UNIT VECTOR SPACES

Structure 3.1 Introduction Objectives 3.2 What Are Vector Spaces? 3.3 Further Properties of a Vector Space 3.4 Subspaces 3.5 Linear Combination 3.6 Algebra of Subspaces Intersection Sum Direct Sum 3.7 Quotient Spaces Cosets The Quotient Space 3.8 Summary 3.9 Solutions/Answers

3.1 INTRODUCTION

In this unit we begin the study of vector spaces and their properties. The concepts that we will discuss here are very important, since they form the core of the rest of the course. In Unit 2 we studied R2and R3. We also defined the two operations of vector addition and scalar multiplication on them along with certain properties. This can be done in a more general setting. That is, we may start with any set V (in place of R2or R" and convert V into a vector space by introducing "addition" and "scalar multiplication" in such a way that they have all the basic properties which vector addition and scalar multiplication have in R2and R3.We will prove a number of results about the general vector space V. These results will be true for all vector spaces -no matter what the elements are. To illustrate the wide applicability of our results, we shall also give several examples of specific vector spaces. We shall also study subsets of a vector space which are vector spaces themselves. They are called subspaces. Finally, using subspaces, we will obtain new vector spaces from given oncs. Since this unit forms part of the backbone of the course, be sure that you understand each concept in it. Objectives After studying this unit, you should be able to define and recognise a vector space; give a wide variety of examples of vector spaces; determine whether a given subset of a vector space is a subspace or not; explain what the linear span of a subset of a vector space is; differentiate between the sum and the d~rectsum of subspaces; define and give examples of cosets and quotient.spaces.

3.2 WHAT ARE VECTOR SPACES?

You have already come across the algebraic structure called a field in Unit 1. We now build another algebraicstructure from a set, by defining on it the opehtions of addition and multiplication by elements of a field. This is a vector space. We give the definition of a vector space now. As you read through it you can keep in mind the example of the vector space R' over R (Unit 2). d F if it has two operations, namely addition (denoted by +) and multiplication of elements of V by elementsof F (denoted by .), such that the following properties hold :

VS1) + is a binary operation, i.e., u + v E V $L u, V-E V. VS~) + is associative. i.e., (U+ v) + w = u + (v + w) + u, v, w E V. VS3) V has an identity element with respect to +, i.e., 3O~VsuchthatOfv = v =v +O$Lv€V. VS4) Every element of V has an inverse with respect to + : For every u E V, 3 v E V such that u +v = 0. v is called the additive inverse of u, and is written as -u. VS5) + is commutative, i.e., u + v = v + u +u, v E V. VS6) . : F x V+V : .(a,v) = a.v is a well defined operation, i.e., $La€Fandv€V,a.v€V. VS7) +a E F and u, v E V, a. (u+v) = a.u + a.v. VS8) $L a,p E F and v E V, (a+P). v = a.v + P.v

VS9) +a,p E F and v E V, (a p). v = a.(P.v) VS10) 1.v = v, for all v E V. When V is a vector space over R, we also call it a real vector space. Similarly, if V is defined over C, it is also called a complex vector space.

The product of a E F and v E V, in the definition, is often denoted by av instead of a.v. Note that this product is a vector. This operation is called scalar multiplication, because the elements of F are called scalars. Elements of V are called vectors. Now that the additive inverse ot a vector is defined (in VS4), we can give another

Definition: If u,v belong to a vector space V, we define their difference u - v to be

For example, in R*we have (3.5) - (1,O) = (33) + (-1,O) = (2,5). After going through Unit 2 and the definition of a vector space it must be clear to you that R~ and R~,with vector addition and scalar multiplication, are vector spaces. We now give some more examples of vector spaces. Example 1: Show that R is a vector space over itself. Solution: '+' is associative and commutative in R. The additive identity is O and the additive inverse of x E R is -'x. The scalar multiplication is the ordinary multiplication in R, and satisfies the properties VS7-VS10. Example 2: For any positive integer n, show that the set R" = {(x,, x2, ...... ,xn) / xi E R} is a vector space over R, if we define vector addition For any field F, and scalar multiplication as : Fn= {(xl ...... x,)jx, E F}. Every element of is called an (xi. xZ...... xn) + (Y,,~2, ...... , yn) = (XI + ~1.~1 + yz, ...... , xn + yn), and n-tuple of elements of F. . a(xl,x?, ...... x,) = (ax,, ax2, ..... ax,), a E F. Solution: The properties VSl - VSlO are easily checked. Since '+' is associative and commutative in R, you can check that '+' is associative and commutative in Rn also. Further, the identity for addition is (0.0, ...... , O), because (x,,X7, ...... , X,) + (0, 0, ...... 0) - (x, + 0, X2 + 0%...... X,, + 0) = (x,,XI, ...., X,).

The additive inverse of (xi, ...... , x,) is (-x , , ...... AX"), Fora, p~R,(a+ p) (x,...... , x,,) = ((a+ P)x,...... (a+ p)~,) = (axI + Px,, ...... axn + ax,,) + (Px,...... px,) = CY(X,,...... XI,) + P(x[...... XIl) Define scalar multiplication as follows: Vector Spaces

For a E R, f E S, let a f be the function given by

(af) (x) - :.f(x) %' x E R. Show that S is a real vector space.

Solution: The properties VSI - VS5 are satisfied. The additive identity is the function o (XI SUC~that o (x) = o tor all xe R. The inverse off is -f where (-f) (x) = - [f(x)] W x E R.

A Example 6: Lt V c R%e given by V = {(x,y) / x.,y E R and y = 5x) We define addition and scalar multiplication on V to be the same as in R2, i.e.,

(xlt~l)+ (~29~2)=(XI + X27Yl + y2)and a (x,y) = (ax, cry), for a E R. Show that V Fs a real vector space.

Solution: First.note that addition is a binary operation on V. This is because (XI,YI) E V, (~2,~2) E V * YI = 5~1.~2= 5x2 3 YI + Y2 = 5(~1+ ~2) - (x,+x,, Y,+Y~)E V. The addition is also associative and commutative, since it is so in R*. Next, the additive I I identity for R2, (0,0), belongs to V and is the additive identity for V. Finally, if 1 (x,y) E V (i.e., y = Sx), then its additive inverse -(x,y) = (-x, -y) E R2. i Also -y = 5(-x). SOthat, -(x,Y) E V. That is, (;i,yj~V --' - (x,y) E V. Thus, VS1 - VS5 are satisfied by addition on V. As for scalar multiplication, if a E R and (x,y) EV,then y = 5x, so that ay = 5(ax). .'., a (x,y) E V. That is, VS6 is satisfied. The properties VS7 - VSlO also hold good, since they do so for R2 Thus V becomes a real vector space. Check your understanding of vector spaces by trying the following exercises.

1 E E3) Let V be the subset of complex numbers given by V = {x + ix( x E R). L Show that, under the usual addition of complex numbers and scalar multiplication defined by a(x + ix) = ax + i (ax), V is a real vector space. I Ver'tor Spaces

E7) Show that.Cn is a complex vector space.

