Vector Spaces Linear Algebra. Session 6 Dr. Marco A Roque Sol 08/01/2017 Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector Spaces
Linear Algebra. Session 6
Dr. Marco A Roque Sol
08/01/2017
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Linear Dependence and Independence
Let S0 and S be subsets of a vector space V
More facts on linear independence
If S0 ⊂ S and S is linearly independent, then so is S0.
If S0 ⊂ S and S0 is linearly dependent, then so is S.
If S is linearly independent in V and V is a subspace of W,then S is linearly independent in W.
The empty set is linearly independent.
Any set containing 0 is linearly dependent.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Vector Spaces
Two nonzero vectors v1 and v2 are linearly dependent if andonly if either of them is a scalar multiple the other.
If S0 is linearly independent and v0 ∈ V − S0 then, S0 ∪ {v0}is linearly independent if and only if v0 /∈ Span(S0).
Theorem
Vectors v1, v2, · · · , vm ∈ Rn are linearly dependent wheneverm > n (i.e., the number of coordinates is less than the number ofvectors).
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Vector Spaces
proof.
Let vj = (a1j , a2j , · · · , anj) for j = 1, 2, · · · ,m Then the vectorequality t1v1 + t2v2 + · · ·+ tmvm = 0 is equivalent to the system
a11t1 + a12t2 + · · ·+ a1mtn = 0a21t1 + a22t2 + · · ·+ a2ntn = 0
...an1t1 + an2t2 + · · ·+ anmtn = 0
Note that vectors v1, v2, · · · , vm are columns of the coefficientmatrix (aij). The number of leading entries in the row echelonform is at most n. If m > n then there are free variables, thereforethe zero solution is not unique. �
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Linear Dependence and Independence
General results on linear independence in Rn
Theorem 1 Given an n ×m matrix the following conditions areequivalent:
i) columns of A are linearly independent (as vectors in Rn )
ii) x = 0 is the only solution of the matrix equation Ax = 0
iii) The row echelon form of A has a leading entry in each column.
Theorem 2 Given a square matrix the following conditions areequivalent:
i) det(A) = 0
ii) Columns of A are linearly independent (as vectors in Rn )
iii) Rows of A are linearly independent (as vectors in Rn )Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Linear Dependence and Independence
Example 6.1
Consider vectors v1 = (1,−1, 1), v2 = (1, 0, 0), v3 = (1, 1, 1), andv4 = (1, 2, 4) in R3
Two vectors are linearly dependent if and only if they are parallel.Hence v1 and v2 are linearly independent.
Vectors v1, v2, and v3 are linearly independent if and only if thematrix A = (v1, v2, v3) is invertible.
det(A) =
∣∣∣∣∣∣1 1 1−1 0 11 0 1
∣∣∣∣∣∣ = −∣∣∣∣ −1 1
1 1
∣∣∣∣ = 2 6= 0
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Linear Dependence and Independence
Therefore v1, v2, v3 are linearly independent.
Four vectors in R3 are always linearly dependent. Thus v1, v2, v3and v4 are linearly dependent.
Example 6.2
Let
A =
(−1 1−1 0
).
Determine whether matrices A,A2,A3 are linearly independent
Solution
We have
A =
(−1 1−1 0
)A2 =
(0 −11 −1
)A3 =
(1 00 1
)Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Linear Dependence and Independence
We need to determine if there exist r1, r2, r3 ∈ R not all zero suchthat r1A + r2A
2 + r3A3 = 0. This matrix equation is equivalent to
a system −r1 + 0r2 + r3 = 0r1 − r2 + 0r3 = 0−r1 + r2 + 0r3 = 00r1 − r2 + r3 = 0
The augmented matrix is
−1 0 1 01 −1 0 0−1 1 0 00 −1 1 0
⇒−1 0 1 00 1 −1 00 0 0 00 0 0 0
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Linear Dependence and Independences
The row echelon form of the augmented matrix shows there is afree variable. Hence the system has a nonzero solution so that thematrices are linearly dependent (one relation is A + A2 + A3 = 0
