Chapter 4 Vector Spaces Linear Algebra. Ch04_2 Definition 1. Let be a sequence of n real numbers. The set of all such sequences is called n-space (or.

Chapter 4

Vector Spaces

Linear Algebra

Ch04_2

Definition 1.Let be a sequence of n real numbers. The set of all such sequences is called n-space (or n-dimensional space) and is denoted Rn.

u1 is the first component of .

u2 is the second component and so on.

) ..., , ,( 21 nuuu

) ..., , ,( 21 nuuu

Example 1

• R2 is the collection of all sets of two ordered real numbers. For example, (0, 0) , (1, 2) and (-2, -3) are elements of R2.

• R3 is the collection of all sets of three ordered real numbers.

For example, (0,0, 0) and (-1,3, 4) are elements of R3.

4.1 The vector Space Rn

Ch04_3

Definition 2. Let be two elements of Rn.

We say that u and v are equal if u1 = v1, …, un = vn.

Thus two elements of Rn are equal if their corresponding components are equal.

) ..., , ,( and ) ..., , ,( 2121 nn vvvuuu vu

Definition 3. Let be elements of Rn

and let c be a scalar.

Addition and scalar multiplication are performed as follows:

Addition:

Scalar multiplication :

) ..., , ,( and ) ..., , ,( 2121 nn vvvuuu vu

) ..., ,() ..., ,(

1

11

n

nn

cucucvuvu

uvu

Ch04_4

► The set Rn with operations of componentwise addition and scalar multiplication is an example of a vector space, and its elements are called vectors. We shall Henceforth interpret Rn to be a vector space.(We say that Rn is closed under addition and scalar multiplication).

► In general, if u and v are vectors in the same vector space, then u + v is the diagonal of the parallelogram defined by u and v.

Figure 4.2

Ch04_5

Example 2

Let u = ( –1, 4, 3) and v = ( –2, –3, 1) be elements of R3.

Find u + v and 3u.

Solution: u + v = (–1, 4, 3) + (– 2, –3, 1) = (-3 ,1 ,4)

3u = 3 (–1, 4, 3) = (-3 ,12 ,9)Example 3

Figure 4.1

In R2 , consider the two elements (4, 1) and (2, 3).Find their sum and give a geometrical

interpretation of this sum.

we get (4, 1) + (2, 3) = (6, 4).

The vector (6, 4), the sum, is the diagonal of the parallelogram.

Ch04_6

Example 4

Figure 4.3

Consider the scalar multiple of the vector (3, 2) by 2, we get

2(3, 2) = (6, 4)

Observe in Figure 4.3 that (6, 4) is a vector in the same direction as (3, 2), and 2 times it in length.

Ch04_7

Zero Vector The vector (0, 0, …, 0), having n zero components, is called the zero vector of Rn and is denoted 0.

Negative Vector

The vector (–1)u is writing –u and is called the negative of u. It is a vector having the same length (or magnitude) as u, but lies in the opposite direction to u.

Subtraction

Subtraction is performed on element of Rn by subtracting corresponding components.

u

u

Ch04_8

Theorem 4.1Let u, v, and w be vectors in Rn and let c and d be scalars.

(a) u + v = v + u

(b) u + (v + w) = (u + v) + w

(c) u + 0 = 0 + u = u

(d) u + (–u) = 0

(e) c(u + v) = cu + cv

(f) (c + d)u = cu + du

(g) c(du) = (cd)u

(h) 1u = u

Figure 4.4 Commutativity of

vector addition u + v = v + u

Ch04_9

Example 5 Let u = (2, 5, –3), v = ( –4, 1, 9), w = (4, 0, 2) in the vector space R3. Determine the vector 2u – 3v + w.

Solution

)31 7, 20,(

)2276 ,0310 4,12(4

2) 0, (4,27) 3, ,12()6 10, ,4(

2) 0, (4,9) 1, ,4(3)3 5, ,2(2

w3v2u

Ch04_10

Column Vectors

nnnn vu

vu

v

v

u

u

1111

We defined additionaddition and scalar multiplicationscalar multiplication of column vectors in Rn in a componentwise manner:

and

nn cu

cu

u

uc

11

Row vector:

Column vector:

) ..., , ,( 21 nuuuu

nu

u

1

Ch04_11

Homework

Exercise set 1.3 page 32:3, 5, 7, 9.

Ch04_12

4.2 Dot Product, Norm, Angle, and Distance

Definition Let be two vectors in Rn. The dot product of u and v is denoted u.v and is defined by .

The dot product assigns a real number to each pair of vectors.

) ..., , ,( and ) ..., , ,( 2121 nn vvvuuu vu

nnvuvu 11vu

Example 1

Find the dot product ofu = (1, –2, 4) and v = (3, 0, 2)

Solution

11

803

)24()02()31(

vu

Ch04_13

Properties of the Dot Product

Let u, v, and w be vectors in Rn and let c be a scalar. Then

1. u.v = v.u

2. (u + v).w = u.w + v.w

3. cu.v = c(u.v) = u.cv

4. u.u 0, and u.u = 0 if and only if u = 0

Proof1.

uv

vuvu

numbers real ofproperty ecommutativ by the

get We). ..., , ,( and ) ..., , ,(Let

11

11

2121

nn

nn

nn

uvuvvuvu

vvvuuu

4.