Note: We often drop the mention of the underlying field of a vector space if it is . understood. For example, we mav say that "Rn is a vector space" when we mean that "Rn is a vector space over R".

Now let us look more closely at vector spaces.

The examples and exercises in the last section illustrate different vector spaces. Elements of a vector space may be directed line segments, or ordered pairs of real numbers, or polynomials, or functions. The one thing that is common in all these examples is that each is a vector space; in each there is an addition and a scalar E10) Prove that -(-u) = u, 3' u in a vector space. Vector Spaces

El 1) Prove that a(u-v) = cxu -av tor all scalars cx and V u,v in a vector space.

Let us now look at some subsets of the underlying sets of vector spaces.

3.4 SUBSPACES

In E3 yousaw that V, a subset of C, was also a vector space. You also saw, in Example 6, that the subset V = {(x,y) E R2Iy = 5x1, of the vector space R', is itself a vector space under the same operations as those in R' In these cases V is a subspace of R2. Let us see what this means. Definition: Let V be a vector space and W E V. If W is also a vector space under the same operations as those in V, we say that W is a subspace of V. The following theorem gives the criterion for a subset to be a subspace. Theorem 2: A non-empty subset W, of a vector space V over a field F, is a subspace of V provided a) w,+w2~W,%'w,,w2~W b) cxw€W~a€FandwEW. c) 0, the additive identity of V, also belongs to W

Proof: We have to show that the properties VS1 - VSlO hold for W. VS1 is true because of (a) given above. VS2 and VS5 are true for elements of W because they are true for elements of V. VS3 is true because of (c) above. VS4 is true because, if w E W then (- 1) w = -w E W, by (b) above. VS6 is true because of (b) above. VS7 to VSlO hold true because they are true for V. Therefore, W is a vector space in its own right, and hence, it is a subspace of V. The next theorem says that condition (c) in Theorem 2 is unnecessary. Theorem 3: A non-empty subset W, of a vector space V over a fieldF, is a subspace of A non-empty subset of a vector V if and only if space is a subspace iff it is closed under vector addition and scalar multiplication.

Proof: If W is a subspace, then obviously (a) and (b) are satisfied. Conversely, suppose (a) and (h) are satisfied. To show that W is a subspace of V, Theorem 2 says that we only need to prove that 0 r W. Since W is non-empty, there is some w E W. Then, by (b), 0. w E W, i.e., 0 E W. This completes the proof of the theorem. Vector Spaces Actually both theconditions in Theorem3 can be merged to glve the following compact result. Theorem 4: A non-empty subset W, of a vector space V over the field F, is a subspace of V if and only if awl +pw2€W~a,p€Fandw,,w,€W.

Proof: Firstly, suppose W isasubspace of V. Then, by Theorem 3. for any a,PE Fand wl,w2~W,wehaveawl~Wand~w2~W,sothataw,+pw2rW.

Conversely, suppose a w, + P w2 E W -tf- a,p E F and w, ,w2 E W. Then, in particular. fora= 1 =p(rememberl~F),w,+w2~W.Also,ifweputp=Oinawl+ pw,,we

get awl E 5 +a E Fand w,E W. . .,by Theorem 3, W isasubspace. Hence, the theorem is proved.

Let us use this theorem to obtain some more examples of vector spaces.

Example 7: Prove that the subset W = {(x, 2x, 3x) (xE d} of R' is a subspace of R3.

Solution: If we take x = 0, we see that (0,0,0) E W, so W + 0. (Remember 0 denotes the empty set.)

Next, w1 E W,w2e W =3 W, = (x,~x,~x),w2 = (y,2y,3y), where XER,~E R. Thus awl = (ax, 2ax, 3ax) and pw2 = (py, 2py, 3py), for a, P E R.

a *a w1 + Pw2 = (ax + Py, 2(ax + py), 3(ax + By)) 3 awl + Pw2 = (z, 22,3z), where z = ax + P y E R. *awl + pw2e W. Hence, by Theorem 4, W is a subspace of R3.

Example 8: Which of the following subsets W of R4 are subspaces of R4? The set of all w = (xl,x2,x3,x4) E R4 such that

a) x, = 0, (b) x, = 1, (c) x3 < 0, (d) 2x1 + 5x4 = 0.

Solution: a) Here, W = {(0,x2 x3,x4)(x2,x3,x4E R } Obviously, W f + as (0;0,0,O) E W. Next,wl, w2€W =3 w, = (0,x2,x3,x4), xi E R, for i = 2,3,4.

w2 = (O,Y~,Y~,Y~),Yi E R, fori = 2,3,4. * awl = (0, ax,, ax3, ax,)>and pw2 = (0, py2, py3, PY~),a,P E R. =s= awl + pw2 = (0, ax2 + PY~,ax3 + PY~,ax4 + PY~)E W. Hence W is a subspace of R4.

Again W # .+,as(l,l,l,l) E W

Now w1 E W, wqe w -T- w1 = (x1,1,x3,x4). w2 = (Y~,~,Y~,Y~) 3 W1 + w;= (xi + Y1,2, x3 + Y3r X4 + ~4) w, + w2fk So W is not a subspace of It4

Note: An easier proof for (b) would be: 0.w = (0,0,0,0) f W; . ., W is not a subspace.

C) Here, W = {(x,,x2,x3,x4)~xiE R; x3< O} ThenW f +,as(O,O,-1,O)r W. NOW,W=(O,O,-i,o)ew,b~t(-i)~=-~=(o,o.i,o)ew. Vector Sparer Therefore, W is not a subspace of R'.

d) Now, W = {(x,, x2, x3. x4)(xiE R, 2xI + 5x4 = 0} Obviously (0,0,0,0) E W. so W # 4. Next, "w,E W, w, E W w, = (x,;x2,x,,x4) with 2xl+5x4 = 0 and w2 = (ylry2,y.3,y4)with 2yl + 5y4 = O WI+WZ = (x1+y1,x2+y2.x3+y3,x4+y4) with 2(xI+y1)+5(x4+y4) = (2x1 + 5x4) + (2y1 + 5y4)'= 0 + 0 = 0. aw,+w,~W. i-k Finally, t a E R, w E W * a E R, w = (xI,x2,x3,x4)with 2x1+5x4= 0. -2 a w = (ax1,ax2,ax3,ax4) with 2(axI)+5(ax4)= a(2xl+5x4) = 0. 3awEW. ':I ii':I . $, { g*, So W is a subspace of R~. ;* , &, 1 Note: We could have also solved @I)by using Theorem 4 as follows i b\bl ' 1 For a,p E R and (xl?x2?~3,x4)?(Y~,YZ,Y~,Y~) in W we have x: L ,~2?~3?~4)+ P(YI,Y~,Y~~Y~) = (axl+p~l ,ax2+P~2?ax3+ il! with 2(ax,+pyl) + 5(ax4+f3y4) = 42x1 + 5x4) + P(2y1+5y4) = 0. C; : $:i Thus, a,p E R and w,,w2 E W 3 awl + pw2 E W. "; j *

Example 9: Let V bc; a vector space over F and v E V. All scalar multiples of a fixed Show that the subset F v = {avla E F) is a subspace of V. vector form a subspace.

