Partial Solutions for Linear Algebra by Friedberg et al.

Chapter 1

John K. Nguyen

December 7, 2011

1.1.8. In any vector space V , show that (a+ b)(x+y) = ax+ay+ bx+ by for any x, y ∈ V and any a, b ∈ F .

Proof. Let x, y ∈ V and a, b ∈ F . Note that (a + b) ∈ F (it is a scalar). By (VS7 ), we have that(a + b)(x + y) = (a + b)x + (a + b)y. By (VS8 ), (a + b)x = ax + bx. Likewise, (a + b)y = ay + by. Thus,(a+ b)x+ (a+ b)y = ax+ bx+ ay + by. Finally, by (VS1 ), we have (a+ b)(x+ y) = ax+ ay + bx+ by.

1.1.9. Prove Corollaries 1 and 2 of Theorem 1.1 and Theorem 1.2(c).a) Corollary 1. The vector 0 described in VS3 is unique.b) Corollary 2. The vector y described in VS4 is unique.c) Theorem 1.2 (c). a0 = 0 for each a ∈ F . Note that 0 denotes the zero vector.

Proof of (a). Suppose 0 ∈ V such that x + 0 = x for all x ∈ V . By (VS3 ), x + 0 = x. It follows thatx+ 0 = x+ 0. By (VS1 ), 0 + x = 0 + x. We apply the Cancellation Law for Vector Addition (Theorem 1.1 )to obtain 0 = 0 as required.

Proof of (b). Assume y ∈ V such that x + y = 0. That is, both y and y are additive inverses of x (VS4 ).Since x + y = 0 and x + y = 0, we have that x + y = x + y. By (VS1 ), we have y + x = y + x. We applythe Cancellation Law to get y = y as required.

Proof of (c). Let a ∈ F . By (VS3 ), a0 = a(0 + 0). By (VS7 ), a0 = a0 + a0. By (VS1 ), 0 + a0 = a0 + a0.By the Cancellation Law, 0 = a0. Then, by (VS1 ), a0 = 0 as required.

1.2.21. Let V and W be vector spaces over a field F . Let Z = {(v, w) : v ∈ V ∧w ∈W}. Prove that Z is avector space over F with the operations (v1, w1) + (v2, w2) = (v1 + v2, w1 + w2) and c(v1, w1) = (cv1, cw1).

Proof. We will show that Z satisfy all of the field axioms.

We note that V and W are vector spaces. Let (v1, x1), (v2, x2), (v3, x3) ∈ V and (y1, w1), (y2, w2), (y3, w3) ∈W .

By (VS1 ), (v1, x1) + (v2, x2) = (v1 + v2, x1 + x2) = (v2, x2) + (v1, x1) = (v2 + v1, x2 + x1). Thus,v1 +v2 = v2 +v1. Likewise, (y1, w1) + (y2, w2) = (y1 +y2, w1 +w2) = (y2, w2) + (y1, w1) = (y2 +y1, w2 +w1).Thus, w1 +w2 = w2 +w1. Since v1 +v2 = v2 +v1 and w1 +w2 = w2 +w1, by definition, (v1, w1)+(v2, w2) =(v1 + v2, w1 + w2) = (v2 + v1, w2 + w1) = (v2, w2) + (v1, w1). We have shown that (VS1 ) holds for Z.

Next, by (VS2 ), ((v1, x1)+(v2, x2))+(v3, x3) = (v1, x1)+((v2, x2)+(v3, x3)) so (v1+v2)+v3 = v1+(v2+v3).Similarly, ((y1, w1)+(y2, w2))+(y3, w3) = (y1, w1)+((y2, w2)+(y3, w3)) so (w1 +w2)+w3 = w1 +(w2 +w3).It follows that ((v1 +v2)+v3, (w1 +w2)+w3) = (v1 +(v2 +v3), w1 +(w2 +w3)). Thus, ((v1, w1)+(v2, w2))+(v3, w3) = (v1, w1) + ((v2, w2) + (v3, w3)) where (v1, w1), (v2, w2), (v3, w3) ∈ Z. We have shown that (VS2 )holds for Z.

By (VS3 ), there exists (0v1, 0v2) ∈ V such that (xv, yv) + (0v1, 0v2) + 0v = (xv, yv) for all (xv, yv) ∈ V .Similarly, there exists (0w1, 0w2) ∈ W such that (xw, yw) + (0w1, 0w2) = (xw, yw) for all (xw, yw) ∈ W . Bydefinition, (0v1, 0w1) ∈ Z. Thus, (v, w) + (0v1, 0w1) = (v, w) and so (VS3 ) holds for Z.

