vector_space.pdf

8/12/2019 Vector_Space.pdf

1/16

LINEAR ALGEBRA

W W L CHEN

c W W L Chen, 1994, 2005.This chapter is available free to all individuals, on the understanding that it is not to be used for financial gain,

and may be downloaded and/or photocopied, with or without permission from the author.

However, this document may not be kept on any information storage and retrieval system without permission

from the author, unless such system is not accessible to any individuals other than its owners.

Chapter 5

INTRODUCTION TO

VECTOR SPACES

5.1. Real Vector Spaces

Before we give any formal definition of a vector space, we shall consider a few concrete examples of suchan abstract object. We first study two examples from the theory of vectors which we first discussed inChapter 4.

Example 5.1.1. Consider the set R2 of all vectors of the form u = (u1, u2), where u1, u2 R. Considervector addition and also multiplication of vectors by real numbers. It is easy to check that we have thefollowing properties:

(1.1) For everyu, v R2, we have u + v R2.(1.2) For everyu, v, w R2, we have u + (v + w) = (u + v) + w.

(1.3) For everyu R2, we have u + 0= 0 + u= u.(1.4) For everyu R2, we have u + (u) =0.(1.5) For everyu, v R2, we have u + v= v + u.(2.1) For everyc Rand u R2, we have cu R2.(2.2) For everyc Rand u, v R2, we have c(u + v) =cu + cv.(2.3) For everya, b R and u R2, we have (a + b)u= au + bu.(2.4) For everya, b R and u R2, we have (ab)u= a(bu).(2.5) For everyu R2, we have 1u= u.

Example 5.1.2. Consider the set R3 of all vectors of the form u = (u1, u2, u3), where u1, u2, u3 R.Consider vector addition and also multiplication of vectors by real numbers. It is easy to check that wehave properties analogous to (1.1)(1.5) and (2.1)(2.5) in the previous example, with reference to R2

being replaced by R3.

We next turn to an example from the theory of matrices which we first discussed in Chapter 2.

Chapter 5 : Introduction to Vector Spaces page 1 of 16


2/16

Linear Algebra c W W L Chen, 1994, 2005

Example 5.1.3. Consider the set M2,2(R) of all 2 2 matrices with entries in R. Consider matrixaddition and also multiplication of matrices by real numbers. Denote byO the 2 2 null matrix. It iseasy to check that we have the following properties:

(1.1) For everyP, Q M2,2(R), we have P+ Q M2,2(R).(1.2) For everyP ,Q ,R M2,2(R), we have P+ (Q + R) = (P+ Q) + R.(1.3) For everyP M2,2(R), we have P+ O= O + P =P.(1.4) For everyP M2,2(R), we have P+ (P) =O.(1.5) For everyP, Q M2,2(R), we have P+ Q= Q + P.(2.1) For everyc Rand P M2,2(R), we have cP M2,2(R).(2.2) For everyc Rand P , Q M2,2(R), we have c(P+ Q) =cP+ cQ.(2.3) For everya, b R and P M2,2(R), we have (a + b)P =aP+ bP.(2.4) For everya, b R and P M2,2(R), we have (ab)P =a(bP).(2.5) For everyP M2,2(R), we have 1P =P.

We also turn to an example from the theory of functions.

Example 5.1.4. Consider the set A of all functions of the form f : R R. For any two functionsf, g A, define the function f+g : R R by writing (f+g)(x) = f(x) + g(x) for every x R. Forevery functionf A and every numberc R, define the functioncf : R R by writing (cf)(x) =cf(x)for every x R. Denote by : R R the function where (x) = 0 for every x R. Then it is easy tocheck that we have the following properties:

(1.1) For everyf, g A, we have f+ g A.(1.2) For everyf , g , h A, we have f+ (g+ h) = (f+ g) + h.(1.3) For everyf A, we have f+ = + f=f.(1.4) For everyf A, we have f+ (f) =.(1.5) For everyf, g A, we have f+ g = g + f.(2.1) For everyc Rand f A, we have cf A.

(2.2) For everyc Rand f , g A, we have c(f+ g) =cf+ cg.(2.3) For everya, b R and f A, we have (a + b)f=af+ bf.(2.4) For everya, b R and f A, we have (ab)f=a(bf).(2.5) For everyf A, we have 1f=f.

There are many more examples of sets where properties analogous to (1.1)(1.5) and (2.1)(2.5) inthe four examples above hold. This apparent similarity leads us to consider an abstract object whichwill incorporate all these individual cases as examples. We say that these examples are all vector spacesover R.

Definition.A vector space V over R, or a real vector space V, is a set of objects, known as vectors,together with vector addition + and multiplication of vectors by element ofR, and satisfying the following

properties:(VA1) For every u, v V, we have u + v V.(VA2) For every u, v, w V, we have u + (v + w) = (u + v) + w.(VA3) There exists an element0 Vsuch that for every u V, we have u + 0= 0 + u= u.(VA4) For every u V, there existsu V such that u + (u) =0.(VA5) For every u, v V, we have u + v= v + u.(SM1) For everyc R and u V, we have cu V.(SM2) For everyc R and u, v V, we have c(u + v) =cu + cv.(SM3) For everya, b R and u V, we have (a + b)u= au + bu.(SM4) For everya, b R and u V, we have (ab)u= a(bu).(SM5) For everyu V, we have 1u= u.

Remark. The elements a,b, c R discussed in (SM1)(SM5) are known as scalars. Multiplication ofvectors by elements ofR is sometimes known as scalar multiplication.