Example 6.3
Show that functions ex , e2x , and e3x are linearly independent inC∞(R)
Solution
Suppose that aex + be2x + ce3x = 0 for all x ∈ R where a, b, c areconstants. We have to show that a = b = c = 0. Differentiate thisidentity twice:
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Linear Dependence and Independence
aex + be2x + ce3x = 0aex + 2be2x + 3ce3x = 0aex + 4be2x + 9ce3x = 0
It follows that A(x)v = 0, where
A(x) =
ex e2x e3x
ex 2e2x 3e3x
ex 4e2x 9e3x
, v =
abc
det(A(x)) =
ex e2x e3x
ex 2e2x 3e3x
ex 4e2x 9e3x
= ex
∣∣∣∣∣∣1 e2x e3x
1 2e2x 3e3x
1 4e2x 9e3x
∣∣∣∣∣∣ =
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Linear Dependence and Independence
exe2x = e3x
∣∣∣∣∣∣1 1 e3x
1 2 3e3x
1 4 9e3x
∣∣∣∣∣∣ =
e3xe3x = e6x
∣∣∣∣∣∣1 1 11 2 31 4 9
∣∣∣∣∣∣ = e6x
∣∣∣∣∣∣1 1 10 1 20 3 8
∣∣∣∣∣∣ = e6x∣∣∣∣ 1 1
3 8
∣∣∣∣ = 2e6x 6= 0
Since the matrix A(x) is invertible we obtainA(x)v = 0⇒ v = 0⇒ a = b = c = 0Hence, the set of functions {ex , e2x , e3x} is linearly independent.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Linear Dependence and Independence
Wronskian
Let f1, f2, ..., fn be smooth functions on an interval [a, b]. TheWronskian W[f1, f2, ..., fn] is a function on [a, b] defined by
W[f1, f2, ..., fn](x) =
∣∣∣∣∣∣∣∣∣f1(x) f2(x) · · · fn(x)f ′1(x) f ′2(x) · · · f ′n(x)
...
f(n−1)1 (x) f
(n−1)2 (x) · · · f
(n−1)n (x)
∣∣∣∣∣∣∣∣∣Theorem
If W[f1, f2, ..., fn](x0) 6= 0 for some x0 ∈ [a, b] then the functionsf1, f2, ..., fn are linearly independent in C [a, b].
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Basis and Dimension
Definition.
Let V be a vector space. Any linearly independent spanning set Sfor V, is called a basis. That means, that any vector v ∈ V can berepresented as a linear combination
v = r1v1 + r2v2 + · · ·+ rkvk
where v1, v2, · · · , vk are distinct vectors from S andr1, r2, · · · , rk ∈ R
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
OBS
“Linearly independent”, in the above definition, implies that theabove representation is unique:
v = r1v1 + r2v2 + · · ·+ rkvk
v = r ′1v1 + r ′2v2 + · · ·+ r ′kvk ⇒
0 = (r1 − r ′1)v1 + (r2 − r ′2)v2 + · · ·+ (rk − r ′k)vk ⇒
(r1 − r ′1) = 0, (r2 − r ′2) = 0 · · · (rk − r ′k) = 0⇒
r1 = r ′1, r2 = r ′2, · · · rk = r ′k
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Example 6.4Standard Basis for Rn
e1 = (1, 0, 0, ..., 0), e2 = (0, 1, ..., 0), · · · , en = (0, 0, ..., 1),
Solution
Indeed,
(x1, x2, ..., xn) = x1e1 + x2e2 + ... + xnen
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Example 6.5Standard Basis for M2,2
E11 =
(1 00 0
),E12 =
(0 10 0
),E21 =
(0 01 0
),E22 =
(0 00 1
)Solution
Indeed,
A =
(a bc d
)= aE12 + bE12 + cE21 + dE22
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Example 6.6Standard Basis for Polinomial Pn
a0, a1x , a2x2, ..., an−1x
n−1Solution
POC ... !!!!
Example 6.7Standard Basis for Polinomials P
a0, a1x , a2x2, ..., anx
n, ...SolutionPOC ... !!!!
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Let v1, v2, · · · vk ∈ Rn and r1, r2, · · · rn ∈ R. The vector
v = r1v1 + r2v2 + · · ·+ rkvk
is equivalent to the matrix equation Ax = v where
A = (v1, v2, · · · vk) x =
r1r2...rk
That is, A is the n × k matrix such that vectors v1, v2, · · · vk areconsecutive columns of A.
Vectors v1, v2, · · · vk span Rn if the row echelon form of A hasno zero rows.Vectors v1, v2, · · · vk are linearly independent if the rowechelon form of A has a leading entry in each column (no freevariables).
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
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Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Dimension
Let v1, v2, · · · vk ∈ Rn
Theorem 1
If k < n then the vectors v1, v2, · · · vk do not span Rn.