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By (VS4 ), there exists (−v,−x) ∈ V such that (v, x) + (−v,−x) = (0v1, 0v2) for all (v, x) ∈ V . Like-wise, there exists (−y,−w) ∈ V such that (y, w) + (−y,−w) = (0w1, 0w2) for all (y, w) ∈ V . It follows thatfor all (v, w) ∈ Z, there exists (−v,−w) ∈ Z such that (v, w) + (−v,−w) = (0z1, 0z2) where (0z1, 0z2) ∈ Zso (VS4 ) holds for Z.

For all (v, x) ∈ V , 1 · (v, x) = (v, x) so 1 · v = v (by (VS5 )). Also, for all (y, w) ∈ W , 1 · (y, w) = (y, w) so1 · w = w. By definition (letting c = 1), 1 · (v, w) = (1 · v, 1 · w) = (v, w) so (VS5 ) holds for Z.

Let a, b ∈ F . By (VS6 ), for all (v, x) ∈ V , (ab)(v, x) = (a(b(v, x)). Likewise, for all (y, w) ∈ W ,(ab)(y, w) = (a(b(y, w)). Accordingly, we have (ab)v = a(bv) and (ab)w = a(bw). By definition, (ab)(v, w) =((ab)v, (ab)w). It follows from above that ((ab)v, (ab)w) = (a(bv), a(bw)) so (VS6 ) holds for Z.

Let c ∈ F . By (VS7 ), c((v1, x1) + (v2, x2)) = c(v1, x1) + c(v2, x2) = (cv1, cx1) + (cv2, cx2) which implies thatc(v1, v2) = (cv1, cv2). Likewise, c((y1, w1) + (y2, w2)) = c(y1, w1) + c(y2, w2) = (cy1, cw1) + (cy2, cw2) whichimplies c(w1, w2) = (cw1, cw2). Then, c((v1, w1)+(v2, w2)) = (c(v1+v2), c(w1+w2)) = (cv1+cv2, cw1+cw2) =c(v1, w1) + c(v2, w2) as required. So (VS7 ) holds for Z.

For all a, b ∈ F , by (VS8 ), (a + b)(v, x) = ((a + b)v, (a + b)x) = (av + bv, ax + bx) where (v, x) ∈ V .This implies (a + b)v = av + bv. Similarly, (a + b)(y, w) = ((a + b)y, (a + b)w) = (ay + by, aw + bw) whichimplies (a+ b)w = aw + bw. It follows that (a+ b)(v, w) = ((a+ b)v, (a+ b)w) = (av + bv, aw + bw) where(v, w) ∈ Z. We have shown that (VS8 ) holds for Z.

We have shown that Z is a vector space.

1.3.5. Prove that A+At is symmetric for any square matrix A.

Proof. Let A ∈Mnxn(F ). Note that a square matrix A is symmetric if At = A where At is the transpose ofA. That is, we wish to show that (A+At)t = A+At. By properties of tranpose, (A+At)t = At + (At)t =At +A = A+At as required.

1.3.6. Prove that tr(aA+ bB) = atr(A) + btr(B) for any A,B ∈Mnxn(F ).

Proof. Let A,B ∈ Mnxn(F ). By definition of trace, tr(A) =

n∑i=0

Aii and tr(B) =

n∑i=0

Bii where i is the row

and column of the matrix (because the matrix is square). Thus, by definition, tr(aA+bB) =

n∑i=0

(aA+bB)ii =

n∑i=0

(aAii+bBii). Then, by property of summation, we have

n∑i=0

aAii+

n∑i=0

bBii = a

n∑i=0

Aii+b

n∑i=0

Bii. Finally,

by definition of trace, atr(A) + btr(B) as required.

1.3.18. Prove that a subset W of a vector space V is a subspace of V if and only if 0 ∈W and ax+ y ∈Wwhenever a ∈ F and x, y ∈W .

Proof. (⇒) Suppose a subset W of a vector space V is a subspace with addition and scalar multiplicationdefined on V . By definition of subspace, we know that there exists 0 ∈ W . We also know that ax ∈ W(closed under scalar multiplication) whenever x ∈ W and a ∈ F . And since W is closed under addition, wehave that ax+ y ∈W for all y ∈W .