!$I$ i Solution: F v # 4 because 0.v = 0 E F v. g Now,ifavandpv~Fvthenav+pv= (a+ p)v~Fv. Also,a~F,andPv~Fv*a(pv)= (aP)v~Fv,sinceap~F. !;i I 2 1 Thus, by Theorem 3, F v is a subspace of V. Note: The subspace Rv, of Rn, represents a line in n-dimensional space.

E E12) Prove that W = {(x, ,x2,x3)E ~~12~~+ x2 - x3 = 0) is a subspace of R~.

E E13) For each of the following subsets W ofa3,determine whether it is a subspace of R~: W is the set of those vectors (x,, x2, x3) in su such that (a) x, = -x2; (b) 4r 0; (c) xlx2 = 0; (d) x,+x2+x3 = 1. E E14) Show that (0) is a subspace of the vector space V over F.

In Example 9 you saw that an element v c V gives rise to a subspace of V. In the next section we look at such subspaces of V, which-grow out of subsets of V that are much smaller than the concerned subspace.

3.5 LINEAR COMBINATIONS

In Unit 2 you'came across the fact that any element of R~ could be written as a-i + b.j + c-k,where a,b,c c R. In this section we will generalise this. Consider the following definition. Detlnition: If v,, v2, ...... , v, are elements of a vector space over F, and al,a2, ...... a, c F, then the vector alv1 ,+ a2v2 + ...... + a,v, is called a linear combination of the vectors vl ,v2,... . . , v,, or of the set {.vl,v2,...... , v d For instance, since (2,4,3) = 2(1,- 1,O) + 3(0,2,1), (2,4,3) is a linear combination of (1 ,- 1,O) and (0,2,1) We are now ready to generalise the result of Example 9. Theorem 5: If v,,v2, .. . . .', v, belong to a vector space V over a field F, then W = {alvl + a2v2+ ...... + a,vnlai are scalars) is a subspace of V. Roof: Firstly, 0 is a scalar and O.vl + O.vz + .. .. . + 0-v,, = 0 + 0 + ...... + 0

SoOc W.andW f Secondly. wl r W, w2-E W n 3wI=alvl+a2vZ+....+ anvn=Xaivi,ai~F. i=l n and~~=~~v~+~~~+~~v~=~'f3,vicF. i= l ~wI+w2=(al+PI)v1+....+(an+Pn)vn*w1+w2aW Finally, if a is a scalar, and w c W, we have w = a lvl + .. . . + anvn,where ai is a scalar +i = 1,. .. . ,n. * a w = (aa,) vl + (aaz) v2 + .... + (aan)vn. *aw~W This proves the theorem. We often denote W (in Theorem 5) by F v., + ... . + F vn. Let us look at the vector space Rn,over R. In this, we see that every vector is a linear combination of then vectorse, = (1,0,. .. ,0), e2 = (0,1,0,. . ...0) ...... , en = (0 ,....,0,l). This.is because (xl,. .. . ,xn) = x,el + xze2 + . .. . + xne,, xir R. In this casetwe say that the set {el,.. .,,en)spans Rn. Let us see what spanning means.

Definition: Let V be a vector space over F, and let S E V. The linear span of S is defined to be the set of all linearcombinations of a finite number of elementsof S. It is denoted by [S]. Thus. n [S] = { ai v,/n positive integer, vi E S, ai scalars} i=1 We also say that S generates [S]. ' Note that S is only a subset of V, and not necessarily a subspace of V. Also note that [S] ~~~isthesetoffinitesumsoftheforma,v,+ ..... +anvn, whereaieF and vies.

Example 10: Suppose ' S C R~,S = {(1,0), (0,l)). What is [S]? Solution: [S] = {a(1,0) + ~(O,l)la,~c R}, i.e., [Sl = {(a,P)la,p E R). In this case, the linear span of S is the whole of R2. Thus, {(1,0), (0,l)) generates R*.,

Example I I :Suppose S S R3, S = {(I ,- 1,O)). What is [S]? Solution: [S] = {a(l ,- 1,O)(a E R). = {(a, -a, 0)la E R) Example 12: Let P be the vector space of real polynomials, and S = {x,x2+1,x3-I)} C P. What is [S]?

~olution:[s] ='{ax + p(x2+ 1) + 7(x3- I) I a, P, T E R). = {7x3 + px2 + ax + (~-~jI a,p, TER) E E15) Let S = {1,x,x2)be a subset of P in the example above. Does 2x+3x% [S]?

. . . Y =lor Spaces In the examplesgivcn above YOLIIllily havc noticccl that [SI is a subspucc ol V. Wc provc this fact now.

Theorem 6: If S is a non-cmpty subset of a vector space V over F, thcn IS1 is a subspacs of V.

Proof: Since S f and S E [S]. [S] f +. Also. since S S V. [S] S V. Now, s, E [SJ,S? E [SJ

3 S, = alvl + a2v2+ .... + anvn,for v, E. S, aleF and s2 = Plwl + P2w2+ .... + Pnlwm. for w, E S, p, E F. Thus,fora,p~F,as,= aal vl + aa2v2+ .... + aa,v,, Ps2 = PPI~I+ PP2w2 + ..... + PPmwm 3 as, + Ps2 = aalvl+ ...... + aa,vn + PPlw1+ ...... + PP;,,w,,,. with vI,W, E S andaa,~F, Pf3, EF.Thisshowsthat asI + Ps2 isa linearcombination of a finite number of elements of S. Thas, as, + ~s,E~s].Therefore, by Theorem 4, [S] is a subspace of V Theorem 6 shows that the linear span of S is a subspace containing S. In fact, it is the smallest subspace of V containing S. as you will see now.

The'linear span of S is the smallest Theorem 7: If S is a subset and T a subspace of the vector space V over F, such that subspace of V containing S. S ST, then [S] ST. Proof: Let s [S], then

s = aivi. where v, c S, ai E F i=l As SST, vi€TW= 1,.... n. AsTisa~ubspaceandv~~Tfor n a11i, 2 aiviE T, i.e., s E T. i= 1 We have proved that s E [S] =+- s E T. Hence, [S] S T. An immediate corollary to Theorem 7 follows.

Corollary I : If S is a subspace of V, then [S] = S. Proof: Since S is a subspace containing S, Theorem 7 gives us [S] c'~.But S 5 [SJ always. Therefore, [S].= S. Tfie theorems above say that we can form subspaces from mere subsets of a space. , Given a subset S of a vector space V, if S is not a subspace of V, what is the 'minimum' that we must add to S to make it a subspace? The answer is - all the finite linear combinations of vectors of S. Look at the following examples.

Example 13: Let S = {(I ,1,0), (2,1,3)) 5 R~.Determine whether the following vectors of R~ are in [S]. (a) (0,O.O); (b) (1,2.3); (c) (413. 1.1).

Solution;[S] = {a(],1.0) + P(2,1,3)la,P E R) L' L' = {(a + 2P, a + P, 3P)la,P E R) (a) (0,0,0) E [S], since [S] is a subspace and (0,0,0) is the additive identity of R'.

