis a lengthy algebraic expression, as long as you explain why it is correct.

The first column can be anything but the zero vector, the second can be anything but multiples of thefirst column, and generally the kth column can be any vector not in the span of the first k−1 columns. Thisgives

(pn − 1)(pn − p) · · · (pn − pn−1)

invertible linear maps. Since the first k − 1 columns are linearly independent, we are not overcounting thenumber of possibilities for the kth column.

Spring 2012 #8. Let A be an n × n complex matrix. Prove that there are two sequences of matricesBi and Li, such that Li are diagonal with distinct eigenvalues, and BiLiB

−1i → A as i → ∞. Here by

convergence of matrices we mean convergence in all entries.

101


First, write A as V UV −1, where U is upper triangular using Jordan canonical form. Then by changingthe diagonal entries of U by less than 1/i, we obtain a matrix Ti with distinct entries along the diagonal.Then Ti has distinct eigenvalues, so there exists some Ci with Ti = CiLiC

−1i , where Li is diagonal. It follows

that(V Ci)Li(V Ci)

−1 → V UV −1 = A

as i→∞.

Spring 2012 #9. Let a1 = 1, a2 = 4, an+2 = 4an+1 − 3an for all n ≥ 1. Find a 2× 2 matrix A such that

An ·(

10

)=

(an+1

an

)for all n ≥ 1. Compute the eigenvalues of A and use them to determine the limit

limn→∞

(an)1/n.

A =

(4 −31 0

).

The eigenvalues of A are 1 and 3. Thus we can diagonalize A and limn→∞(an)1/n = 3 since it picks out thelargest eigenvalue.

Spring 2012 #10. Let A be a complex n × n matrix. State and prove under which conditions on A thefollowing identity holds:

det(eA) = exp(tr(A)).

Here the matrix exponentiation is defined via the Taylor series. You can assume known that this sumconverges (entrywise) for all complex matrices A.

This holds for any matrix A. Using Jordan normal form, there exists an invertible matrix V such thatU = V AV −1 is upper triangular. It is straightforward to show eV AV

−1

= V eAV −1 using the definitions.Then

det(eA) = det(eV UV−1

) = det(V eUV −1) = det(eT )

andetr(A) = etr(V UV −1) = etr(UV V −1) = etr(T ).

Thus it suffices to show that det(eT ) = etr(T ) for upper triangular T . But this follows immediately since thedeterminant is the product of the diagonal entries of eT , which are eλi .

Spring 2012 #11. (a) Find a polynomial P (x) of degree 2, such that P (A) = 0, for

A =

(1 34 2

).

(b) Prove that such P (x) is unique, up to multiplication by a constant.

(i) P (x) = x2 − 3x− 10.

(ii) Suppose there exist two monic quadratic polynomials P and Q with P (A) = 0 = Q(A). Then(P −Q)(A) = 0, and P −Q is either 0 or a first degree polynomial. But clearly no first degree polynomialevaluated at A is 0, hence P −Q = 0, so P = Q.

102


Spring 2012 #12. Recall that the quadratic forms Q1(x, y) and Q2(x′, y′) are said to be equivalent ifthey are related by a non-singular change of coordinates (x, y) 7→ (x′, y′). Decide whether Q1 = xy andQ2 = x2 + y2 are equivalent over C and whether they are equivalent over R. If not, give a proof. If yes, findthe matrix for change of coordinates.

Send x to x′ = x + iy and y to y′ = x − iy. Then x′y′ = x2 + y2. This is a non-singular changeof coordinates, so the quadratic forms are equivalent over C. They are not equivalent over R. Supposea, b, c, d ∈ R with (ax+ by)(cx+dy) = acx2 + bdy2 +(ad+ bc)xy. If this equals x2 +y2, then ad+ bc = 0, andac = bd = 1. Thus a and c are either both positive or both negative, and b and d are either both positiveor both negative. But this implies ad and bc are both positive or both negative, so they cannot sum to 0, acontradiction.

Spring 2013 #6. (a) Prove that diagonalizable matrices are dense in the set of all n × n matrices withcomplex entries.

(b) Are diagonalizable matrices with real entries dense in the set of all n× n matrices with real entries?

(a) See Spring 2012 #8.

(b) No. For example,

(0 1−1 0

), which has eigenvalues ±i cannot be approximated by diagonalizable

matrices. The roots of a polynomial are continuous functions of its coefficients, and the coefficients of thecharacteristic polynomial are continuous functions of the entries of the matrix, thus the eigenvalues of amatrix are continuous functions of its entries, so any far-enough-along approximating matrices cannot bediagonalizable in the reals, because the diagonal matrix would need to have complex entries.

Spring 2013 #7. (a) Show that the series exp(A) = I +A+A2/2! + · · · converges to a limit in the usualsense of convergence of matrices (converge entry by entry).

(b) Show that the series ln(I+A) = A−A2/2+A3/3+ · · ·+(−1)n+1An/n+ · · · converges if the operatornorm of A is less than one.

(c) Show that exp(ln(I +A)) = I +A if the operator norm of A is less than 1.

(a) Since ||AB|| ≤ ||A||||B||, we have ||Ak|| ≤ ||A||k. By the ratio test,∑∞k=0

1k! ||A||

k converges, and thepartial sums of eA have norm less than the partial sums of this series, so eA converges. This relies on thecompleteness of the normed space of matrices.

(b) Note ∑k≥1

|| (−1)k+1

kAk|| ≤

∑k≥1

1

k||Ak|| ≤

∑k≥1

||A||k <∞,

where the final series is geometric since ||A|| < 1. Technically we should estimate the partial sums, thenconclude that the series converges.