. ifonly and if 0 Thus.0 , ,0 ifonly and if ,0

.0 thus,0

122

1

221

22111

0uuu

uu

uu

nn

n

nnn

uuuu

uu

uuuuuu

Ch04_14

Norm of a Vector in Rn

Definition The norm (length or magnitude) of a vector u = (u1, …, un) in Rn

is denoted ||u|| and defined by

Note: The norm of a vector can also be written in terms of the dot product

221 nuu u

uuu

Figure 4.5 length of u

Ch04_15

Definition A unit vector is a vector whose norm is 1.

If v is a nonzero vector, then the vector

is a unit vector in the direction of v.

This procedure of constructing a unit vector in the same direction as a given vector is called normalizing the vector.

vv

u1

Find the norm of each of the vectors u = (1, 3, 5) of R3 and v = (3, 0, 1, 4) of R4.

Solution 352591)5()3()1( 222 u

2616109)4()1()0()3( 2222 v

Example 2

Ch04_16

Example 3

Solution

(a) Show that the vector (1, 0) is a unit vector.

(b) Find the norm of the vector (2, –1, 3). Normalize this vector.

(a) Thus (1, 0) is a unit vector. It can be similarly shown that (0, 1) is a unit vector in R2.

.1010) ,1( 22

(b) The norm of (2, –1, 3) is

The normalized vector is

The vector may also be written

This vector is a unit vector in the direction of (2, –1, 3).

.143)1(23) ,1 ,2( 222 .14

)3 ,1 ,2(141

.143

,141

,142

Ch04_17

Angle between Vectors ( in R2)

← Figure 4.6

The law of cosines gives:

Ch04_18

Definition Let u and v be two nonzero vectors in Rn.

The cosine of the angle between these vectors is

0 cosvuvu

Example 4Determine the angle between the vectors u = (1, 0, 0) and v = (1, 0, 1) in R3.

Solution 11) 0, (1,0) 0, 1,( vu

2101 1001 222222 vu

Thus the angle between u and v is 45.,2

1cos

vuvu

Angle between Vectors (in R n)

Ch04_19

Definition Two nonzero vectors are orthogonal if the angle between them is a

right angle .

Two nonzero vectors u and v are orthogonal if and only if u.v = 0.

Theorem 4.2

Proof

0 0cos orthogonal are , vuvu

Orthogonal Vectors

Ch04_20

Solution

Example 5Show that the following pairs of vectors are orthogonal.

(a) (1, 0) and (0, 1).

(b) (2, –3, 1) and (1, 2, 4).

(a) (1, 0).(0, 1) = (1 0) + (0 1) = 0.

The vectors are orthogonal.

(b) (2, –3, 1).(1, 2, 4) = (2 1) + (–3 2) + (1 4) = 2 – 6 + 4 = 0.

The vectors are orthogonal.

Ch04_21

Note

(1, 0), (0,1) are orthogonal unit vectors in R2.

(1, 0, 0), (0, 1, 0), (0, 0, 1) are orthogonal unit vectors in R3.

(1, 0, …, 0), (0, 1, 0, …, 0), …, (0, …, 0, 1) are orthogonal unit vectors in Rn.

Ch04_22

Example 6

Determine a vector in R2 that is orthogonal to (3, –1). Show that there are many such vectors and that they all lie on a line.

Solution

Let the vector (a, b) be orthogonal to (3, –1). We get

abba

baba

3030))1(()3(01) ,3() ,(

Thus any vector of the form (a, 3a)is orthogonal to the vector (3, –1). Any vector of this form can be written

a(1, 3)The set of all such vectors lie on the line defined by the vector (1, 3).

Figure 4.7

Ch04_23

Theorem 4.3

Let u and v be vectors in Rn.

(a) Triangle Inequality:

||u + v|| ||u|| + ||v||.

(a) Pythagorean theorem : If u.v = 0 then ||u + v||2 = ||u||2 + ||v||2.

Figure 4.8(a) Figure 4.8(b)

Ch04_24

Distance between PointsLet be two points in Rn.

The distance between x and y is denoted d(x, y) and is defined by

Note: We can also write this distance as follows.

) ..., , ,( and ) ..., , ,( 2121 nn yyyxxx yx

2211 )()() ,( nn yxyxd yx

yxyx ) ,(d

Example 7. Determine the distance between the points x = (1,–2 , 3, 0) and y = (4, 0, –3, 5) in R4.

Solution

74

253649

)50()33()02()41(),( 2222

yxd

x

yxy

Note: ( , ) ( , ) ( )It is clear that d d the symmetric propertyx y y x

Ch04_25

Homework

Exercise set 1.5 pages 47 to 48:2, 6, 10, 16, 20, 33, 35.

Exercise 36

Let u and v be vectors in Rn. Prove that ||u|| = ||v|| if and only if u + v and u v are orthogonal.

Ch04_26

4.3 General Vector Spaces

Definition A vector space is a set V of elements called vectors, having operations of addition and scalar multiplication defined on it that satisfy the following conditions.

Let u, v, and w be arbitrary elements of V, and c and d are scalars.

• Closure Axioms

1. The sum u + v exists and is an element of V. (V is closed under addition.)

2. cu is an element of V. (V is closed under scalar multiplication.)