(b) (1,2,3) E [S] ~f we can find a, p E R, such that (a + 28, a + p, 38) = (1,2,3) , i.e.,a+2P=l,a+p=2,3P=3. Now3P=3*p= l,andthen,a+p=2=+a= 1. But then a + 2P = 1 + 2 = 3 # 1. Hence, (1,2,3) + [S]. (c) (4/3,1,1)~[S]ifa+2P=4/3,a+P= l,3P= lforsomea,PeR. These equations are satisfied if P = 113, a = 213. So (413,1,1) c [S]. 'E16) IfS = {(1,2,1), (2,1,0))cR3, determine whether the fo~lowin~vectorsofR' are in [S]. (a) (531 (b) = (2,1.0), (c) (4,521.

E17) Let P be the vector space of polynomials over R and S = {x,x2+ 1, x3- 1). Determine whether the following polynomials are in [S]. (a)x2+x+1, (b)2x3+x2+3x+2.

Now that you have got used to the concept of subspaces we go on to construct new vector spaces from existi:lg ones.

3.6 .ALGEBRA OF SUBSPACES

In this section we will consider the union, intersection, sum and direct sum of vector spaces. 3.6.1 Intersection If U and W are subspaces of a vtctorspace V over a field F, then the set UnW is a subset of V. We will prove that it is actually a subzpace of V. Theorem 8: 'The intersection of two, subspaces is a subspace.

'Proof: Let U and W be two subspaces of a vector space V 'Then 0 E U and 0 E W ' Therefore, 0 E UnW; hence UnW f O. Next, if v, E U nW, and v, E UnW, then v, E U, v2 E U, v,, E W, v, E W. Thus. for any a,p E F, avl + pv2 ti U, aV1 + Bv~E W (as U and W are subspaces).

:., avl + pv2 E unw. This proves that UnW is a subspace of V.

Example 14: U = {(x,2x,3x)(xtiR) and W = {(O,y,(3/2)yly E R) are subspaces of R~.What is UnW? Solution: Any element of UnW is of the form (x,2x,3x) and of the form (O,y;(3/2)y). Thus, the only possibility is (0,0,0). Therefore, UnW = {(0,0,0)). By E 14 you know that this is a vector space.

Example 15: U = {(x,y,O)lx,y E R) and W = {(O,y,z)ly,z E R) are subspaces of R~.What is U n W? Solution: U n W is the set {(O,y,O)(yE R). In this example note that U is the xy-plane, W is the yz-plane and U n W is the y-axis. E E18) If U = {(x,y,2x)lx,y E R) and W = {(x,2x,y)lx,y E R), what is U n W?

Note: It can be shown that the intersection of any finite or infinite family of subspaces is a subspace. In particular, if V,,V2 ,...... ,V, are all subspaces of V, then v1nV2n..... nVnis a subspace of V. Let us now look at what happens to the union of two or more subspaces. 3.6.2 Sum Consider the subspaces U and w of R~ given in bxample 15. Here v, = (1,2,0) E U and v2 = (0,2,3) W. Therefore, v, and v2 belong to U U W. But v, + v2 = (1,4,3) is neither in U nor in W, and hence, not in U U W. So U U W is not a subspace of R~ Thus, we see that, while the intersection of two subspaces is a subspace, the union of two subspaces may not be a subspace. However, if we take two subspaces U and W, of a vector space V, then [U U W], the linear span of U U W, is a subspace of V. What are the elements of [UUW]? They are linear combinations of elements of U U W. So, for each VE .[U U W], there are vectorsv,,~~,.... .,v, E U U W of which visa linear combination. Now some (or all) of the v, ,. .. .. ,vn are in U and the rest in W. We rename those that are in U as ul,u,,....., u, and those in W as wl,w2,....., wk (jrO,kzO,j+k=n). Then, there are scalars a],.. . .. ,aJ,PI,. . .. ,Pk such that v = alul + .... + aJuJ+ Plwl + .... + Pkwk =u+w, where u = alul + .... + aJuJE U, since each u, E U, and w = P,wl + ..... + PkwkeW,sinceeachw,~ W. (Ifj = 0,wetakeu =O;ifk= 0, we take w = 0.) So what we have proved is that every element of [U U W] is of the type u+w, u E U, w E W. This motivates the following definition. Definition: If A and B are subsets of a vector space, we define the set A + B by A + B = {a+b(a~A.beB}. Thus, each element of A + B is the sum of an element of A and an element of B.

Example 16: If A = {(0,0), (1 ,I), (2,-3)) and B = {(-3,l)) are subsets of R*, find A + B. 28; ,I Solution: A+B = {(-3,1), (-2,2), (- 1,-2)) because, for example, (0,O) + (-3,lj = (-3,l) (1,l) + (-3,l) = (-2,2), etc. p' ( ;)I Exampk 17: Let A = {(O,y,z)(y,zE R ) and B = {(x,O,z)(x,zE R }. k 1 Prove that A+B = R3. So1ution:Since ASR3, BcR3, so A+BcR3. Itis,therefore,enoughtoprovethat R3 s A + B. Let (a,b,c) E R3. Then (a,b,c) = (O,b,c/2) + (a,O,cQ), where (O,b,cQ) E A and (a,O,c/2) EB. So (a,b,c) E A+ B. Thus,R3c A+B. Hence, A + B = R3. Note that in the discussion preceding the definition of a sumof subsets, we have actually proved that if U and W are subspaces of a vector space \I, then IU U W] S U -! W. Indeed, we have the following theorem.

Theorem 9: If A and B are subspaces of a vector space V, then [A U Q] = A + B. Proof: We have already proved (see above) ihat [A U B] S A + B. So it only remains to prove that A + B 5 [A U B]. Letv~A+g,thenv=a+b,a~A,b~B.Nowa~A=+aEAUB =+a€ [AUB]. Similarly, b E B =+ b E A UB + b E [A U B]. As [A U B] is a vector space and a,bE [AUB],weseethata+b~[AUB],i.e.,v~[AUB].Thiscompletestheproof f of the theorem. Since [A U B] is the smallest subspace containing A U B, we see, from Theorem 9 that A+B is the smallest'subspace of Vcontaining both A and B.

E E19) For the subspaces A = {(x,O,O)(xE R} and B = {(0,y,O)ly~R}of R3, find [A U B].

I

We consider a special kind of sum of subsets now. 3.6.3 Direct Sum If A and B are subspaces of a vector space, you know that eve-y vector v in A+B is of the form a+b, where a EA, b e B. But in how many ways can a given v E A+B be expressed in the form, a + b? In Example 17 we have expressed (a,b,c) E R3 in the form a+b by writing (a,b,c) = (O,b,a) + (a,O,d2). But we could also write

(a,b,c) = (O,b,O) + (a,O,c) or (a,b,c) = (O,b,c) + (a,O,O) or (a ,b,c) = (0, b ,d3) + (a,0,2d3). Indeed, for any real number 6 we can write (a,b,c) = (O,b,6) + (a,O,c-6). Note that, in each case, we have expressed (a,b,c) as a sum of vector from A and a vector from B. So, in this case, there are infinitely many ways of writing v e A + B in the form a+b, with a E A, b E B Vector Spaces But there are some cases in which every vector v E A+ H can hc writtcn in one and only one wav as a+b, a E A. b c B. For example, supposc A = {(x,y.O)(x.y E R} and B = {(O,O,z)lz E R). Then, for any (p,q,r) E R\ve can write

It follows tMat A + B = R-'. But here (p,q.r) can be written in only one way as a+b, namely (p,q,O) t. CO,O,r), because, if we write (p.q.r) = (x.y.0) + (0.0.z), then (p,q,r) = (x,y,z),sothatx = p,y =q.z = r.Thismeansthat(x.y.0) = (p,q,O)and (0,0,z) = (O,O,r).