(c) Too long to work out - not worth it.

Spring 2013 #8. Let T be a linear transformation from a finite-dimensional vector space V with an innerproduct to a finite dimensional vector space W also with an inner product (the dimension of W can bedifferent from the dimension of V here).

(a) Define the adjoint T ∗ : W → V .(b) Show that if matrices are written relative to orthonormal bases of V and W , then the matrix of T ∗

is the transpose of the matrix of T .

103


(a) Petersen takes a basis ei for V and sets

T ∗y =

n∑i=1

(y|T (ej))W ej .

Then it follows that(Lx|y) = (x|L∗y).

Uniqueness: suppose (x|K1y) = (x|K2y) for all x, y. Then

0 = (x|K1y −K2y).

Then taking x = K1y −K2y, we get K1y = K2y.(b) Let e1, . . . , en be an orthonormal basis of V and f1, . . . , fm be an orthonormal basis of W . Suppose

T with respect to these bases has entries aij . Thus T (ei) =∑mi=1 aijfi. It follows that

T ∗fj =

n∑i=1

(fi|T (ei))ei =

n∑i=1

aijei.

Hence the matrix representation of T ∗ with respect to these bases has entries aji, so the matrix of T ∗ is thetranspose of the matrix of T .

Spring 2013 #10. Denote by G the set of real 4 × 4 upper triangular matrices with 1’s on the diagonal.Fix

M =

1 1 1 10 1 1 10 0 1 10 0 0 1

.

Denote by C the set of matrices in G commuting with M .(a) Prove that C is an affine subspace in the space R16 of all 4×4 real matrices. S is an “affine subspace”

of a vector space V if there is a vector w ∈ V such that S0 = v − w : v ∈ S is a subspace of V . Thedimension of S is defined to be the dimension of S0.

(b) Find the dimension of C.

By brute force, we check that matrices in C take the form1 a b c0 1 a b0 0 1 a0 0 0 1

,

where a, b, c are arbitrary. Taking w = I4, C − w is a subspace of R16, so C is an affine subspace of R16.

(b) Since a, b, c are arbitrary, C − w has dimension 3, so C is an affine subspace of dimension 3.

Fall 2011 #9. Let V be a finite dimensional inner product space, and let L : V → V be a self-adjoint linearoperator. Let µ and ε be given. Suppose there is a unit vector x ∈ V such that

||L(x)− µx|| ≤ ε.

Prove that L has an eigenvalue λ so that |λ− µ| ≤ ε.

By the spectral theorem, there exists an orthonormal basis (ei) consisting of eigenvectors of L witheigenvalues λi. Write

x =

n∑i=1

(x · ei)ei.

104


ThenL(x)− µx =

∑((x · ei)λi − µ(x · ei))ei =

∑(λi − µ)(x · ei)ei.

Now ∑|λi − µ||x · ei|2 =

∑||(λi − µ)(x− ei)||2 = ||L(x)− µx||2 ≤ ε2.

Since |x| = 1,∑|x · ei|2 = 1, hence there must exist i with

|λi − µ|2 ≤ ε2.

Thus |λi − µ| ≤ ε.

Spring 2010 #2. Let A be an n× n real symmetric matrix and let λ1 ≥ . . . ≥ λn be the eigenvalues of A.Prove that

λk = maxU :dim(U)=k

minx∈U :||x||=1

(Ax|x),

where (·, ·) denotes the usual scalar product in Rn and the maximum is taken over all k-dimensional subspacesof Rn.

As in Spring 2008 #12, we can show minx∈U :||x||=1(Ax|x) is an eigenvalue for any subspace U . In fact itis the least eigenvalue with an eigenvector in U . Now any k-dimensional subspace contains eigenvectors fork eigenvalues λi. Hence the max over all U is λk.

Spring 2010 #4. (i) Let A = (ai,j) be an n × n real symmetric matrix such that∑i,j ai,jxixj ≤ 0 for

every vector (x1, . . . , xn) ∈ Rn. Prove that if tr(A) = 0, then A = 0.(ii) Let T be a linear transformation in the complex finite dimensional vector space V with a positive

definite Hermitian inner product. Suppose that TT ∗ = 4T − 3I, where I is the identity transformation.Prove that T is positive definite Hermitian and find all possible eigenvalues of T .

(i) Suppose tr(A) = 0. Using x = ei, we see aii ≤ 0. Thus the diagonal entries are 0. Then using ei + ejand ei − ej for i 6= j, as well as the symmetry of A, we see aij ≥ 0 and aij ≤ 0, so A = 0.

(ii) ?

Spring 2011 #5. Let A be an n by n matrix with real entries, and let b be an n by 1 column vector withreal entries. Prove that there exists an n by 1 column vector solution x to the equation Ax = b if and onlyif b is in the orthocomplement of the kernel of the transpose of A.

There exists x such that Ax = b holds if and only if b is in the column space of A, which holds if andonly if b is in the row space of AT . Now b is in the row space of AT if and only if every element in the kernelof AT is orthogonal to b, which holds if and only if b is in the orthocomplement of the kernel of AT .

Winter 2006 #9. Let A ∈ M3(R) be invertible and satisfy A = At and detA = 1. Prove that A has oneas an eigenvalue.

?

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UCLA Basic Exam Problems and Solutions Brent Woodhouse ...bwoodhouse729/Basic Exam... · UCLA Basic Exam Problems and Solutions Brent Woodhouse This document covers nearly all ...

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