3/16


Example 5.1.5. Let n N. Consider the set Rn of all vectors of the form u = (u1, . . . , un), whereu1, . . . , un R. For any two vectors u = (u1, . . . , un) and v = (v1, . . . , vn) in R

n and any number c R,write

u + v= (u1+ v1, . . . , un+ vn) and cu= (cu1, . . . , c un).

To check (VA1), simply note that u1+v1, . . . , un+vn R. To check (VA2), note that ifw= (w1, . . . , wn),then

u + (v + w) = (u1, . . . , un) + (v1+ w1, . . . , vn+ wn) = (u1+ (v1+ w1), . . . , un+ (vn+ wn))

= ((u1+ v1) + w1, . . . , (un+ vn) + wn) = (u1+ v1, . . . , un+ vn) + (w1, . . . , wn)

= (u + v) + w.

If we take 0 to be the zero vector (0, . . . , 0), then u+ 0 = 0+ u = u, giving (VA3). Next, writingu= (u1, . . . , un), we have u + (u) =0, giving (VA4). To check (VA5), note that

u + v= (u1+ v1, . . . , un+ vn) = (v1+ u1, . . . , vn+ un) =v + u.

To check (SM1), simply note that cu1, . . . , c un R. To check (SM2), note that

c(u + v) =c(u1+ v1, . . . , un+ vn) = (c(u1+ v1), . . . , c(un+ vn))

= (cu1+ cv1, . . . , c un+ cvn) = (cu1, . . . , c un) + (cv1, . . . , c vn) =cu + cv.

To check (SM3), note that

(a + b)u= ((a + b)u1, . . . , (a + b)un) = (au1+ bu1, . . . , a un+ bun)

= (au1, . . . , a un) + (bu1, . . . , b un) =au + bu.


(ab)u= ((ab)u1, . . . , (ab)un) = (a(bu1), . . . , a(bun)) =a(bu1, . . . , b un) =a(bu).

Finally, to check (SM5), note that

1u= (1u1, . . . , 1un) = (u1, . . . , un) =u.

It follows that Rn is a vector space over R. This is known as the n-dimensional euclidean space.

Example 5.1.6. Let k N. Consider the set Pk of all polynomials of the form

p(x) =p0+ p1x + . . . +pkxk, where p0, p1, . . . , pk R.

In other words, Pk is the set of all polynomials of degree at most k and with coefficients in R. For anytwo polynomialsp(x) =p0 +p1x + . . . +pkx

k andq(x) =q0 + q1x + . . . + qkxk inPk and for any number

c R, write

p(x) + q(x) = (p0+ q0) + (p1+ q1)x + . . . + (pk+ qk)xk and cp(x) =cp0+ cp1x + . . . + cpkx

k.

To check (VA1), simply note that p0 +q0, . . . , pk +qk R. To check (VA2), note that if we writer(x) =r0+ r1x + . . . + rkx

k, then we have

p(x) + (q(x) + r(x)) = (p0+ p1x + . . . +pkxk) + ((q0+ r0) + (q1+ r1)x + . . . + (qk+ rk)x

k)

= (p0+ (q0+ r0)) + (p1+ (q1+ r1))x + . . . + (pk+ (qk+ rk))xk

= ((p0+ q0) + r0) + ((p1+ q1) + r1)x + . . . + ((pk+ qk) + rk)xk

= ((p0+ q0) + (p1+ q1)x + . . . + (pk+ qk)xk) + (r0+ r1x + . . . + rkxk)= (p(x) + q(x)) + r(x).



4/16


If we take 0 to be the zero polynomial 0+0x + . . . + 0xk, thenp(x) + 0= 0 +p(x) =p(x), giving (VA3).Next, writing p(x) = p0 p1x . . . pkx

k, we have p(x) + (p(x)) = 0, giving (VA4). To check(VA5), note that

p(x) + q(x) = (p0+ q0) + (p1+ q1)x + . . . + (pk+ qk)xk

= (q0+ p0) + (q1+ p1)x + . . . + (qk+ pk)xk =q(x) + p(x).

To check (SM1), simply note that cp0, . . . , c pk R. To check (SM2), note that

c(p(x) + q(x)) =c((p0+ q0) + (p1+ q1)x + . . . + (pk+ qk)xk)

=c(p0+ q0) + c(p1+ q1)x + . . . + c(pk+ qk)xk

= (cp0+ cq0) + (cp1+ cq1)x + . . . + (cpk+ cqk)xk

= (cp0+ cp1x + . . . + cpkxk) + (cq0+ cq1x + . . . + cqkx

k)

=cp(x) + cq(x).


(a + b)p(x) = (a + b)p0+ (a + b)p1x + . . . + (a + b)pkxk

= (ap0+ bp0) + (ap1+ bp1)x + . . . + (apk+ bpk)xk

= (ap0+ ap1x + . . . + apkxk) + (bp0+ bp1x + . . . + bpkx

k)

=ap(x) + bp(x).


(ab)p(x) = (ab)p0+ (ab)p1x + . . . + (ab)pkxk =a(bp0) + a(bp1)x + . . . + a(bpk)x

k

=a(bp0+ bp1x + . . . + bpkxk) =a(bp(x)).

Finally, to check (SM5), note that

1p(x) = 1p0+ 1p1x + . . . + 1pkxk =p0+ p1x + . . . +pkx

k =p(x).