Now, note that in this case AnB = {(0,0,0)}, whereas in the earlier example AnB = {(O,O,z)lz E R} f {(0,0,0)} It is this difference in AnB that is reflected in a unique or a multiple representation o v in the form a+b. Definition: Let A ahd B be subspaces of a vector space. The sum A + B is said to be the dirkct sum of A and B (and is denoted by A @ B) if An0 = (0). We have the following result. Theorem 10: A sum A + B, of subspaces A and B, is a direct sum A @ B if and only if every v r A+B is uniquely expressible in the form a + b, a E A, b E B. Proof: First suppose A + B is a direct sum i.e., AnB = (0). If possible, suppose v has two representations, v = a, + b, and v = a, + b2, aiE A, b, E B. Then a, + b, = a, + b2, i.e., a,-a, = b2-b,. Now a,,a2 t A =z= a,-a, E A. Similarly, b2-b, c B, that is, a,-a, E B (since a, - a2 = b2 - bl). Thus, a, - a, E AnE =S a,-a, = 0 =Z= a, = a,. And then, b, = b,. This means that a, + b, and a, + b2 are the same representations of v as a + b. Conversely, suppose every v E A+B has exactly one representatiorl as a+b. We must prove that A n B = (0). Since A and B are subspaces, 0 E A, 0 c B. :..{ 0) c A fl B. If AnB # {0}, then there must be some v # 0 such that v c A n B. Then, v has two distinct representations as a+b, namely, v + 0 (v E A, 0 E B) and O+v (0 E A, v E B). This is a contradiction. So A n B = (0). Hence A+B is a direct sum. Example 18: Let A and B be subspaces of R3 defined by A = {(x,y,z) E R3Ix = y = z}, B = {(O,y,z)(y,z~Rl. Prove that R3 = A @ B. Solution: First note that A + B CR~.Secondly, if (a,b,c) E A n B, thena = b = c, an< a = 0; so a = 0 = b = c, i.e., (a,b,c) = (0,0,0). Hence, the sum A + B is the direct sum A@B. Next given any (a,b,c) E R3, we have (a,b,c) = (a,a,a) + (0,b-a, c-a), where (a,a,a) E A and (0,b-a, c-a) c B; this proves that R3C A @ B. Therefore, R3 = A@B. Example 19: Let V be the space of all functions from R to R, and A and B be the A function f : R -r R is called an subspaces of V defined by even hdonif f(x) = f(-x) YX.E Rrmirn odd bactim if A = {flf(x) = f(-x), +x) f(-x) = -f(x) y x r R. B = {fJf(-X) = -f(x), 4x1 i.e., A is the subspace of all even functions and B is the subspace of all odd functions. = 66 . Show that V A@ B. Solution: First, suppose f E A n B. then + x E R, f(-x) = f(x) and f(-x) = -f(x). So, + X, f(x) = -f(x), i.e.. Vx, f(x) = 0. Thus, f is the zero function, and ~n 13 = (0).

Next, let f E V, define

g(x) = .-{f(x)1 f(-x)}, and 2 +

Then, (i) f(x) = g(x) + h(x)? x E R, i.e., f = g + h,

I 1 (ii) g(-x) = -{f(-x)2 + f(x)) = g(x), . . g E A.

(iii) h(-x) = $f(-x) - f(x)} = -h(x). . . h E B.

Thus, for each f E V. f = g+h, for some g E A, h E B. =+- V = A+B, and, as AnB = {0), we get V=A@B Note: Example 19 says that every function from R to R can be uniquely written as the sum of an even function and an odd function. E E20) Let A,B,C be the subspaces of R~ given by A = {(x,y.z)) E ~-'lx+~+z= 0). B = {(x,y,z) E R~X= z}, C = {(O,O,z)lz E R) Prove that R" A+C and R" B+C. Which bf these sums islare direct?

E E21) Considerthe real vector space C, of all complex numbers (Example 3). If A and B are the subspaces of C given by A =-{a+i.O(a~R),B = {iblb E R), prove that C=A@B.

Now, we will look at vector spaces that are obtained by "taking the quotient" of a vector space bv a subspace. ------Vector Spaces I 3.7 OUO'I'IENT SPACES

From a vcctor \pace V. and it4 SU~S~;ICL'W, we ~illno~ create a new vector space For this, we first definc the concept of a co4et.

3.7.1 Cosets Let W be a subspace of V. If v E V the set v + W. defined by v+W={v+w~wEW} is called a coset of W in V.

Example 20: Consider the subspace W = {(a,O.O)/a E R} of R'. Let v = (1,0,2). Find the coset v + W. Is it a subspace of R~?

Solution: v + W = {v+wl'w 6 W} = {(I,0,2) + (0,0)) a E R) = {(a+ l,O.2)la E R} Thus, v + W = {(a,0,2)la E R}, (because, as a takes all the real values, a + 1 also takes all the real values. so that we may replace a+ 1 by a). v+W is not a subspace of R~ as (0.0.0) $ v+W. Observe that each element v of V yields a coset v + W of W. Every coset of W in V is a subset of V, but it may not be a subspace of V, as you have seen in Example 20.

Example 21: With W as in Example 20, and v = (2,O.O). prove 'hat v+ W is a subspace and, in fact, v+W = W. Solution: Here v + W = ((2,O.O) + (a.O,O)la E R) = {(a + 2,O.O. I a E R) = {(p,O,O)lpE Rl = W Observe that. in the example above v E W whereas in the previous example, v $ W. In the next theorem we substantiate this observation.

Theorem 11: Let W be a subspace of a vector space V. Then v E W if and only if v + W = W. Also. if v $ W, then v + W is not a subspace of V.

Proof: We first prove that VE W *v+W = W. For thislet UEV + W. Then. forsome w E W, u = v + w. This implies that u E W, as both v,w E W. This means that v+wcw. Also, w E W w-v E W, since v E W. w = v+(w-V).EV + W, so that W Sv+W. This proves that v + W = W.

Now let us prove the converse, namely, v+ W = W => v E W. For this we use the fact that 0 E W. Then we have v = v+O EV+W = W =z=- v E W. Lastly, we prove that, ifv $ W then v+W is not a subspace of V. If v + W is a subspace = v + w is a subspace of of V, then 0 E v + W. Therefme, for some w E W, v + w 0, i.e., w = -v. Hence. v=> v • W. -v E W and, as W is a subspace, v E W. Thus,v+WisasubspaceofV~v~W.So,v~W~v+W~snotasubspaceofV. E E22) Let W = {f(x) E Plf(1) = 0) be a subspace of P, the real vector space of all polynomials in one variable x. a) If v = (x-1) (x2+1), what is v + W? b) If v = (x-2) (x2+1), what is v + W? Now we ask : Given a vector space V and a subsr;~ceW. cpn we get V back if we know Veclor Spaces a11 tlie cosets of W? The answer is given in the i'ollov.,ing thf:orem.