Our aim in this section will be to focus on the algebraic properties of Rn.

Ch04_27

Example 1

(1) V={ …, 3, 1, 1, 3, 5, 7, …} V is not closed under addition because 1+3=4 V.

(2) Z={ …, 2, 1, 0, 1, 2, 3, 4, …} Z is closed under addition because for any a, b Z, a + b Z. Z is not closed under scalar multiplication because ½ is a scalar, for any odd a Z, (½)a Z.

Ch04_28

• Addition Axioms

3. u + v = v + u (commutative property)

4. u + (v + w) = (u + v) + w (associative property)

5. There exists an element of V, called the zero vector, denoted 0, such that u + 0 = u.

6. For every element u of V there exists an element called the negative of u, denoted u, such that u + (u) = 0.

• Scalar Multiplication Axioms

7. c(u + v) = cu + cv

8. (c + d)u = cu + du

9. c(du) = (cd)u

10. 1u = u

Definition of Vector Space (continued)

Ch04_29

W. Thus W is closed under scalar multiplication.

A Vector Space in R3

Let , for some a, bR.Wba )1,0,1( and )1,0,1( vu

Axiom 1:

u + v W. Thus W is closed under addition.

)1,0,1)(()1,0,1()1,0,1( baba vu

Axiom 3,4 and 7~10: trivial

}. | 1) 0, (1, {Let R aaW Prove that W is a vector space.

Proof

)1,0,1(cac uAxiom 2:

Axiom 5: Let 0 = (0, 0, 0) = 0(1,0,1), then 0 W and 0+u = u+0 = u for any u W.

Axiom 6: For any u = a(1,0,1) W. Let u = a(1,0,1), then u W and (u)+u = 0.

Ch04_30

Vector Spaces of Matrices (Mmn)

Let . and 22Mhg

fe

dc

ba

vu

Axiom 1:

u + v is a 2 2 matrix. Thus M22 is closed under addition.

► Question: Prove Axiom 2 and Axiom 7 .

hdgcfbea

hgfe

dcba

vu

Axiom 3 and 4:

From our previous discussions we know that 2 2 matrices are commutative and associative under addition (Theorem 2.2).

}.,,, | {Let 22 R

srqp

sr

qpM Prove that M22 is a vector space.

Proof

Ch04_31

{ | , , , 0} a vector space?p q

Is the set W p q r sr s

Axiom 5:

The 2 2 zero matrix is , since

00

000

u0u

dcba

dcba

0000

Axiom 6:

0uu

uu

0000

ddccbbaa

dcba

dcba

dcba

dcba

)(

since , then , If

In general: The set of m n matrices, Mmn, is a vector space.

Ch04_32

Vector Spaces of Functions

Axiom 1:

f + g is defined by (f + g)(x) = f(x) + g(x). f + g : R R

f + g F. Thus F is closed under addition.

Axiom 2:

cf is defined by (cf)(x) = c f(x).

cf : R R

cf F. Thus F is closed under scalar multiplication.

Prove that F = { f | f: R R } is a vector space.

Let f, g F, c R. For example: f: R R, f(x)=2x, g: R R, g(x)=x2+1.

Ch04_33

Vector Spaces of Functions (continued)Axiom 5:

Let 0 be the function such that 0(x) = 0 for every xR.

0 is called the zero function.

We get (f + 0)(x) = f(x) + 0(x) = f(x) + 0 = f(x) for every xR.

Thus f + 0 = f. (0 is the zero vector.)

Axiom 6:Let the function –f defined by (f )(x) = f (x).

Thus [f + (f )] = 0, f is the negative of f.)(

0

)]([)(

))(()())](([

x

xfxf

xfxfxff

0

Is the set F ={ f | f(x)=ax2+bx+c for some a,b,c R} a vector space?

Ch04_34

Theorem 4.4 (useful properties)

Let V be a vector space, v a vector in V, 0 the zero vector of V,

c a scalar, and 0 the zero scalar. Then

(a) 0v = 0

(b) c0 = 0

(c) (1)v = v(d) If cv = 0, then either c = 0 or v = 0.

Ch04_35

Homework

Exercise set 4.1, pages 206-207:5, 6, 13, 14.

Ch04_36

4.4 Subspaces

Figure 4.9

Ch04_37

Definition Let V be a vector space and U be a nonempty subset of V.

U is said to be a subspace of V if it is closed under addition

and under scalar multiplication.

Note:► In general, a subset of a vector space may or may not satisfy the closure axioms. ► However, any subset that is closed under both of these operations satisfies all the other vector space properties.

Ch04_38

Example 1Let U be the subset of R3 consisting of all vectors of the form (a, 0, 0) (with zeros as second and third components and aR ), i.e., U = {(a, 0, 0) R3 }.Show that U is a subspace of R3.

Solution

Let (a, 0, 0), (b, 0, 0) U, and let k R. We get

(a, 0, 0) + (b, 0, 0) = (a + b, 0, 0) Uk(a, 0, 0) = (k a, 0, 0) U

The sum and scalar product are in U. Thus U is a subspace of R3. #

Geometrically, U is the set of vectors that lie on the x-axis.

Ch04_39

Example 2Let V be the set of vectors of of R3 of the form (a, a2, b), namely V = {(a, a2, b) R3 }.Show that V is not a subspace of R3.