(⇐) Suppose 0 ∈ W and ax + y ∈ W for any x, y ∈ W and a ∈ F . Since we already know that 0 ∈ W , weonly need to show that W is closed under addition and scalar multiplication. Since a ∈ F , setting a = 1 ∈ F ,we have that 1 ·x+y = x+y ∈W so W is closed under addition. Since y ∈W and 0 ∈W , we let y = 0 ∈W .This means that ax+ 0 = ax ∈W so W is closed under scalar multiplication.

1.3.19. Let W1 and W2 be subspaces of a vector space V . Prove that W1∪W2 is a subspace of V if and onlyif W1 ⊆W2 or W2 ⊆W1.

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Proof. (⇒) Suppose W1 and W2 are subspaces of a vector space V and assume W1 ∪W2 is a subspace. Wewish to prove by contradiction so assume W1 6⊆W2 and W2 6⊆W1. Since W1 6⊆W2, there exists a ∈W1 buta 6∈ W2. Also, since W2 6⊆ W1, there exists b ∈ W2 but b 6∈ W1. Since W1 ∪W2 is a subspace, it must bethat (a+ b) ∈W1 or (a+ b) ∈W2. We have two cases.

Case I: Assume (a + b) ∈ W1. By the additive inverse property, a + b + (−a) = b ∈ W1 which contra-dicts our assumption that b 6∈W1.Case II: Assume (a + b) ∈ W2. By the additive inverse property, a + b + (−b) = a ∈ W2 which contradictsour assumption that a 6∈W2.

We have shown that if W1 ∪W2 is a subspace, then W1 ⊆W2 or W2 ⊆W1.

(⇐) Suppose W1 ⊆W2 or W2 ⊆W1. We have two cases.

Case I: Assume W1 ⊆W2. Then, by definition of union, W1 ∪W2 = W2 which is a subspace of V .Case II: Assume W2 ⊆W1. Then, by definition of union, W1 ∪W2 = W1 which is a subspace of V .

In either cases, we have that W1 ∪W2 is a subspace if W1 ⊆W2 or W2 ⊆W1.

1.3.23. Let W1 and W2 be subspaces of a vector space V .a) Prove that W1 +W2 is a subspace of V that contains both W1 and W2.b) Prove that any subspace of V that contains both W1 and W2 must also contain W1 +W2.

Proof of (a). We define W1 + W2 = {w1 + w2|w1 ∈ W1 ∧ w2 ∈ W2}. We first show that W1 + W2 is asubspace. First, since 0 ∈ W1 and 0 ∈ W2, we have that 0 = 0 + 0 ∈ W1 +W2. Next, if w1, w2 ∈ W1 +W2,there exists x ∈ W1 and y ∈ W2 such that w1 = x+ y. Similarly, there exists z ∈ W1 and t ∈ W2 such thatw2 = z+ t. Then, w1 +w2 = (x+y)+ (z+ t) = (x+z)+ (y+ t) ∈W1 +W2 (since W1 and W2 are subspaces,x+ z ∈W1 and y+ t ∈W2) so W1 +W2 is closed under addition. Lastly, since W1 and W2 are subspaces, wehave that qw1 ∈W1 and qw2 ∈W2 for any q ∈ F by definition. Thus, aw1 + aw2 = a(w1 + w2) ∈W1 +W2

so W1 +W2 is closed under scalar multiplication. Thus, W1 +W2 is a subspace of V .

We now show W1 ⊆ W1 + W2 and W2 ⊆ W1 + W2. By definition of subspace, we know that everyvector in W1 +W2 has an additive inverse which means that with any arbitrary vector we can always obtain0 ∈ W1 +W2. And since, 0 ∈ W1 and 0 ∈ W2 by definition of subspace, it follows that W1 ⊆ W1 +W2 andW2 ⊆W1 +W2.

Proof of (b). Suppose W is a subspace of V . Assume W1 ⊆ W and W2 ⊆ W . We wish to prove thatW1 + W2 ⊆ W . Let u ∈ W1 + W2. From part (a), we define u = x + y where x ∈ W1 and y ∈ W2. SinceW1 ⊆W , x ∈W . Similarly, since W2 ⊆W , y ∈W . And since W is a subspace, we know that u = x+y ∈Wwhere x, y ∈W . Since W was arbitrary, we have shown that any subspace of V that contains both W1 andW2 must also contain W1 +W2.