1 Theorem 12: If W is a subspace of V, the union :)f nll the LX)Y:\ of W in V is V.

Proof: Sir- t e-crycoser of W in V is a scbset of V the union i4 certainly a subset of V. Conversel;,. piven v E V, v ;= vi-1J E v+W (::s OE W). Thus, every v E V bclon'gs to some co.:et or W in V. Hr::ice. V is contained in the union of all tile coscts of W in V. Hence, the th~orernis c!.t::t;!ishcd.

I We may write the statement of Theorem 12 as

I V=U(v+w)

V€V p A very special property of cosets is given in th: iollowing tileorem.

he or em 13: Two cosets v, + W = .v, + W in V are ehher equal or disjoint. In fact, f v,+W =v2+ Wiiv, - v2~W,forvl,v,~V.

I Proof: We have toprove that,forv,, v2€Veither(v,f W)ll(vi+W) = {O)orv,+W = v2 + W. Now, suppose (v,+ W) n (v2+ W) f {0 ). The11they have a common non-zero element v, say. That is, v = v, tw, = v,+w,. for some w,,w2 E W. Then v, - v, = w2 -- w, c W ...... (1) We want to prove that v, + W = vi+ W. For this we prove that v, +W 5 v2 + W and v,+WCv, +W. Now, u E v1 + W * u = v1 + w,, where w, E W * u = v, + (w,-wl) + w,, by (1) = v, + w,, where w, E W * UEV2+W. Hence, v, + W 5 v2 + W. We can similarly show that v2 + W 5 v, + W. t:crtce, v, + W = v2 + .W. Note that we have shown that V:-v2Ew-=v,+w=v2+w.

E E23) Eh the proof above. we have essentially proved that v, - v2 E W =s v; + W = v2 + W. The converseof this is also true. Prove it.

Note: The last two theorems tell us that if W is a subspace of V, then W partitions V inro mutually disjoint subsets (namely, the cosets of W in V). Consider the following example in which we show how a vector space can be partitioned by cosets of a subspace.

Example 22: Consider the subspace W = {a(l,O,O)Ja E R) of R~ HOWcan you write R~ as the union of disjoint cosets of W? I Solution: Not. that W is just the x-axis in 3-dimensional space. Any coset of W is of the form (a,b,c) + W = {(a,b,c) + (a,O,O)laER) = {(a+a, b,c)la e R). Now, for any (a,b,c) E R3, (a,b,c) - (O,b,c) = (a,O,O) E W. Therefore, (a,b,c) + W = (O,b,c) + W. Also, the cosets (0.b.c) + W and (O.b'.c') + W are the same iff b = b' and c = c'. Thus, {(O,b,c) + Wlb,c ER} is the set of disjoint cogts of W in R3. And R3 = U {(o,~,c)+ Wlb,~ER }. Geometrica!ly, the coset (O,b,c') + W represents a line (in the plane determined by the point (0,b.c; and thex-axis) which is parallel to the x-axis and passes through the point (O,b,c): Thus, R3 is the union of all such distinct lines. Vector Spaces E E24) Write R~ as a disjoint union of the casets of

a) the subspace {(0.0)), *' b) the subspace R~.

. Betore we proceed, let us stress that our notation for a coset of W in V has a peculiarity. A coset v, + W can also be written as v2 + W provided v, - v2 c W. So the same coset can be written in many different wsys. Indeed, if C is a coset of W in V, then C = v + W, for an9 vector v in C. Let us now see how the set of all coseu of W in V can form a vector space. 3.7.2 The Quotient Space We have pointedout that generally a coset v + W of a subspace W of a vector space V is nat itself a subspace of V. We shall now prove that if we take the set of all cosets of W in V, this set can be made into a vector space by defining addition and scalar multiplication suitably. Notation: Let W be a subspace of the vector space V. We denote the set of all cosetsof W i~ V by V/W. Thus, V/W = {v+Wlv E V). Consider the following ekample.

Example 23: Let P be the vector space of real polynomials id x and W = {flf E P, f(Q) = 0) be the subspace of P consisting of all those polynomials whose constant term is zero. Show that P/W =-{a+Wla E R}. Solution: Since a E P4a E R, certainly a,) W is a coset of W in P. So a+W 4 a c R. Conversely, take an element of P/W, say f(x) + W, where f(x) is a polynomial.. Suppose f(x) = anxn4 a,-,xn-' + .... + a2x2 +.alx + %,ai E R. Then f(x) = ao+g(x), where g(x) = a,x+a2x2 + .... + a,xn. Since g(0) = 0, g E W: Hence, f = a,, + g, where g E W. Thus, f+W = q, + W ('lheorem 13). Hence, f+W ~{a+WlaE R). 'fhis completes the proof that P/W = {a+W(ac R).

E E25) If Pn de~otesthe vector spacelof all polynomials of.degree 5 n, prove that P@, = {ax" P21a 4 R ). (Hint : For any f(x) E P,, 3 a E R such that f(x) - ax3 E P2 .) We now proceed to introduce two operations on the set VIW, namely, addition and scalar ,multiplication. Definition: Let W be a subspace of V. We define addition on VIW by (v,+W) + (v2+W) = (v1+v2) + W. If a E R, v+W E VIW, then we define scalar multiplication on VIW by a-(v+W) = (av) + W. Note that our definitions of addition and scalar multiplication seem to depend on the way in which we write a coset. Let us explain this. Suppose C1 and C2 are two cosets. What is C1 + C2? To find C1 + C2 we must express C, as v, + W and C2 as v2 + W. Having done this we can then say that

But C, can be written in the form v + W in many ways and the same is true for C2. So the question arises : Is C, + C2 dependent on the particular way of writing C1 and C2, or is it independent of it? In other words, suppose C1 = vl + W = v,' + W and C2 = v2+W = v2'+W. Then we may say that C1+C2 = (v1+W) + (v2+W) = (vl+v2) + W; but we may also say that c, + c, = (v,'+W) +(v2#+W) = (v,' + v2') + W. Are these two answers the same? If they are not, which one is to be C1+C2? A similar question'can arise in the case of aC where a is a scalar and C a coset. These are important questions. Fortunately, they have simple answers as shown by the following theorem.