Solution

Let (a, a2, b), (c, c2, d) V. (a, a2, b) + (c, c2, d) = (a+ c, a2 + c2, b + d)

(a + c, (a + c)2, b + d) ,

since a2 + c2 (a + c)2 .Thus (a, a2, b) + (c, c2, d) V. V is not closed under addition. V is not a subspace.

Ch04_40

Example 3Prove that the set W of 2 2 diagonal matrices is a subspace of the vector space M22 of 2 2 matrices.

Solution

(+) Let W.

qp

ba

00

and 0

0vu

We get

u + v W. W is closed under addition.

qbpa

qp

ba

00

00

00

vu

() Let c R. We get

cu W. W is closed under scalar multiplication. W is a subspace of M22.

cbca

ba

cc0

0

00

u

Ch04_41

The vector space of polynomials (Pn)

Example 5. Let Pn denoted the set of real polynomial functions of degree n. Prove that Pn is a vector space if addition and scalar multiplication are defined on polynomials in a pointwise manner.

Example 5. Let Pn denoted the set of real polynomial functions of degree n. Prove that Pn is a vector space if addition and scalar multiplication are defined on polynomials in a pointwise manner.

SolutionLet f and g Pn, where

11 1 0

11 1 0

( ) ... and

( ) ...

n nn n

n nn n

f x a x a x a x a

g x b x b x b x b

►(+)

(f + g)(x) is a polynomial of degree n. Thus f + g Pn. Then Pn is closed under addition.

)()(...)()(

]...[]...[

)()(

))((

00111

11

011

1011

1

baxbaxbaxba

bxbxbxbaxaxaxa

xgxf

xgf

nnn

nnn

nn

nn

nn

nn

Example 4.

Ch04_42

►() Let c R,

(cf )(x) is a polynomial of degree n. So cf Pn. Then Pn is closed under scalar multiplication.

In conclusion : By (+) and (), Pn is a subspace of the vector space F of functions. Therefore Pn is itself a vector space.

011

1

011

1

...

]...[

)]([))((

caxcaxcaxca

axaxaxac

xfcxcf

nn

nn

nn

nn

Ch04_43

Theorem 4.5 (Very important condition)

Let U be a subspace of a vector space V. U contains the zero vector of V.

Note. Let 0 be the zero vector of V. If 0 U U is not a subspace of V. If 0 U (+)() hold U is a subspace of V. (+)() failed U is not a subspace of V.Caution. This condition is necessary but not sufficient. (See, for instance, Example 2 above and Example 5 below)

Ch04_44

Example 5

Let W be the set of vectors of the form (a, a, a+2). Show that W is not a subspace of R3.

Solution

If (a, a, a+2) = (0, 0, 0), then a = 0 and a + 2 = 0 .

This system is inconsistent it has no solution.

Thus (0, 0, 0) W. (The necessary condition does not hold)

W is not a subspace of R3.

Ch04_45

Homework

Exercise set 1.3, page 32: 11, 12, 13.

Let F ={ f | f: R R } the vector space of functions on R.Which of the following are subspaces of F ?(a) W1={ f | f: R R, f(0)=0 }. (b) W2={ f | f: R R, f(0)=3 }. (c) W3={ f | f: R R, for some cR, f(x)=c for every x}.

Exercise

Ch04_46

4.5 Linear Combinations of Vectors

W={(a, a, b) | a,b R} R3

(a, a, b) = a (1,1,0) + b (0,0,1)

W is generated by (1,1,0) and (0,0,1). e.g., (2, 2, 3) = 2 (1,1,0) + 3 (0,0,1) (1, 1, 7) = 1 (1,1,0) + 7 (0,0,1).

Definition Let v1, v2, …, vm be vectors in a vector space V.

We say that v, a vector of V, is a linear combination of

v1, v2, …, vm , if there exist scalars c1, c2, …, cm such that

v can be write v = c1v1 + c2v2 + … + cmvm .

Ch04_47

The vector (5, 4, 2) is a linear combination of the vectors

(1, 2, 0), (3, 1, 4), and (1, 0, 3), since it can be written

(5, 4, 2) = (1, 2, 0) + 2(3, 1, 4) – 2(1, 0, 3)

Example 1

Ch04_48

Example 2

Determine whether or not the vector (1, 1, 5) is a linear combination of the vectors (1, 2, 3), (0, 1, 4), and (2, 3, 6).

Solution

Suppose 5) 1, 1,(6) 3, (2,4) 1, (0,3) 2, (1, 321 ccc

)5 ,1 ,1()643 ,32 ,2()5 ,1 ,1()6 ,3 ,2()4 , ,0()3 ,2 ,(

32132131

33322111

cccccccccccccccc

1,2,1

5 643

1 3 2

12

321

321

321

31

ccc

ccc

ccc

cc

Thus (1, 1, 5) is a linear combination of (1, 2, 3), (0, 1, 4), and (2, 3, 6), where 6). 3, (2,14) 1, (0,23) 2, (1,5) 1, 1,(

Ch04_49

Example 3

Express the vector (4, 5, 5) as a linear combination of the vectors (1, 2, 3), (1, 1, 4), and (3, 3, 2).