Theorem 2
If k > n then the vectors v1, v2, · · · vk are linearly dependent.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Vector Spaces
Theorem 3
If k = n then the following conditions are equivalent:
{v1, v2, · · · vn} is a basis for Rn
{v1, v2, · · · vn} is a spanning set for Rn
{v1, v2, · · · vn} is a linearly independent set.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Vector Spaces
Example 6.8
v1 = (1,−1, 1), v2 = (1, 0, 0), v3 = (1, 1, 1), v4 = (1, 2, 4),
Vectors v1 and v2 are linearly independent (as they are notparallel), but they do not span R3
Vectors v1, v2 and v3 are linearly independent since∣∣∣∣∣∣1 1 1−1 0 11 0 1
∣∣∣∣∣∣ =
∣∣∣∣ −1 11 1
∣∣∣∣ = 2 6= 0
Therefore, {v1, v2, v3} is a basis for R3
The set {v1, v2, v3, v4} span R3 ( because {v1, v2, v3} already spanR3 ), but they are linearly dependent.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Vector Spaces
Dimension
Theorem 1
Any vector space has a basis.
Theorem 2
If a vector space V has a finite basis, then all bases for V are finiteand have the same number of elements.
Definition
The dimension of a vector space V, denoted dim(V), is thenumber of elements in any of its bases.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Vector Spaces
Example 6.9
Rn : n-dimensional space. dim(Rn) = n
M2,2 : the space of 2× 2 matrices. dim(M2,2) = 4
Mm,n : the space of m × n matrices. dim(Mm,n) = mn
Pn : the space of polynomials of degree less than n.dim(Pn) = n
P : the space of all polynomials. dim(Pn) =∞
{0} : the trivial space . dim({0}) = 0
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Vector Spaces
Example 6.10Find the dimension of the plane x + 2z = 0 in R3
Solution
The general solution of the equation isx = 2sy = tz = s
(t, s,∈ R)
That is, (x , y , z) = (−2s, t, s) = t(0, 0, 1) + s(−2, 0, 1)
Hence, the plane is the span of vectors v1 = (0, 0, 1) andv2 = (−2, 0, 1). These vectors are linearly independent as they arenot parallel. Thus, {v1, v2} is a basis so that the dimension of theplane is 2.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Basis
Theorem
Let S be a subset of a vector space V. Then, the followingconditions are equivalent:
i) S is a basis, i.e., is a linearly independent spanning set for V.
ii) S is a minimal spanning set for V.
iii) S is a maximal linearly independent subset of V.
’Minimal spanning set’ means: remove any element from this set,and it is no longer a spanning set.
’ Maximal linearly independent subset’ means: add any element ofV to this set, and it will become linearly dependent.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Theorem
Let V be a vector space. Then
i) Any spanning set for V can be reduced to a minimal spanningset.
ii) any linearly independent subset of V can be extended to amaximal linearly independent set.
Corollary 1Any spanning set contains a basis while any linearly independentset is contained in a basis.
Corollary 2
A vector space is finite-dimensional if and only if it is spanned by afinite set.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
How to find a basis?
Approach 1.
Get a spanning set for the vector space, then reduce this set to abasis dropping one vector at a time.
Theorem
Let v0, v1, ..., vk be a spanning set for a vector space V . If v0 is alinear combination of vectors v1, v2, ..., vk then v1, v2, ..., vk is alsoa spanning set for V
Example 6.11
Find a basis for the vector space V spanned by vectorsw1 = (1, 1, 0),w2 = (0, 1, 1),w3 = (2, 3, 1) and w4 = (1, 1, 1)
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Solution
Since we have four vector in R3, then they are linearly dependent,and satisfy an equation of the form
r1w1 + r2w2 + r3w3 + r4v4 = 0
where ri ∈ R are not all equal to zero. Equivalently,
1 0 2 11 1 3 10 1 1 1
r1r2r3r4
=
0000
to solve this system of linear equations for r1, r2, r3, r4 we apply rowreduction
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
1 0 2 11 1 3 10 1 1 1
→ 1 0 2 1
0 1 1 00 1 1 1
→ 1 0 2 1
0 1 1 00 0 0 1
→ 1 0 2 0
0 1 1 00 0 0 1
→
r1 + 2r3r2 + r3r4 = 0
→
r1 = −2r3r2 = −r3r4 = 0
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Thus, the general solution is(r1, r2, r3, r4) = (−2t,−t, t, 0), t ∈ R and a particular solution is(2, 1,−1, 0). We have obtained that2w1 + w2 −w3 = 0⇒ 2w1 + w2 = w3. Hence, we can dropp w3
and take V = Span(w1,w2,w4)
Let us check whether vectors w1,w3,w4 are linearly independent
∣∣∣∣∣∣1 0 11 1 10 1 1
∣∣∣∣∣∣ =
∣∣∣∣∣∣1 0 10 1 00 1 1
∣∣∣∣∣∣ =
∣∣∣∣ 1 01 1
∣∣∣∣ = 1 6= 0
They are!!!. Hence {w1,w3,w4} is a basis for V . Besides, itfollows that V = R3
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Approach 2.