1.3.30. Let W1 and W2 be subspaces of a vector space V . Prove that V is the direct sum of W1 and W2 ifand only if each vector in V can be uniquely written as x1 + x2 where x1 ∈W1 and x2 ∈W2.

Proof. (⇒) Suppose W1 and W2 are subspaces of a vector space V . Assume that V is a direct sum of W1

and W2 (that is, V = W1 ⊕W2). Then, each vector in V can be written as x1 + x2 where x1 ∈ W1 andx2 ∈W2. We wish to show that x1 + x2 is unique so assume x1 + x2 = y1 + y2 where y1 ∈W1 and y2 ∈W2.It follows that x1 − y1 = y2 − x2. Because V = W1 ⊕W2, we know that W1 ∩W2 = {0}. Then, in knowingthat W1 and W2 are disjoint and that x1 − y1 ∈ W1 and y2 − x2 ∈ W2 (W1 and W2 are subspaces of V soboth are closed under addition), we have that x1 − y1 = 0 and y2 − x2 = 0. Thus, x1 = y1 and x2 = y2 sox1 + x2 is unique.

(⇐) Let W1 and W2 be subspaces of a vector space V . Suppose each vector in V can be uniquely writtenas x1 + x2 where x1 ∈ W1 and x2 ∈ W2. Then, it follows that W1 + W2 = V . Next, we let t be in theintersection of W1 and W2 (i.e. t ∈ W1 ∩W2). Since W1 and W2 are subspaces, we know that 0 ∈ W1 and

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0 ∈W2. Then, t = t+ 0 where t ∈W1 and 0 ∈W2. Also, t = 0 + t where 0 ∈W1 and t ∈W2. It must followthat t = 0 so W1 ∩W2 = {0} since t ∈ W1 ∩W2. Since W1 +W2 = V and W1 ∩W2 = {0}, by definition ofdirect sum, we have V = W1 ⊕W2 as required.

We conclude that V is the direct sum of W1 and W2 if and only if each vector in V can be uniquelywritten as x1 + x2.

1.3.31. Let W be a subspace of a vector space V over a field F . For any v ∈ V the set {v}+W = {v +w :w ∈ W} is called the coset of W containing v. It is customary to denote this coset by v + W rather than{v}+W .

(a) Prove that v +W is a subspace of V if and only if v ∈W .(b) Prove that v1 +W = v2 +W if and only if v1 − v2 ∈W .(c) Prove that the preceding operations are well defined; that is, show that if v1 + W = v1 + W andv2 + W = v2 + W , then (v1 + W ) + (v2 + W ) = (v1 + W ) + (v2 + W ) and a(v1 + W ) = a(v1 + W )for all a ∈ F .(d) Prove that the set S is a vector space with the operations defined in (c).This vector space is called thequotient space of V modulo W and is denoted by V/W .

Proof of (a). (⇒) Suppose W is a subspace of a vector space V over a field F . Assume v+W is a subspaceof V . Because v +W is a subspace, we have that v ∈ v +W since 0 ∈ v +W . Since v ∈ v +W and v +Wis a subspace, it follows that v + v ∈ v + W (since v + W is closed under addition). We wish to prove bycontradiction so suppose v 6∈ W . Since v 6∈ W , we have by definition that v + v 6∈ v + W , a contradiction.Thus, v ∈W as required.

(⇐) Assume v ∈W . We wish to prove that v+W is a subspace of V so we check the three parts of the def-inition of subspace. Since W is a subspace, it is closed under scalar multiplication and so −v = (−1)v ∈W .Then, 0 = v+ (−v) ∈ v+W by definition of v+W . Next, suppose x = v+a ∈ v+W and y = v+ b ∈ v+Wfor a, b ∈W . Then, x+ y = (v + a) + (v + b) = v + (a+ v + b). Since we had assumed v ∈W and a, b ∈W ,it follows that (a+ v+ b) ∈W . So by definition, x+ y = v+ (a+ v+ b) ∈ v+W where (a+ v+ b) ∈W andv ∈W so v+W is closed under addition. Finally, assume f ∈ F . Then, fx = fv+ fa which we can rewriteas fx = v + (−v) + fv + fa. Since a ∈ W and W is a subspace, fa ∈ W . Likewise, since v ∈ W , fv ∈ W .And since we have already stated that −v ∈W , we know that (−v)+cv+ca ∈W because W is closed underaddition. Thus, fx ∈ v+W so v+W is closed under scalar multiplication. We have shown that 0 ∈ v+W ,v+W is closed under addition and v+W is closed under scalar multiplication and so v+W is a subspace of V .