Solution

Suppose 5) 5, (4,2) 3, (3,4) 1,1(3) 2, (1, 321 c, cc

)5 ,5 ,4()243 ,32 ,3()5 ,5 ,4()2 ,3 ,3()4 , ,()3 ,2 ,(

321321321

333222111

cccccccccccccccccc

rcrcrc

ccc

ccc

ccc

321

321

321

321

,1,32

5 243

53 2

43

Thus (4, 5, 5) can be expressed in many ways as a linear combination of (1, 2, 3), (1, 1, 4), and (3, 3, 2):

2) 3, ,3(4) 1, ,1()1(3) 2, (1,)32(5) 5, ,4( rrr

Ch04_50

Example 4

Show that the vector (3, 4, 6) cannot be expressed as a linear combination of the vectors (1, 2, 3), (1, 1, 2), and (1, 4, 5).

Solution

Suppose

)6 ,4 ,3(5) 4, (1,2) ,11(3) 2, (1, 321 c, cc

6 523

44 2

3

321

321

321

ccc

ccc

ccc

This system has no solution. Thus (3, 4, 6) is not a linear combination of the vectors (1, 2, 3), (1, 1, 2), and (1, 4, 5).

Ch04_51

Example 5

Determine whether the matrix is a linear combination

of the matrices in the vector space

M22 of 2 2 matrices.

1871

0210

and ,2032

,1201

SolutionSuppose

1871

0210

2032

1201

321 ccc

18

71

222

32

2121

3221

cccc

ccccThen

Ch04_52

12

822

73

12

21

31

32

21

cc

cc

cc

cc

This system has the unique solution c1 = 3, c2 = 2, c3 = 1. Therefore

0210

2032

21201

318

71

Ch04_53

Example 6Determine whether the function is a linear combination of the functions and

710)( 2 xxxf

Solution

Suppose .21 fhcgc

710)42()13( 222

21 xxxxcxxc

Then

13)( 2 xxxg

.42)( 2 xxxh

7104)3()2( 22121

221 xxccxccxcc

74

103

12

21

21

21

cc

cc

cc

1 ,3 21 cc .3 hgf

Ch04_54

Definition The vectors v1, v2, …, vm are said to span a vector space if every vector in the space can be expressed as a linear combination of these vectors.

In this case {v1, v2, …, vm} is called a spanning set.

Spanning Sets

Ch04_55

Show that the vectors (1, 2, 0), (0, 1, 1), and (1, 1, 2) span R3.

SolutionLet (x, y, z) be an arbitrary element of R3. Suppose )2 ,1 ,1()1 ,1 ,0()0 ,2 ,1() , ,( 321 ccczyx

)2 ,2 ,() , ,( 3232131 ccccccczyx

Example 7

zcc

yccc

xcc

32

321

31

2

2

zyxc

zyxc

zyxc

2

24

3

3

2

1

The vectors (1, 2, 0), (0, 1, 1), and (1, 1, 2) span R3.

)2 ,1 ,1)(2()1 ,1 ,0)(24()0,2,1)(3() , ,( zyxzyxzyxzyx

Ch04_56

Example 8Show that the following matrices span the vector space M22 of 2 2 matrices.

1000

0100

0010

0001

Solution

dcba

Let M22 (an arbitrary element).

We can express this matrix as follows:

proving the result.

1000

0100

0010

0001

dcbadcba

Ch04_57

Theorem 4.6

Let v1, …, vm be vectors in a vector space V. Let U be the set consisting of all linear combinations of v1, …, vm . Then U is a subspace of V spanned the vectors v1, …, vm . U is said to be the vector space generated by v1, …, vm .Proof

(+) Let u1 = a1v1 + … + amvm and u2 = b1v1 + … + bmvm U.

Then u1 + u2 = (a1v1 + … + amvm) + (b1v1 + … + bmvm) = (a1 + b1) v1 + … + (am + bm) vm

u1 + u2 is a linear combination of v1, …, vm .

u1 + u2 U.

U is closed under vector addition.

Ch04_58

()Let c R. Then

cu1 = c(a1v1 + … + amvm)

= ca1v1 + … + camvm

cu1 is a linear combination of v1, …, vm . cu1 U.

U is closed under scalar multiplication.

Thus U is a subspace of V.

By the definition of U, every vector in U can be written as a linear combination of v1, …, vm .

Thus v1, …, vm span U.

Ch04_59

Example 9

Consider the vector space R3.

The vectors (1, 5, 3) and (2, 3, 4) are in R3.

Let U be the subset of R3 consisting of all vectors of the form

c1(1, 5, 3) + c2(2, 3, 4)

Then U is a subspace of R3 spanned by (1, 5, 3) and (2, 3, 4).

The following are examples of some of the vectors in U, obtained by given c1 and c2 various values.

)81 ,1 ,4(vector ;3 ,2)0 ,0 ,0(vector ;0 ,0

)4 ,3 ,2(vector ;1 ,0)3 ,5 ,1(vector ;0 ,1

21

21

21

21

cccccccc

We can visualize U. U is made up of all vectors in the plane defined by the vectors (1, 5, 3) and (2, 3, 4).

Ch04_60

Figure 4.10

Ch04_61

We can generalize this result. Let v1 and v2 be vectors in the space R3.

The subspace U generated by v1 and v2 is the set of all vectors of the form c1v1 + c2v2.