It follows that Pk is a vector space over R. Note also that the vectors are the polynomials.

There are a few simple properties of vector spaces that we can deduce easily from the definition.

PROPOSITION 5A.Suppose thatV is a vector space overR, and thatu V andc R.(a) We have0u= 0.(b) We havec0= 0.(c) We have(1)u= u.

(d) Ifcu= 0, thenc= 0 oru= 0.

Proof.(a) By (SM1), we have 0u V. Hence

0u + 0u= (0 + 0)u (by (SM3)),

= 0u (since 0 R).

It follows that

0u= 0u + 0 (by (VA3)),

= 0u + (0u + ((0u))) (by (VA4)),

= (0u + 0u) + ((0u)) (by (VA2)),

= 0u + ((0u)) (from above),=0 (by (VA4)).



5/16


(b) By (SM1), we have c0 V. Hence

c0 + c0= c(0 + 0) (by (SM2)),

=c0 (by (VA3)).

It follows that

c0= c0 + 0 (by (VA3)),

=c0 + (c0 + ((c0))) (by (VA4)),

= (c0 + c0) + ((c0)) (by (VA2)),

=c0 + ((c0)) (from above),

=0 (by (VA4)).

(c) We have

(1)u= (1)u + 0 (by (VA3)),

= (1)u + (u + (u)) (by (VA4)),

= ((1)u + u) + (u) (by (VA2)),

= ((1)u + 1u) + (u) (by (SM5)),

= ((1) + 1)u + (u) (by (SM3)),

= 0u + (u) (since 1 R),

=0 + (u) (from (a)),

= u (by (VA3)).

(d) Suppose that cu= 0 and c = 0. Then c1 R and

u= 1u (by (SM5)),

= (c1c)u (since c R \ {0}),

=c1(cu) (by (SM4)),

=c10 (assumption),

=0 (from (b)),

as required.

5.2. Subspaces

Example 5.2.1. Consider the vector space R2 of all points (x, y), where x, y R. Let L be a linethrough the origin 0 = (0, 0). Suppose that L is represented by the equation x+ y = 0; in otherwords,

L= {(x, y) R2 :x + y = 0}.

Note first of all that 0 = (0, 0) L, so that (VA3) and (VA4) clearly hold in L. Also (VA2) and (VA5)clearly hold in L. To check (VA1), note that if (x, y), (u, v) L, then x + y = 0 and u + v = 0, sothat (x + u) + (y+ v) = 0, whence (x, y) + (u, v) = (x + u, y+ v) L. Next, note that (SM2)(SM5)clearly hold in L. To check (SM1), note that if (x, y) L, then x + y = 0, so that (cx) + (cy) = 0,whence c(x, y) = (cx, cy) L. It follows that L forms a vector space over R. In fact, we have shownthat every line in R2 through the origin is a vector space over R.



6/16


Definition.Suppose thatVis a vector space over R, and thatWis a subset ofV. Then we say thatWis a subspace ofV ifW forms a vector space over Runder the vector addition and scalar multiplicationdefined in V.

Example 5.2.2. We have just shown in Example 5.2.1 that every line in R2 through the origin is asubspace ofR2. On the other hand, if we work through the example again, then it is clear that we havereally only checked conditions (VA1) and (SM1) for L, and that 0= (0, 0) L.

PROPOSITION 5B.Suppose that V is a vector space overR, and thatW is a non-empty subset ofV. ThenW is a subspace ofV if the following conditions are satisfied:(SP1) For everyu, v W, we haveu + v W.(SP2) For everyc R and u W, we havecu W.

Proof. To show that W is a vector space over R, it remains to check that W satisfies (VA2)(VA5)and (SM2)(SM5). To check (VA3) and (VA4) for W, it clearly suffices to check that 0 W. Since Wis non-empty, there exists u W. Then it follows from (SP2) and Proposition 5A(a) that 0 = 0u W.

The remaining conditions (VA2), (VA5) and (SM2)(SM5) hold for all vectors in V, and hence also forall vectors in W.

Example 5.2.3.Consider the vector space R3 of all points (x,y,z), where x, y,z R. LetPbe a planethrough the origin 0 = (0, 0, 0). Suppose that P is represented by the equation x+y+ z = 0; inother words,

P = {(x,y,z) R2 :x + y + z = 0}.

To check (SP1), note that if (x,y,z), (u,v ,w) P, then x+y+ z = 0 and u+v+ w = 0, sothat (x+u) + (y+ v) + (z+ w) = 0, whence (x,y,z) + (u,v ,w) = (x+u, y+ v, z+ w) P. Tocheck (SP2), note that if (x,y,z) P, thenx + y + z = 0, so that (cx) + (cy) + (cz) = 0, whence

c(x,y,z) = (cx, cy, cz) P. It follows that P is a subspace ofR3

. Next, let L be a line through theorigin 0 = (0, 0, 0). Suppose that ( , , ) R3 is a non-zero point on L. Then we can write

L= {t(, ,) :t R}.

Suppose that u = t( , , ) L and v = s( , , ) L, and that c R. Then u+ v = t( , , ) +s( , , ) = (t + s)( , , ) L, giving (SP1). Also, cu= c(t(, ,)) = (ct)( , , ) L, giving (SP2).It follows thatL is a subspace ofR3. Finally, it is not difficult to see that both {0} and R3 are subspacesofR3.

Example 5.2.4. Note that R2 is not a subspace ofR3. First of all, R2 is not a subset ofR3. Note alsothat vector addition and scalar multiplication are different in R2 and R3.