Theorem 14: Let W be a subspace of a vector space V. If v,+ W = v,'+ W and v, + W = v2' + W, then a) (v,+v2) + W = (vll + v2') + W Also, if a is any scalars,then b)(avl) + W = (av,') + W

Proof: .a) For v,, v,', v,, v2' E V, v, + W = v,' + W, v2 + W = v2' + W * v1 - vl' E W, v2 - v2' E W (by E 23) 3 (v, - v,') + (v2 - v2') E W - (v, + v2) - (v11+v2')E W =+ (vl+v2) + W = (v,' + v,') + W (by Theorem 13). Thus, (a) is true.

b) Foranyscalaraandvl,vl'~V,vl+ W = v,' + W *vl-v,'~ W =+a (v, - vll)E W + a vl - avll E W =3av1+W=av1'+W i Thus (b) is also proved. I Theorem 14 assures us that the sum and scalar multiplication of cosets is independent i of the particular way in which a coset is written. We express this by saying that addition t and scalar multiplication of cosets are well defined by the way we have defined them. This also means that when adding two cosets or when multiplying a scalar and a coset we are free to use any representation for the cosets involved. We now come to the actual proof that VIW is a vector space. Theorem 15: Let V be a vector space over a field F, and W be a subspace. Then VIW is a vector space over F.

Proof: We will show that VS1- VSlO hold for VIW where the operations are addition and scalar multiplication as defined above. i) VS1 is true since the sum of two cosets is a coset. Vector Spaces ii) For vl, v,, v3 in V we know that (v, + v,) Therefore, {(v, +W)+(v2+ W)) + (vg + W) = {(vl +v2) + W) + (v3 + W) = {(v, + vz) + v3) + W = {v, + (v2 + v3)) + W = (v, + W) + {(v2 + v3) + W) = (v1 + W) + {(v, + W) + (v3 + W)) Thus, VS2 is true.

iii) We claim that the coset 0 + W = W (since 0 E W) is the identity element for V/W. Forv~V,W+ (v+ W) = (O+W) + (v+W) = (O+v)+W = v+W. Similarly, (v+W) + W = (v+W) + (O+W) = v+W. Hence W is the 'zero' of V/W, and VS3 is true.

iv) The additive inverse of v + W is (-v) + W, because (v+W) + {(-v) + W) = (v+(-v)) + W = 0 + W = Wand (-v) + W + (v+W) = (-v+v) + W = O+ W = W. ThisprovesthatVS4is true.

v) We note that addition in V is already commutative because V is a vector space. So *v~,v~EV,V~+ ~2 =V2 + V1. Hence(vl+W) + (v2+W) = (vl+v2) + W = (v2+ vl) + W = (v, + W) + (vl + W) Thus VS5 holds for V/W.

vi) VS6 is true because, for a E F and v + W E V/W, a (v + W) = av + W E V/W.

vii) To prove that VS7 holds, let a E F and u,v E V. Then a{(u+ W) + (v+ W)) = a {(u+v)+ W) = a(u+v) + W = (au+av) + W = (au+ W) + (av + W) = a(u+W) + a(v+W).

viii) For any a,P E F and v E V you can show, as above, that (a+p) (v+W) = a (v+W) + p (v+W). Thus, VS8 holds. ix) For any a,p E F and v E V we have a(p (u+ W))= a (pu + W) = (ap) u + W = (4)(u+W) Thus VS9 is true for V/W.

x) For UEV, wehavel.(u+W)=(l.u)+W=u+W. Thus, VSlO holds for V/W.

The vector space we have just obtained has a name. Definition: If W is a subspace of V, then the vector space VIW is called the quotient space of V by W.

The name quotient is very apt because, in a sense, we quotient out the elements of W from those of V. Example 24: Let V be a vector space over F and W = (0). What is V/W? Solution:V/W= {v+W(VEV)={v + {O)JVEV) ={vlv€V)=V Vector Spaces E E26) Let W = {u(O.l)/uE R}. What is R'/w?

E E27) For any vector space V. show that V/V hasonly 1 element. namely. the coset V.

And now, let us see what we have done in this unit.

3.8 SUMMARY

Let us conclude the unit by summarising what we have covered i.n it. In this unit we have 1) defined a general vector space. 2) given several examples of vector spaces. 3) .proved some important properties of a general vector space. 4) defined the notion of a subspace and given criteria to identify subspaces. 5) introduced the idea of the linear span of a set of vectors. 6) shown that the intersection of subspaces of a vector space is a subspace. 7) defined the sum and direct sum of subspaces of a vector space and shown lhat they are subspaces also. 8) defined cosets and a quotient space.

. 1-r

El) $La E R and (xl ,.... .,xn), (yl ,.....,y,) E Rn, ~[(XI,...,X~)+ (~lr..+,~n)] = Q(x~+Y~, XZ+Y~~....,X~+Y~) = (~(XI+YI),~(xz+Yz), ..., a(xn+Yn)) = (axl + ayl,ax2 + ay2, ..., ax, + ay,) = (ax,, ax2,. . .. ,axn) t(ayl, ay2,.. .. ,ay,) = a(x1,x2,.... ,xn) + a(yI,y2,...., y,), which proves VS7. Also, for a,P E R, (4)(XI, .... ,xn) = ((41x1, (~P)xz,..., (aP)xn) = (a(P XI),a(Px~),.....,a(P~n)) = a(pxl, Px2,....., Pxn) = a{P(xl ,.....,x,)}, which proves VS9. Finally, l.(xl ,. .. . ,x,) = (laxl ,loxZ,... ,l.x,) = (x,,. . .. ,xn), which proves VS 10. tor Spaces E2) For any a E R and f E S, af is a function from R to R. Thus, VS6 is satisfied. To show that VS7 - VSlO are satisfied take any a,P E R and f,g E S. Then, for any x E R, [a(f+g)](x) = a(f+g)(x) = a(f(x) + g(x)) = af(x) + ag(x) = (af)(x) + (as) (XI= (af + ag) (x). Therefore, a(f+g) = af + ag, that is, VS7 is true. You can simitarlv show that (a+ p) f = af + pf, (ap) f = a(f3f) and 1. f = 1, thus showing that VS8 - VS10 are also true.

E3) Since (x+ix) + (y+iy) = (x+y)+i(x+y) E V, and a(x+ix) = (ax)+i(ax) +a E R, and +(x+ix), (y+iy) E V, we see that VS1 and VS6 are true. VS2 and VS5 follow from the same properties in R. 0 = O+iO is the additive identity for V, and (-x) + i(-x) is the additive inverse of x + ix, x E R. Also, for any a, P E R, and (x + ix), (y + iy) in V the properties VS7 - VSlO can be easily shown to be true. Thus VS1 - VSlO are all true for V.

E4) Addition is a binary operation on Q, since (ax2+bx+c) + (dx2+ex+f) = (a+d) x2 + (b+e) x + (c+f), +a,b,c,d,e,f E C. Scalar multiplication from C # Q to Q is also well defined since a (ax2+ bx+c) = aax2 + abx + ac+ a, a,b,c-~~CNow on the lines of Example 4, you can show that Q is a complex vector space. E5) Note that Q' is a subset of Q in E4. Addition is closed on Q, but not on Q', because, for example, 2x2 E Q' and (-2)x2e Q', but 2x2 + (-2)x2$ Q'. Thus, Q' can't be a vector space under the usual operations.