If v1 and v2 are not colinear, then U is the plane defined by v1 and v2 .

Figure 4.11

Ch04_62

Figure 4.12

If v1 and v2 are vectors in R3 that are not colinear, then we can visualize U as a plane in three dimensions.

k1v1 and k2v2 will be vectors on the same lines as v1 and v2.

Ch04_63

Example 10Let U be the subspace of R3 generated by the vectors (1, 2, 0) and (3, 1, 2). Let V be the subspace of R3 generated by the vectors (1, 5, 2) and (4, 1, 2). Show that U = V.

(U V) Let u be a vector in U. Let us show that u is in V. Since u is in U, there exist scalars a and b such that

Let us see if we can write u as a linear combination of (1, 5, 2) and (4, 1, 2).

Such p and q would have to satisfy

Solution

)2 ,2 ,3()2 ,1 ,3()0 ,2 ,1( bbababa u

)22 ,5 ,4()2- ,1 ,4()2 ,5 ,1( qpqpqpqp u

bqpbaqpbaqp

22225

34

Ch04_64

This system of equations has unique solutionThus u can be written

Therefore, u is a vector in V.

.32

,3

baq

bap

)2 ,1 ,4(32

)2 ,5 ,1(3

babau

(V U) Let v be a vector in V. Let us show that v is in U. Since v is in V, there exist scalars c and d such that

It can be shown that)2 14()2 ,5 ,1( ,, dcv

)2 ,1 ,3)(()0 ,2 ,1)(2( dcdcv

Therefore, v is a vector in U and hence U=V.

Ch04_65

Figure 4.13

Ch04_66

Example 11Let U be the vector space generated by the functions f(x) = x + 1 and g(x) = 2x2 – 2x + 3. Show that the function h(x) = 6x2–10x+5 lies in U.Solutionh will be in the space generated by f and g if there exist scalars a and b such that

a(x + 1) + b(2x2 – 2x + 3) = 6x2 – 10x + 5 This given

2bx2 + (a – 2b)x + a + 3b = 6x2 – 10x + 5

2b = 6 a – 2b = – 10

a + 3b = 5This system has the unique solution a = – 4, b = 3. Thus – 4(x + 1) + 3(2x2 – 2x + 3) = 6x2 – 10x + 5

Ch04_67

Homework

Exercise set 4.2 pages 214-215:2, 4, 6, 7, 15, 17.

Ch04_68

4.6 Linear Dependence and Independence

Definition(a) The set of vectors { v1, …, vm } in a vector space V is said to

be linearly dependent if there exist scalars c1, …, cm, not all zero, such that c1v1 + … + cmvm = 0

(b) The set of vectors { v1, …, vm } is linearly independent if c1v1 + … + cmvm = 0 can only be satisfied when

c1 = 0, …, cm = 0.

The concepts of dependence and independence of vectors are useful tools in constructing “efficient” spanning sets for vectorspaces – sets in which there are no redundant vectors.

Ch04_69

Example 1

Show that the set {(1, 2, 3), (2, 1, 1), (8, 6, 10)} is linearly dependent in R3.

SolutionSuppose

0 )10 ,6 ,8()1 ,1 ,2()3 ,2 ,1( 321 ccc

00

)103 ,62 ,82()10 ,6 ,8() , ,2()3 ,2 ,(

321321321

333222111

cccccccccccccccccc

Thus The set of vectors is linearly dependent.

0 )10 ,6 ,8()1 ,1 ,2(2)3 ,2 ,1(4

c1 = 4c2 = 2c3 = 1

0103

062

082

321

321

321

ccc

ccc

ccc

Ch04_70

Example 2

Show that the set {(3, 2, 2), (3, 1, 4), (1, 0, 5)} is linearly independent in R3.

Suppose

0 )5 ,0 ,1()4 ,1 ,3()2 ,2 ,3( 321 ccc

00

)542 ,2 ,33()5 ,0 ,()4 , ,3()2 ,2 ,3(

32121321

33222111

cccccccccccccccc

Solution

This system has the unique solution c1 = 0, c2 = 0, and c3 = 0. Thus the set is linearly independent.

0542

02

033

321

21

321

ccc

cc

ccc

Ch04_71

Example 3Consider the functions f(x) = x2 + 1, g(x) = 3x – 1, h(x) = – 4x + 1 of the vector space P2 of polynomials of degree 2. Show that the set of functions { f, g, h } is linearly independent.Solution

Supposec1f + c2g + c3h = 0

Since for any real number x,

Consider three convenient values of x. We get

0 )14()13()1( 322

1 xcxcxc

0542 :10322 :1

0 :0

321

321

321

cccxcccx

cccx

Ch04_72

It can be shown that this system of three equations has the unique solution

c1 = 0, c2 = 0, c3 = 0

Thus c1f + c2g + c3h = 0 implies that c1 = 0, c2 = 0, c3 = 0. The set { f, g, h } is linearly independent.

Ch04_73

Theorem 4.7

A set consisting of two or more vectors in a vector space is linearly dependent if and only if it is possible to express one of the vectors as a linearly combination of the other vectors.