Example 5.2.5. Suppose that A is an m n matrix and 0 is the m 1 zero column matrix. Considerthe system Ax= 0 ofmhomogeneous linear equations in the n unknowns x1, . . . , xn, where

x=

x1...

xn

is interpreted as an element of the vector space Rn, with usual vector addition and scalar multiplication.Let Sdenote the set of all solutions of the system. Suppose that x, y Sand c R. Then A(x + y) =Ax +Ay= 0 + 0= 0, giving (SP1). Also,A(cx) =c(Ax) =c0= 0, giving (SP2). It follows that S isa subspace ofRn. To summarize, the space of solutions of a system ofmhomogeneous linear equationsin n unknowns is a subspace ofRn.

Example 5.2.6.As a special case of Example 5.2.5, note that if we take two non-parallel planes in R3

through the origin 0 = (0, 0, 0), then the intersection of these two planes is clearly a line through the



7/16


origin. However, each plane is a homogeneous equation in the three unknowns x,y, z R. It follows thatthe intersection of the two planes is the collection of all solutions (x,y,z) R3 of the system formed bythe two homogeneous equations in the three unknowns x, y,z representing these two planes. We have

already shown in Example 5.2.3 that the line representing all these solutions is a subspace of R3

.

Example 5.2.7.We showed in Example 5.1.3 that the set M2,2(R) of all 2 2 matrices with entries inR forms a vector space over R. Consider the subset

W =

a11 a12a21 0

:a11, a12, a21 R

ofM2,2(R). Since

a11 a12a21 0

+

b11 b12b21 0

=

a11+ b11 a12+ b12a21+ b21 0

and c

a11 a12a21 0

=

ca11 ca12ca21 0

,

it follows that (SP1) and (SP2) are satisfied. Hence W is a subspace ofM2,2(R).

Example 5.2.8.We showed in Example 5.1.4 that the set A of all functions of the formf : R R formsa vector space over R. Let C0 denote the set of all functions of the form f : R Rwhich are continuousat x = 2, and let C1 denote the set of all functions of the form f : R R which are differentiable atx= 2. Then it follows from the arithmetic of limits and the arithmetic of derivatives that C0 andC1 areboth subspaces ofA. Furthermore, C1 is a subspace ofC0 (why?). On the other hand, let k N. Recallfrom Example 5.1.6 the vector space Pk of all polynomials of the form

p(x) =p0+ p1x + . . . +pkxk, where p0, p1, . . . , pk R.

In other words, Pk is the set of all polynomials of degree at most k and with coefficients in R. ClearlyPk

is a subspace ofC1

.

5.3. Linear Combination

In this section and the next two, we shall study ways of describing the vectors in a vector space V. Ourultimate goal is to be able to determine a subset B of vectors in Vand describe every element ofV interms of elements ofB in a unique way. The first step in this direction is summarized below.

Definition.Suppose that v1, . . . , vr are vectors in a vector space V over R. By a linear combinationof the vectors v1, . . . , vr, we mean an expression of the type

c1v1+ . . . + crvr,

where c1, . . . , cr R.

Example 5.3.1. In R2, every vector (x, y) is a linear combination of the two vectors i = (1, 0) andj= (0, 1), for clearly (x, y) =xi + yj.

Example 5.3.2. In R3, every vector (x,y,z) is a linear combination of the three vectors i = (1, 0, 0),j= (0, 1, 0) and k= (0, 0, 1), for clearly (x,y,z) =xi + yj + zk.

Example 5.3.3. In R4, the vector (1, 4, 2, 6) is a linear combination of the two vectors (1, 2, 0, 4) and(1, 1, 1, 3), for we have (1, 4, 2, 6) = 3(1, 2, 0, 4) 2(1, 1, 1, 3). On the other hand, the vector (2, 6, 0, 9)is not a linear combination of the two vectors (1, 2, 0, 4) and (1, 1, 1, 3), for

(2, 6, 0, 9) =c1(1, 2, 0, 4) + c2(1, 1, 1, 3)



8/16


would lead to the system of four equations

c1+ c2= 2,

2c1+ c2= 6,c2= 0,

4c1+ 3c2= 9.

It is easily checked that this system has no solutions.

Example 5.3.4. In the vector space A of all functions of the form f : R R described in Example5.1.4, the function cos 2x is a linear combination of the three functions cos2 x, cosh2 x and sinh2 x. It isnot too difficult to check that

cos2x= 2cos2 x + sinh2 x cosh2 x,

noting that cos 2x= 2 cos2 x 1 and cosh2 x sinh2 x= 1.

We observe that in Example 5.3.1, every vector in R2 is a linear combination of the two vectors i andj. Similarly, in Example 5.3.2, every vector in R3 is a linear combination of the three vectors i, j and k.On the other hand, we observe that in Example 5.3.3, not every vector in R4 is a linear combination ofthe two vectors (1, 2, 0, 4) and (1, 1, 1, 3).

Let us therefore investigate the collection of all vectors in a vector space that can be represented aslinear combinations of a given set of vectors in V.

Definition.Suppose that v1, . . . , vr are vectors in a vector space V over R. The set

span{v1, . . . , vr} = {c1v1+ . . . + crvr :c1, . . . , cr R}

is called the span of the vectors v1, . . . , vr. We also say that the vectors v1, . . . , vr span V if

span{v1, . . . , vr} =V;

in other words, if every vector in Vcan be expressed as a linear combination of the vectors v1, . . . , vr.