E6) Now (x,y,z), (~l,~l,Zl)E V 3 ax + by + cz= Oandax, + byl + czl = 0. 3 a(x+xl) + b(y+yl) + c(z+zl) = 0. =%- (x+x,, y+y,, z+zl) E V 3 VS1 is true for V. Also, for a E R and (x, y ,z) E V, a(x,y ,z) = (ax, ay, az) E V. This is because ax + by + cz = 0 * a(ax + by + cz) = 0. 3 a(ax) + b(ay) + c(az) = 0. Thus, VS6 is true for V. (0,0,0) E V and is the additive identity for V: Thus, VS3 is true. For (x,y,z) E V, (-x,-y,-z) E V, and is the additive inverse of (x,y,z). Thus, VS4 is true. VS2 and VS5 are true for V, since they are true for R3. VS7-VS9 are true for V, since they are true for R3. VSlO is true, by definition of scalar multiplication.

E7) C" = {(xl ,xz,. .. . . ,x,)/xi E C): This problem can be solved on the lines of Example 2.

E8) From Theorem 1 you know that (-a)u = -(au) +a E F. In particular, (-l)u = -u. Therefore, (- 1) (-u) = (- 1) [(- l)u] = [(- 1) (-I)] U,by VS9.

E9) Now, (u+v) + (-u-v) = (v+u)+(-u-v), by VS5 = [V+(U+(-u))] + (-v), by VS2 = (v+O) + (-v) = v + (-v) = 0 Thus, by VS4, -(u+v) = -u -v.

E10) -(-u) = (- 1) (-u) by Theorem 1. = u, by E8. Ell) a(u-v) = a(u+(-v)) = au + a(-v) = au + a(-1)v = au + (-a)~ Vector Spaces = au - av.

E12) This is a particular case of the vector space in E6 (with a = 2, b = 1, c = - 1).

If a9.P E Rand (~19x29 ~3)-(~17 Y2, ~3)E W, then ~(xI,xz,x~)+ P(~lr~27~3) = (axl+P~17"x~+PY~, ax3+P~3). 2(axl + PY~)+ (ax2 + PY~)- (ax3+P~3) = a (2x1+x2-x3) + p(2yl+y2-y3) = 0, since 2x1 + x2 - x3 = 0 and 2yl + y2 - y3 = 0. Thus, a (~1,~2,~3)+ P(Y~,Y~,Y~) W. Hence, W is a subspace of R~.

E13) a) W = {(xl, -xl, x2)(xl,x2E R}, W f 4 , since (0,0,0) E W. For a,P E R and (~1,-~17~2),(yl,-yI,y2) EW, we have a(xl,-xl,x2) +

P (yl,-y1,y2) = (axl+pyl, -(axl+Pyl), ax2+Py2)E W.: . .., W is avector space.

b) W = {(xl,x2,x3)E ~~142 0) Since 6 r 0 Y xl E R, we see that W = R~,and hence is a vector space.

C)W= {(x1,x2,x3)E = 0). W # Ql , since (0,0,O) E W. Now, (1,O.O) E W and (0,1,O) E W, but (1,0,0) + (0,1,0) = (1,1,0) f W. .'. , W is not a subspace of R~.

d) W = {(x~,x~,x~)E R~IX~+X~+X~ = 1). Now, (1,0,0) and (0,1,O) E W, but (1,0,0) + (0,1,0) = (1,1,0) f W, since 1+1+0 = 2 # 1. .'., W isnot asubspaceof~~. E14) Firstly, {0) is non-empty. Secondly, 0 + 0 = 0 E {O} and a.0 = 0 E {O}, for any a E F. Thus, by Theorem 3, (0) is a subspace of V.

E15) [S] = {a+bx+cx21a,b,c E R). . .. 2x+3x3 f [S]. E16) [S] = {a(1,2,1) + P(2,1,0)(a,P E Rl = {(a+2P, ~~+P,Q)IQ,PE RI

a) (5,3,1) E [S] 3 a,@E R such that a+2P = 5,2a+P = 3, a = 1. Now,a=landa+2P=5-p=2.Butthen2a+P=2+2=4#3.

E17) [S] = {ax + b(x2+1) + c(x3-l))a,b,c E R) = {cx3 + bx2 + ax+ (b-c)la,b,c E R} a)x2+x+.lE[~],takinga=1,b= l,c=O. b)2x3 + x2 + 3x + 2f [S],since b = 1,c = 2andb-c # 2.

E18) If (x,y,z) E UCtW then (x,y,z) E U and (x,y,z) E W. Now. (x,y,z) E U =s z = 2x, and (x,y4) E W - y = 2x .'., Any element of UnW is of the form (x,2x,2x)-, x E R. That is, UnW = {(x,2x,2x)Ix E R}. Vector Spaces E20) Each of A + C and B + .C are subspaces of R3. NOW,for any (a,b,c) E R3, (a,b,c) = (a,b,-a-b) + (O,O,a+b+c) E A + C, and, (a,b,c) = (a,b,a) + (O,O,c-a) E B + C. Therefore, R3 = A+C and R3 = B+C Now AnC = {(x,y,z) E ~~lx+~+z= 0 and x = 0 = y) = {(0,0,0)), .'. A + C is a direct sum. AlsoBllC = {(x,y,z) E R31x = zandx = 0 = y)

= {(0,0,0)), .'. B+C is also a direct sum. E21) Firstly, A+BSC. Secondly, AnB = {x+iyly = 0 and x = 0) = (0). This means that the sum, A+B, is a direct sum. Finally, take any element x + iy E C. Thenx + iy = (x+iO) + iy~A+ B Therefore, C t A @ B.

E22) a) Since v E W, v + W = W b) v + W = {(x-2)(x2+1) + f(x)lf(x) E P and f(1) = 0) E23) vl + W= v2+W +- v~Ev~+ W=V~+ W =s vl E v2 + W =s v1 = v2 + W,for some w E W +-v, -v,=w~W+-v, -v,EW. E24) a) ~n~ coset of ((0,o))in R2 is (a,b) + {(0,0)) = {(a,b)). Thus two cosets (a,b) and (c,d) are disjoint, iff (a,b) # (c,d), i.e., iff (a,b) and (c,d) are distinct elements of R2. Thus, R2 = U{(a,b) + {(0,0)) I a,b E R) =~{(a,b)( a, b E R) b) Any coset (a,b) + R2 = R2, since (a,b) E R2. Thus, the only coset of R2 in R2 is R2 itself. So the disjoint union is only R2.

E25) P3/P2 = {(ax3+bx2+cx+d) + p21a,b,c,d E R). Now, {ax3+p21aE R) G PJP2 Conversely, any element of Pf12 is (ax3 + bx2+cx+d) + P,, where a,b,c,d E R. Now (ax3+bx2+cx+d) -ax3 = bx2+cx+d E P2. Therefore, (ax3+bx2+cx+d) + P2 = ax3 + P2 (by Theorem 13) E {ax3 + P2 a E R). Thus, P3/P2 = {ax3 + P21a E R).. E26) Firstly, note that W is a subspace of R2, and hence R2/W is meaningful. Now R2/W = {(a,b) + Wla,b E Rl For any (a,b) E R~,we have (a,b) - (a,O) = (0,b) E W .'., (a,b) + W = (a,O) + W Therefore, R2/W = {(a,O) + Wla E R)