Example 4The set of vectors {v1=(1, 2, 1) , v2 =(-1, -1, 0) , v3 = (0, 1,1)}

is linearly dependent, since v3 = v1 + v2 .Thus, v3 is a linear combination of v1 and v2.

m

Ch04_74

Linear Dependence of {v1, v2}

{v1, v2} linearly dependent;vectors lie on a line

{v1, v2} linearly independent;vectors do not lie on a line

Figure 4.14 Linear dependence and independence of {v1, v2} in R3.

Ch04_75

Linear Dependence of {v1, v2, v3}

{v1, v2, v3} linearly dependent;vectors lie in a plane

{v1, v2, v3} linearly independent;vectors do not lie in a plane

Figure 4.15 Linear dependence and independence of {v1, v2, v3} in R3.

Ch04_76

Theorem 4.8

Let V be a vector space. Any set of vectors in V that contains the zero is linearly dependent.

Proof

Consider the set { 0, v2, …, vm }, which contains the zero vectors. Let us examine the identity

0vv0 mmccc 221

We see that the identity is true for c1 = 1, c2 = 0, …, cm = 0 (not all zero). Thus the set of vectors is linearly dependent, proving the theorem.

Ch04_77

Theorem 4.9Let the set {v1, …, vm} be linearly dependent in a vector space V. Any set of vectors in V that contains these vectors will also be linearly dependent.

Example 5The set of vectors

{v1=(1, 2, 1) , v2 =(-1, -1, 0) , v3 = (0, 1,1) , v4 =( 1, 1, 1) } is linearly dependent, since it contains the vectors v1 , v2 , v3 which are linearly dependent.

Ch04_78

Example 6

Let the set {v1, v2} be linearly independent. Prove that {v1 + v2, v1 – v2} is also linearly independent.

Solution

Suppose(1)

We get0)()( 2121 vvvv ba

0)()(0

21

2121

vvvvvv

bababbaa

Since {v1, v2} is linearly independent

Thus system has the unique solution a = 0, b = 0. Returning to identity (1) we get that {v1 + v2, v1 – v2} is linearly independent.

00

baba

Ch04_79

Homework

Exercise set 4.3 pages 219-220:1, 3, 8, 9, 12, 16.

Ch04_80

4.7 Bases and Dimension

DefinitionA finite set of vectors {v1, …, vm} is called a basis for a vector space V if the set spans V and is linearly independent.

Standard BasisThe set of n vectors

{(1, 0, …, 0), (0, 1, …, 0), …, (0, …, 1)}

is a basis for Rn. This basis is called the standard basis for Rn.

How to prove it?

Ch04_81

Example 1

Show that the set {(1, 0, 1), (1, 1, 1), (1, 2, 4)} is a basis for R3.

Solution

(span)Let (x1, x2, x3) be an arbitrary element of R3. Suppose

)4 ,2 ,1()1 ,1 ,1()1 ,0 ,1() , ,( 321321 aaaxxx

Thus the set spans the space.

3321

232

1321

4

2

xaaa

xaa

xaaa

3213

3212

3211

2

252

32

xxxa

xxxa

xxxa

Ch04_82

04020

321

32

321

bbbbbbbb

(linearly independent) Consider the identity

The identity leads to the system of equations)0 ,0 ,0()4 ,2 ,1()1 ,1 ,1()1 ,0 ,1( 321 bbb

b1 = 0, b2 = 0, and b3 = 0 is the unique solution. Thus the set is linearly independent. {(1, 0, 1), (1, 1, 1), (1, 2, 4)} spans R3 and is linearly independent. It forms a basis for R3.

Ch04_83

Example 2

Show that { f, g, h }, where f(x) = x2 + 1, g(x) = 3x – 1, and h(x) = –4x + 1 is a basis for P2.

Solution(linearly independent) see Example 3 of the previous section. (span). Let p be an arbitrary function in P2. p is thus a polynomial of the form

dcxbxxp 2)(

Suppose )()()()( 321 xhaxgaxfaxp

for some scalars a1, a2, a3.

This gives

)()43(

)14()13()1(

321322

1

322

12

aaaxaaxa

xaxaxadcxbx

Ch04_84

Thus the polynomial p can be expressed

The functions f, g, and h span P2.

They form a basis for P2.

)()()()( 321 xhaxgaxfaxp

daaacaaba

321

32

1

43cdba

cdba

ba

33

44

3

2

1

Ch04_85

Theorem 4.10 Let {v1, …, vn } be a basis for a vector space V. If {w1, …, wm} is a set of more than n vectors in V, then this set is linearly dependent.Proof

Suppose(1)

We will show that values of c1, …, cm are not all zero.

011 mmcc ww

The set {v1, …, vn} is a basis for V. Thus each of the vectors w1, …, wm can be expressed as a linear combination of v1, …, vn. Let

nmnmmm

nn

aaa

aaa

vvvw

vvvw

2211

12121111

Ch04_86

Substituting for w1, …, wm into Equation (1) we get

0vvvvvv )()( 221112121111 nmnmmmnn aaacaaac

0vv nmnmnnmm acacacacacac )()( 221111212111

Since v1, …, vn are linear independent,

0

0

2211

1221111

mmnnn

mm

cacaca

cacaca

Since m > n, there are many solutions in this system.

Rearranging, we get

Thus the set {w1, …, wm} is linearly dependent.