Example 5.3.5. The two vectors i= (1, 0) and j= (0, 1) span R2.

Example 5.3.6. The three vectors i= (1, 0, 0), j= (0, 1, 0) and k= (0, 0, 1) span R3.

Example 5.3.7. The two vectors (1, 2, 0, 4) and (1, 1, 1, 3) do not span R4.

PROPOSITION 5C.Suppose thatv1, . . . , vr are vectors in a vector spaceV overR.(a) Thenspan{v1, . . . , vr} is a subspace ofV.(b) Suppose further thatW is a subspace ofV and v1, . . . , vr W. Thenspan{v1, . . . , vr} W.

Proof.(a) Suppose that u, w span{v1, . . . , vr} and c R. There exist a1, . . . , ar, b1, . . . , br Rsuchthat

u= a1v1+ . . . + arvr and w= b1v1+ . . . + brvr.

Then

u + w= (a1v1+ . . . + arvr) + (b1v1+ . . . + brvr)= (a1+ b1)v1+ . . . + (ar+ br)vr span{v1, . . . , vr}



9/16


and

cu= c(a1v1+ . . . + arvr) = (ca1)v1+ . . . + (car)vr span{v1, . . . , vr}.

It follows from Proposition 5B that span{v1, . . . , vr} is a subspace ofV.

(b) Suppose that c1, . . . , cr R and u = c1v1+ . . .+ crvr span{v1, . . . , vr}. If v1, . . . , vr W,then it follows from (SM1) for W that c1v1, . . . , crvr W. It then follows from (VA1) for W thatu= c1v1+ . . . + crvr W.

Example 5.3.8. In R2, any non-zero vector v spans the subspace {cv : c R}. This is clearly a linethrough the origin. Also, try to draw a picture to convince yourself that any two non-zero vectors thatare not on the same line span R2.

Example 5.3.9. In R3, try to draw pictures to convince yourself that any non-zero vector spans a

subspace which is a line through the origin; any two non-zero vectors that are not on the same line spana subspace which is a plane through the origin; and any three non-zero vectors that do not lie on thesame plane span R3.

5.4. Linear Independence

We first study two simple examples.

Example 5.4.1. Consider the three vectors v1 = (1, 2, 3), v2 = (3, 2, 1) and v3 = (3, 3, 3) in R3. Then

span{v1, v2, v3} = {c1(1, 2, 3) + c2(3, 2, 1) + c3(3, 3, 3) :c1, c2, c3 R}

= {(c1+ 3c2+ 3c3, 2c1+ 2c2+ 3c3, 3c1+ c2+ 3c3) :c1, c2, c3 R}.

Write (x,y,z) = (c1+ 3c2+ 3c3, 2c1+ 2c2+ 3c3, 3c1+ c2+ 3c3). Then it is not difficult to see that

xy

z

=

1 3 32 2 3

3 1 3

c1c2

c3

,

and so (do not worry if you cannot understand why we take this next step)

( 1 2 1 )

x

y

z

= ( 1 2 1 )

1 3 32 2 3

3 1 3

c1c2c3

= ( 0 0 0 )

c1c2c3

= ( 0 ) ,

so that x 2y+ z = 0. It follows that span{v1, v2, v3} is a plane through the origin and not R3. Note,

in fact, that 3v1+ 3v2 4v3 = 0. Note also that

det

1 3 32 2 3

3 1 3

= 0.

Example 5.4.2.Consider the three vectors v1 = (1, 1, 0), v2= (5, 1, 3) and v3 = (2, 7, 4) in R3. Then

span{v1, v2, v3} = {c1(1, 1, 0) + c2(5, 1, 3) + c3(2, 7, 4) :c1, c2, c3 R}= {(c1+ 5c2+ 2c3, c1+ c2+ 7c3, 3c2+ 4c3) :c1, c2, c3 R}.



10/16


Write (x,y,z) = (c1+ 5c2+ 2c3, c1+ c2+ 7c3, 3c2+ 4c3). Then it is not difficult to see that

x

yz =

1 5 2

1 1 70 3 4

c1

c2c3 ,

so that 25 26 334 4 5

3 3 4

xy

z

=

25 26 334 4 5

3 3 4

1 5 21 1 7

0 3 4

c1c2

c3

=

1 0 00 1 0

0 0 1

c1c2

c3

=

c1c2

c3

.

It follows that for every (x,y,z) R3, we can find c1, c2, c3 Rsuch that (x,y,z) =c1v1+ c2v2+ c3v3.

Hence span{v1, v2, v3} =R

3

. Note that

det

1 5 21 1 7

0 3 4

= 0,

and that the only solution for

(0, 0, 0) =c1v1+ c2v2+ c3v3

is c1 = c2 = c3 = 0.

Definition.Suppose that v1, . . . , vr are vectors in a vector space V over R.(LD) We say thatv

1, . . . , v

rare linearly dependent if there exist c

1, . . . , c

r R, not all zero, such that

c1v1+ . . . + crvr =0.(LI) We say thatv1, . . . , vr are linearly independent if they are not linearly dependent; in other words,

if the only solution ofc1v1+ . . . + crvr =0 in c1, . . . , cr R is given by c1 = . . .= cr = 0.

Example 5.4.3. Let us return to Example 5.4.1 and consider again the three vectors v1 = (1, 2, 3),v2 = (3, 2, 1) and v3 = (3, 3, 3) in R

3. Consider the equation c1v1 + c2v2 + c3v3 = 0. This can berewritten in matrix form as

1 3 32 2 33 1 3

c1c2

c3

=

00

0

.