Build a maximal linearly independent set adding one vector at atime.
If the vector space V is trivial, it has the empty basis.
If V 6= 0, pick any vector v1 6= 0, if v1 spans V, it is a basis.Otherwise pick any vector v2 ∈ V that is not in the span of v1. Ifv1, v2 span V they constitute a basis. Otherwise pick any vectorv3 ∈ V that is not in the span of v1, v2. And so on.
Modifications.
Instead of the empty set, we can start with any linearlyindependent set (if we are given one). If we are given a spanningset S, it is enough to pick new vectors only in S
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Remark. This inductive procedure works for finite-dimensionalvector spaces. There is an analogous procedure forinfinite-dimensional spaces (transfinite induction ).
Example 6.12
Vectors v1 = (0, 1, 0) and v2 = (−2, 0, 1) are linearly independent.Extend the set {v1, v2} to a basis for R3
Solution
Our task is to find a vector v3 that is not a linear combination ofv1, v2. Now, the vectors v1, v2, span the plane x + 2z = 0 and thevetor v3 = (1, 1, 1) does not lie in that plane. Hence, it is not alinear combination of v1, v2. Thus, {v1, v2, v3} is a basis for R3.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Example 6.13
Vectors v1 = (0, 1, 0) and v2 = (−2, 0, 1) are linearly independent.Extend the set {v1, v2} to a basis for R3
Solution
Since vectors e1 = (1, 0, 0), e2 = (0, 1, 0), e3 = (0, 0, 1) form aspanning set for R3, at least one of them can be chosen as v3
Let us check that {v1, v2, e3} and {v1, v2, e2} form a basis for R3
∣∣∣∣∣∣0 −2 11 0 00 1 0
∣∣∣∣∣∣ = 1 6= 0
∣∣∣∣∣∣0 −2 11 0 00 1 1
∣∣∣∣∣∣ = 2 6= 0
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
Row space of a matrix
The row space of an m × n matrix A is the subspace of Rn
spanned by the rows of A
The dimension of the row space is called the rank of the matrix A.
Theorem 1
The rank of a matrix A is the maximal number of linearlyindependent rows in A.
Theorem 2
Elementary row operations do not change the row space of amatrix.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
Theorem 3
If a matrix A is in row echelon form, then the nonzero rows of Aare linearly independent.
Corollary
The rank of a matrix is equal to the number of nonzero rows in itsrow echelon form.
Theorem 4
Elementary row operations do not change the row space of amatrix.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
proofSuppose that A and B are m × n matrices such that B is obtainedfrom A by an elementary row operation. Let a1, ..., am be the rowsof A and b1, ...,bm, be the rows of B. We have to show thatSpan(a1, ..., am) = Span(b1, ...,bm)
However, we have to observe that any row bi of B belongs toSpan(a1, ..., am). Indeed, either bi = aj for some 1 ≤ j ≤ m (interchange of two rows ), or bi = rai for some sclar r 6= 0 (multiplying any row by a scalar different from zero ), orbi = ai + raj for some i 6= j and r ∈ R ( adding to a row, amultiple of another row )
It follows that Span(b1, ...,bm) ⊂ Span(a1, ..., am)
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
Now, the matrix A can also be obtained from B by an elementaryrow operation. By the above,
Span(a1, ..., am) ⊂ Span(b1, ...,bm) �
Example 6.14
Find the rank of the matrix
1 1 00 1 12 3 11 1 1
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
Solution
Elementary row operations do not change the row space. Let usconvert A to row echelon form:
1 1 00 1 12 3 11 1 1
⇒
1 1 00 1 10 1 11 1 1
⇒
1 1 00 1 10 1 10 0 1
⇒
1 1 00 1 10 0 00 0 1
⇒
1 1 00 1 10 0 10 0 0
⇒Thus, the rank of A is 3.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
Column space of a matrix
Definition.The column space of an m × n matrix A is the subspace of Rm,spanned by columns of A is the subspace of Rm spanned bycolumns of A
Theorem 1
The column space of a matrix A coincides with the row space ofthe transpose matrix AT
Theorem 2
Elementary row operations do not change linear relations betweencolumns of a matrix.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
Theorem 5
Elementary row operations do not change the dimension of thecolumn space of a matrix (however they can change the columnspace).