Ch04_87

Theorem 4.11Any two bases for a vector space V consist of the same number of vectors.

Proof

Let {v1, …, vn} and {w1, …, wm} be two bases for V.

Thus n = m.

By Theorem 4.11, m n and n m

Ch04_88

DefinitionIf a vector space V has a basis consisting of n vectors, then the dimension of V is said to be n. We write dim(V) for the dimension of V.

• V is finite dimensional if such a finite basis exists.• V is infinite dimensional otherwise.

Ch04_89

Example 3

Consider the set {{1, 2, 3), (-2, 4, 1)} of vectors in R3. These vectors generate a subspace V of R3 consisting of all vectors of the form

The vectors (1, 2, 3) and (-2, 4, 1) span this subspace.

)1 ,4 ,2()3 ,2 ,1( 21 ccv

Furthermore, since the second vector is not a scalar multiple of the first vector, the vectors are linearly independent.Therefore {{1, 2, 3), (-2, 4, 1)} is a basis for V. Thus dim(V) = 2. We know that V is, in fact, a plane through the origin.

Ch04_90

Theorem 4.12

(a) The origin is a subspace of R3. The dimension of this subspace is defined to be zero.

(b) The one-dimensional subspaces of R3 are lines through the origin.

(c) The two-dimensional subspaces of R3 are planes through the origin.

Figure 4.16 One and two-dimensional subspaces of R3

Ch04_91

Proof

(a) Let V be the set {(0, 0, 0)}, consisting of a single elements, the zero vector of R3. Let c be the arbitrary scalar. Since

(0, 0, 0) + (0, 0, 0) = (0, 0, 0) and c(0, 0, 0) = (0, 0, 0)

V is closed under addition and scalar multiplication. It is thus a subspace of R3. The dimension of this subspaces is defined to be zero.

(b) Let v be a basis for a one-dimensional subspace V of R3. Every vector in V is thus of the form cv, for some scalar c. We know that these vectors form a line through the origin.

(c) Let {v1, v2}be a basis for a two-dimensional subspace V of R3. Every vector in V is of the form c1v1 + c2v2. V is thus a plane through the origin.

Ch04_92

Theorem 4.13

Let {v1, …, vn} be a basis for a vector space V. Then each vector in V can be expressed uniquely as a linear combination of these vectors.

ProofLet v be a vector in V. Since {v1, …, vn} is a basis, we can express v as a linear combination of these vectors. Suppose we can write

nnnn bbaa vvvvvv 1111 and Then

givingnnnn bbaa vvvv 1111

0vv nnn baba )()( 111 Since {v1, …, vn} is a basis, the vectors v1, …, vn are linearly independent. Thus (a1 – b1) = 0, …, (an – bn) = 0, implying that a1 = b1, …, an = bn. There is thus only one way of expressing v as a linear combination of the basis.

Ch04_93

Theorem 4.14

Let V be a vector space of dimension n.

(a) If S = {v1, …, vn} is a set of n linearly independent vectors in V, then S is a basis for V.

(b) If S = {v1, …, vn} is a set of n vectors V that spans V, then S is a basis for V.

Let V be a vector space, S = {v1, …, vn} is a set of vectors in V.

(a) dim(V) = |S|.

(b) S is a linearly independent set.

(c) S spans V.S is a basis of V.

Ch04_94

Example 4Prove that the set B={(1, 3, -1), (2, 1, 0), (4, 2, 1)} is a basis for R3.

SolutionSince dim(R3)=|B|=3. It suffices to show that this set is linearly independent or it spans R3.Let us check for linear independence. Suppose

)0 ,0 ,0()1 ,2 ,4()0 ,1 ,2()1 ,3 ,1( 321 ccc

This identity leads to the system of equations

This system has the unique solution c1 = 0, c2 = 0, c3 = 0. Thus the vectors are linearly independent.The set {(1, 3, -1), (2, 1, 0), (4, 2, 1)} is therefore a basis for R3.

0023042

31

321

321

cccccccc

Ch04_95

Theorem 4.15

Let V be a vector space of dimension n. Let {v1, …, vm} be a set of m linearly independent vectors in V, where m < n.

Then there exist vectors vm+1, …, vn such that

{v1, …, vm, vm+1, …, vn } is a basis of V.

Ch04_96

Example 5State (with a brief explanation) whether the following statements

are true or false.

(a) The vectors (1, 2), (1, 3), (5, 2) are linearly dependent in R2.

(b) The vectors (1, 0, 0), (0, 2, 0), (1, 2, 0) span R3.

(c) {(1, 0, 2), (0, 1, -3)} is a basis for the subspace of R3 consisting of vectors of the form (a, b, 2a 3b).

(d) Any set of two vectors can be used to generate a two-dimensional subspace of R3.

Solution

(a) True: The dimension of R2 is two. Thus any three vectors are linearly dependent.

(b) False: The three vectors are linearly dependent. Thus they cannot span a three-dimensional space.

Ch04_97

(c) True: The vectors span the subspace since

(a, b, 2a, -3b) = a(1, 0, 2) + b(0, 1, -3)

The vectors are also linearly independent since they are not colinear.

(d) False: The two vectors must be linearly independent.

Homework

Exercise set 4.4 pages 226-227:5, 7, 11, 15, 18, 20, 21, 31.

Ch04_98