Since

det

1 3 32 2 3

3 1 3

= 0,

the system has non-trivial solutions; for example, (c1, c2, c3) = (3, 3, 4), so that 3v1+ 3v2 4v3 =0.Hence v1, v2, v3 are linearly dependent.

Example 5.4.4. Let us return to Example 5.4.2 and consider again the three vectors v1 = (1, 1, 0),v2 = (5, 1, 3) and v3 = (2, 7, 4) in R3. Consider the equation c1v1+ c2v2 + c3v3 = 0. This can berewritten in matrix form as

1 5 21 1 70 3 4

c1c2c3

= 000

.



11/16


Since

det1 5 2

1 1 70 3 4 = 0,

the only solution is c1 = c2= c3 = 0. Hence v1, v2, v3 are linearly independent.

Example 5.4.5.In the vector space A of all functions of the form f : R R described in Example 5.1.4,the functionsx, x2 and sin x are linearly independent. To see this, note that for everyc1, c2, c3 R, thelinear combination c1x + c2x

2 + c3sin x is never identically zero unless c1 = c2 = c3 = 0.

Example 5.4.6. In Rn, the vectors e1, . . . , en, where

ej = (0, . . . , 0

j1, 1, 0, . . . , 0

nj) for every j = 1, . . . , n ,

are linearly independent (why?).

We observe in Examples 5.4.35.4.4 that the determination of whether a collection of vectors in R3

are linearly dependent is based on whether a system of homogeneous linear equations has non-trivialsolutions. The same idea can be used to prove the following result concerning Rn.

PROPOSITION 5D. Suppose that v1, . . . , vr are vectors in the vector space Rn. If r > n, then

v1, . . . , vr are linearly dependent.

Proof.For every j = 1, . . . , r, writevj = (a1j , . . . , anj).

Then the equation c1v1+ . . . + crvr =0 can be rewritten in matrix form as

a11 . . . a1r... ...

an1 . . . anr

c1...

cr

=

0...

0

.

Ifr > n, then there are more variables than equations. It follows that there must be non-trivial solutionsc1, . . . , cr R. Hence v1, . . . , vr are linearly dependent.

Remarks.(1) Consider two vectors v1 = (a11, a21) and v2 = (a12, a22) in R2. To study linear indepen-

dence, we consider the equation c1v1+ c2v2 = 0, which can be written in matrix form as

a11 a12a21 a22

c1c2

=

00

.

The vectors v1 and v2 are linearly independent precisely when

det

a11 a12a21 a22

= 0.

This can be interpreted geometrically in the following way: The area of the parallelogram formed bythe two vectors v1 and v2 is in fact equal to the absolute value of the determinant of the matrix formedwith v1 and v2 as the columns; in other words,

det a11 a12

a21 a22

.



12/16


It follows that the two vectors are linearly dependent precisely when the parallelogram has zero area;in other words, when the two vectors lie on the same line. On the other hand, if the parallelogram haspositive area, then the two vectors are linearly independent.

(2) Consider three vectors v1 = (a11, a21, a31), v2 = (a12, a22, a32), and v3 = (a13, a23, a33) in R3. To

study linear independence, we consider the equation c1v1+ c2v2+ c3v3 = 0, which can be written inmatrix form as

a11 a12 a13a21 a22 a23a31 a32 a33

c1c2

c3

=

00

0

.

The vectors v1, v2 and v3 are linearly independent precisely when

det

a11 a12 a13a21 a22 a23

a31 a32 a33

= 0.

This can be interpreted geometrically in the following way: The volume of the parallelepiped formed bythe three vectors v1, v2 and v3 is in fact equal to the absolute value of the determinant of the matrixformed with v1, v2 and v3 as the columns; in other words,

det

a11 a12 a13a21 a22 a23

a31 a32 a33

.

It follows that the three vectors are linearly dependent precisely when the parallelepiped has zero volume;in other words, when the three vectors lie on the same plane. On the other hand, if the parallelepipedhas positive volume, then the three vectors are linearly independent.

(3) What is the geometric interpretation of two linearly independent vectors in R3? Well, note thatif v1 and v2 are non-zero and linearly dependent, then there exist c1, c2 R, not both zero, such thatc1v1 + c2v2 = 0. This forces the two vectors to be multiples of each other, so that they lie on thesame line, whence the parallelogram they form has zero area. It follows that if two vectors in R3 forma parallelogram with positive area, then they are linearly independent.

5.5. Basis and Dimension

In this section, we complete the task of describing uniquely every element of a vector space V in termsof the elements of a suitable subset B. To motivate the ideas, we first consider an example.

Example 5.5.1.Let us consider the three vectors v1 = (1, 1, 0),v2 = (5, 1, 3) andv3 = (2, 7, 4) in R3,

as in Examples 5.4.2 and 5.4.4. We have already shown that span{v1, v2, v3} = R3, and that the vectorsv1, v2, v3 are linearly independent. Furthermore, we have shown that for every u = (x,y,z) R

3, wecan write u= c1v1+ c2v2+ c3v3, where c1, c2, c3 Rare determined uniquely by

c1c2c3

=

25 26 334 4 5

3 3 4

xy

z

.