Theorem 4
If a matrix is in row echelon form, then the columns with leadingentries form a basis for the column space.
Corollary
For any matrix, the row space and the column space have thesame dimension.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
Example 6.15
Find a basis for the column space of the matrix
A =
1 1 00 1 12 3 11 1 1
Solution
The column space of A coincides with the row space of AT . Tofind a basis, we convert AT to row echelon form:
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
AT =
1 0 2 11 1 3 10 1 1 1
⇒ 1 0 2 1
0 1 1 00 1 1 1
⇒ 1 0 2 0
0 1 1 00 0 0 1
⇒Vectors (1, 0, 2, 1), (0, 1, 1, 0), and (0, 0, 0, 1) form a basis for thecolumn space of A.
Nullspace of a matrix
Let A = (aij) be an m × n
Definition
The nullspace of the matrix A, denoted by N(A) is the set of alln− dimensional column vectors x such that Ax = 0
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
a11 a12 · · · a1na21 a22 · · · a2n
...am1 am2 · · · amn
x1x2...xn
=
00...0
OBS The nullspace N(A) is the solution set of a system of linearhomogeneous equations (with A as the coefficient matrix)
Theorem
N(A) is a subspace of the vector space of Rn
Definition
The dimension of the nullspace N(A) is called the nullity of thematrix A
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
Rank + Nullity
TheoremThe rank of a matrix A plus the nullity of A equals the number ofcolumns in A.
rank(A) + N(A) = n
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
Example 6.16
Let B given by
B =
0 −1 4 11 1 2 −1−3 0 −1 02 −1 0 1
Find the rank and the nullity of the matrix B.
Find a basis for the row space of B, then extend this basis toa basis for R4.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
Solution
The rank (= dimension of the row space) and the nullity (=dimension of the nullspace) of a matrix are preserved underelementary row operations. We apply such operations toconvert the matrix B into its row echelon form
B =
0 −1 4 11 1 2 −1−3 0 −1 02 −1 0 1
⇒Interchange the 1st row with the 2nd row:
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
1 1 2 −10 −1 4 1−3 0 −1 02 −1 0 1
⇒Add 3 times the 1st row to the 3rd row, then subtract 2 times the1st row from the 4th row
1 1 2 −10 −1 4 10 3 5 −32 −1 0 1
⇒
1 1 2 −10 −1 4 10 3 5 −30 −3 −4 3
⇒
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
Multiply the 2nd row by−11 1 2 −10 −1 −4 −10 3 5 −30 −3 −4 3
⇒Add the 4th row to the 3rd row:
1 1 2 −10 −1 −4 −10 0 1 00 −3 −4 3
⇒Add 3 times the 2nd row to the 4th row:
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
1 1 2 −10 −1 −4 −10 0 1 00 0 −16 0
⇒Add 16 times the 3rd row to the 4th row:
1 1 2 −10 −1 −4 −10 0 1 00 0 0 0
⇒
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
Now that the matrix is in row echelon form, its rank equals thenumber of nonzero rows, which is 3. Since
(rank of B ) + (nullity of B ) = (the number of columns of B ) =4,
it follows that the nullity of B equals 1.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
The row space of a matrix is invariant under elementary rowoperations. Therefore the row space of the matrix B is thesame as the row space of its row echelon form:
B =
0 −1 4 11 1 2 −1−3 0 −1 02 −1 0 1
⇒
1 1 2 −10 1 −4 −10 0 1 00 0 0 0
The nonzero rows of the latter matrix are linearly independentso that they form a basis for its row space:
v1 = (1, 1, 2,−1), v2 = (0, 1,−4,−1), v3 = (0, 0, 1, 0)
To extend the basis v1, v2, v3, to a basis for R4, we need av4 ∈ R4 that is not a linear combination of v1, v2, v3.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
It is known that at least one of the vectorse1 = (1, 0, 0, 0), e2 = (0, 1, 0, 0), e3 = (0, 0, 1, 0), ande4 = (0, 0, 0, 1), can be chosen as v4
In particular, the vectors v1, v2, v3, e4 form a basis for R4
Dr. Marco A Roque Sol Linear Algebra. Session 6