Definition.Suppose that v1, . . . , vr are vectors in a vector space V over R. We say that {v1, . . . , vr}is a basis for V if the following two conditions are satisfied:

(B1) We have span{v1, . . . , vr} =V.(B2) The vectorsv1, . . . , vr are linearly independent.



13/16


Example 5.5.2. Consider two vectors v1 = (a11, a21) and v2 = (a12, a22) in R2. Suppose that

deta11 a12

a21 a22 = 0;in other words, suppose that the parallelogram formed by the two vectors has non-zero area. Then itfollows from Remark (1) in Section 5.4 that v1 and v2 are linearly independent. Furthermore, for everyu = (x, y) R2, there exist c1, c2 R such that u= c1v1+ c2v2. Indeed, c1 and c2 are determined asthe unique solution of the system

a11 a12a21 a22

c1c2

=

x

y

.

Hence span{v1, v2} = R2. It follows that {v1, v2} is a basis for R

2.

Example 5.5.3.Consider three vectors of the type v1 = (a11, a21, a31), v2 = (a12, a22, a32) and v3 =

(a13, a23, a33) in R3. Suppose that

det

a11 a12 a13a21 a22 a23

a31 a32 a33

= 0;

in other words, suppose that the parallelepiped formed by the three vectors has non-zero volume. Thenit follows from Remark (2) in Section 5.4 that v1, v2 andv3 are linearly independent. Furthermore, forevery u= (x,y,z) R3, there exist c1, c2, c3 R such that u= c1v1+c2v2+ c3v3. Indeed, c1, c2 andc3 are determined as the unique solution of the system

a11 a12 a13a

21 a

22 a

23a31 a32 a33

c1c

2c3 =

x

y

z .

Hence span{v1, v2, v3} = R3. It follows that {v1, v2, v3} is a basis for R

3.

Example 5.5.4. In Rn, the vectors e1, . . . , en, where

ej = (0, . . . , 0 j1

, 1, 0, . . . , 0 nj

) for every j = 1, . . . , n ,

are linearly independent and span Rn. Hence {e1, . . . , en} is a basis for Rn. This is known as thestandard basis for Rn.

Example 5.5.5. In the vector space M2,2(R) of all 2 2 matrices with entries in R as discussed inExample 5.1.3, the set

1 00 0

,

0 10 0

,

0 01 0

,

0 00 1

is a basis.

Example 5.5.6. In the vector space Pk of polynomials of degree at most k and with coefficients in Ras discussed in Example 5.1.6, the set {1, x , x2, . . . , xk} is a basis.

PROPOSITION 5E. Suppose that{v1, . . . , vr} is a basis for a vector space V over R. Then everyelementu V can be expressed uniquely in the form

u= c1v1+ . . . + crvr, wherec1, . . . , cr R.



14/16


Proof. Since u V = span{v1, . . . , vr}, there exist c1, . . . , cr R such that u = c1v1+ . . .+ crvr.Suppose now that b1, . . . , br Rsuch that

c1v1+ . . . + crvr =b1v1+ . . . + brvr.

Then

(c1 b1)v1+ . . . + (cr br)vr = 0.

Since v1, . . . , vr are linearly independent, it follows that c1 b1 = . . . = cr br = 0. Hence c1, . . . , crare uniquely determined.

We have shown earlier that a vector space can have many bases. For example, any collection of threevectors not on the same plane is a basis for R3. In the following discussion, we attempt to find out someproperties of bases. However, we shall restrict our discussion to the following simple case.

Definition.A vector space V over R is said to be finite-dimensional if it has a basis containing onlyfinitely many elements.

Example 5.5.7. The vector spaces Rn, M2,2(R) and Pk that we have discussed earlier are all finite-dimensional.

Recall that in Rn, the standard basis has exactly n elements. On the other hand, it follows fromProposition 5D that any basis for Rn cannot contain more than n elements. However, can a basis forR

n contain fewer than n elements?

We shall answer this question by showing that all bases for a given vector space have the same numberof elements. As a first step, we establish the following generalization of Proposition 5D.

PROPOSITION 5F.Suppose that{v1, . . . , vn} is a basis for a vector spaceV overR. Suppose furtherthatr > n, and that the vectorsu1, . . . , ur V. Then the vectorsu1, . . . , ur are linearly dependent.

Proof.Since {v1, . . . , vn} is a basis for the vector space V, we can write

u1 = a11v1+ . . . + an1vn,

...

ur =a1rv1+ . . . + anrvn,

where aij R for every i = 1, . . . , nand j = 1, . . . , r. Let c1, . . . , cr R. Since v1, . . . , vn are linearly

independent, it follows that if

c1u1+ . . . + crur =c1(a11v1+ . . . + an1vn) + . . . + cr(a1rv1+ . . . + anrvn)

= (a11c1+ . . . + a1rcr)v1+ . . . + (an1c1+ . . . + anrcr)vn

=0,

thena11c1 + . . . + a1rcr =. . .= an1c1 + . . . + anrcr = 0; in other words, we have the homogeneous system

a11 . . . a1r... ...

an1 . . . anr

c1...

cr

=

0...

0

.

Ifr > n, then there are more variables than equations. It follows that there must be non-trivial solutionsc1, . . . , cr R. Hence u1, . . . , ur are linearly dependent.



15/16


PROPOSITION 5G.Suppose that V is a finite-dimensional vector space V over R. Then any twobases forVhave the same number of elements.

Proof. Note simply that by Proposition 5F, the vectors in the basis with more elements must belinearly dependent, and so cannot be a basis.

We are now in a position to make the following definition.

Definition. Suppose that V is a finite-dimensional vector space over R. Then we say that V is ofdimension n if a basis for V contains exactly n elements.

Example 5.5.8. The vector space Rn has dimension n.

Example 5.5.9. The vector space M2,2(R) of all 2 2 matrices with entries in R, as discussed inExample 5.1.3, has dimension 4.

Example 5.5.10. The vector space Pk of all polynomials of degree at most k and with coefficients inR, as discussed in Example 5.1.6, has dimension (k+ 1).

Example 5.5.11. Recall Example 5.2.5, where we showed that the set of solutions to a system of mhomogeneous linear equations inn unknowns is a subspace ofRn. Consider now the homogeneous system

1 3 5 1 51 4 7 3 21 5 9 5 90 3 6 2 1

x1x2x3x4x5

=

0000

.

The solutions can be described in the form

x= c1

12100

+ c2

130

51

,

wherec1, c2 R (the reader must check this). It can be checked that (1 , 2, 1, 0, 0) and (1, 3, 0, 5, 1)are linearly independent and so form a basis for the space of solutions of the system. It follows that thespace of solutions of the system has dimension 2.

Suppose that V is an n-dimensional vector space over R. Then any basis for Vconsists of exactly n

linearly independent vectors in V. Suppose now that we have a set ofn linearly independent vectors inV. Will this form a basis for V?

We have already answered this question in the affirmative in the cases when the vector space is R2 orR

3. To seek an answer to the general case, we first establish the following result.

PROPOSITION 5H.Suppose thatV is a finite-dimensional vector space overR. Then any finite setof linearly independent vectors inV can be expanded, if necessary, to a basis forV.

Proof.Let S = {v1, . . . , vk} be a finite set of linearly independent vectors in V. IfS spans V, thenthe proof is complete. IfSdoes not span V, then there exists vk+1 V that is not a linear combinationof the elements ofS. The set T = {v1, . . . , vk, vk+1}is a finite set of linearly independent vectors in V;for otherwise, there existc1, . . . , ck, ck+1, not all zero, such that

c1v1+ . . . + ckvk+ ck+1vk+1 = 0.



16/16


Ifck+1 = 0, then c1v1+ . . .+ckvk = 0, contradicting the assumption that S is a finite set of linearlyindependent vectors in V. Ifck+1 = 0, then

vk+1

=

c1

ck+1v

1 . . .

ck

ck+1vk

,

contradicting the assumption that vk+1 is not a linear combination of the elements ofS. We now studythe finite set Tof linearly independent vectors in V. IfT spansV, then the proof is complete. IfT doesnot span V, then we repeat the argument. Note that the number of vectors in a linearly independentexpansion ofScannot exceed the dimension ofV, in view of Proposition 5F. So eventually some linearlyindependent expansion ofSwill span V.

PROPOSITION 5J. Suppose that V is an n-dimensional vector space over R. Then any set of nlinearly independent vectors inV is a basis forV.

Proof.Let Sbe a set ofn linearly independent vectors in V. By Proposition 5H, Scan be expanded,if necessary, to a basis for V. By Proposition 5F, any expansion ofSwill result in a linearly dependent

set of vectors inV

. It follows thatS

is already a basis forV

. Example 5.5.12. Consider the three vectors v1 = (1, 2, 3), v2 = (3, 2, 1) and v3 = (3, 3, 3) in R

3, asin Examples 5.4.1 and 5.4.3. We showed that these three vectors are linearly dependent, and span theplane x 2y+ z = 0. Note that

v3 = 3

4v1+

3

4v2,

and that v1 and v2 are linearly independent. Consider now the vector v4 = (0, 0, 1). Note that v4 doesnot lie on the plane x 2y+ z = 0, so that {v1, v2, v4} form a linearly independent set. It follows that{v1, v2, v4} is a basis for R

3.

Problems for Chapter 5

1. Determine whether each of the following subsets ofR3 is a subspace ofR3:a) {(x,y,z) R3 :x = 0} b) {(x,y,z) R3 :x + y = 0}c) {(x,y,z) R3 :xz = 0} d) {(x,y,z) R3 :y 0}e) {(x,y,z) R3 :x = y = z}

2. For each of the following collections of vectors, determine whether the first vector is a linear com-bination of the remaining ones:

a) (1, 2, 3); (1, 0, 1), (2, 1, 0) in R3

b) x3 + 2x2 + 3x + 1; x3, x2 + 3x, x2 + 1 in P4c) (1, 3, 5, 7); (1, 0, 1, 0), (0, 1, 0, 1), (0, 0, 1, 1) in R4

3. For each of the following collections of vectors, determine whether the vectors are linearly indepen-dent:

a) (1, 2, 3), (1, 0, 1), (2, 1, 0) in R3 b) (1, 2), (3, 5), (1, 3) in R2

c) (2, 5, 3, 6), (1, 0, 0, 1), (4, 0, 9, 6) in R4 d) x2 + 1, x + 1, x2 + x in P3

4. Find the volume of the parallelepiped in R3 formed by the vectors (1, 2, 3), (1, 0, 1) and (3, 0, 2).

5. Let Sbe the set of all functions y that satisfy the differential equation

2d2y

dx2 3

dy

dx+ y= 0.

Show that Sis a subspace of the vector space A described in Example 5.1.4.

6. For each of the sets in Problem 1 which is a subspace ofR3, find a basis for the subspace, and thenextend it to a basis for R3.


vector_space.pdf

Documents