We conclude that v +W is a subspace of V if and only if v ∈W .

Proof of (b). (⇒) Suppose W is a subspace of a vector space V over a field F . Assume v1 +W = v2 +W .Suppose x ∈ v1 +W and y ∈ v1 +W . By definition of coset, we have that x = v1 +a where a ∈W . Likewise,we have that y = v2 + b where b ∈W . Since v1 +W = v2 +W , we have that x = y. That is, v1 + a = v2 + bwhich can be rewritten as v1−v2 = b−a. Since a ∈W , b ∈W and W is a subspace, we have that b−a ∈W .Thus, since v1 − v2 = b− a, it follows that v1 − v2 ∈W as required.

(⇐) Now suppose v1 − v2 ∈ W . We wish to show that v1 + W = v2 + W . That is, v1 + W ⊆ v2 + W andv2 +W ⊆ v1 +W .

(⊆) Let x ∈ v1 + W and y ∈ v2 + W . By definition of coset, we have that x = v1 + a where a ∈ W .Likewise, we have that y = v2 + b where b ∈ W . Since v1 − v2, a ∈ W , we set b = (v1 − v2) + a. Then,y = v2 + (v1 − v2) + a = v1 + a. Since x = v1 + a, we have that x = y so x ∈ v2 +W . Because x ∈ v1 +Wand x ∈ v2 +W , v1 +W ⊆ v2 +W .

(⊇) Let x ∈ v1 + W and y ∈ v2 + W . Again, by definition of coset, we have that x = v1 + a wherea ∈ W . Likewise, we have that y = v2 + b where b ∈ W . Since v1 − v2 ∈ W and W is a subspace, wehave that v2 − v1 ∈ W (it is clear that v2 + (−1)v1 ∈ W because W is closed under addition and scalar

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multiplication). Now, since v2 − v1, b ∈W , we set a = (v2 − v1) + b. Then, x = v1 + (v2 − v1) + b = v2 + b.Since y = v2 + b, x = y so y ∈ v1 +W . Since y ∈ v2 +W and y ∈ v1 +W , v2 +W ⊆ v1 +W .

Because v1 +W ⊆ v2 +W and v2 +W ⊆ v1 +W , we conclude that v1 +W = v2 +W as required.

Proof of (c). Assume W is a subspace of a vector space V over a field F . Suppose v1 + W = v1 + W andv2 +W = v2 +W . We wish to show (v1 +W ) + (v2 +W ) = (v1 +W ) + (v2 +W ). However, from part(b)we know this is equivalent to (v1 + v2)− (v1 + v2) ∈W . Since v1 +W = v1 +W and v2 +W = v2 +W , bypart (b), we have that v1 − v1 ∈ W and v2 − v2 ∈ W . Since W is a subspace, it is closed under addition so(v1 − v1) + (v2 − v2) ∈W . Rearranging terms, (v1 + v2)− (v1 + v2) ∈W as required.

Now we wish to prove that a(v1 + W ) = a(v1 + W ) for all a ∈ F . However, we apply part (b) whichmeans showing av1 − av1 ∈ W is equivalent. From above, we already have v1 − v1 ∈ W . Since W is asubspace, it is closed under scalar multiplication so we have that a(v1 − v1) ∈ W for all a ∈ F . By thedistribution law, we conclude that av1 − av1 ∈W as required.

Proof of (d). To show that set S is a vector space with the operations defined in (c), we must verify the fieldaxioms.

For (V S1), let v1 +W ∈ S and v2 +W ∈ S where v1, v2 ∈ V . Then, (v1 +W ) + (v2 +W ) = (v1 + v2) +Wand (v2 + W ) + (v1 + W ) = (v2 + v1) + W by the definition in part (c). Since v1, v2 ∈ V , by (V S1), weknow that v1+v2 ∈ V and v2+v1 ∈ V so v1+v2 = v2+v1. Thus, (v1+v2)+W = (v2+v1)+W so (V S1) holds.

Now suppose v1 + W ∈ S, v2 + W ∈ S and v3 + W ∈ S where v1, v2, v3 ∈ V . Then, by the definitionin part (c), ((v1 + W ) + (v2 + W )) + (v3 + W ) = ((v1 + v2) + W ) + (v3 + W ) = ((v1 + v2) + v3) + W =(v1 + (v2 + v3)) +W = (v1 +W ) + ((v2 + v3) +W ) = (v1 +W ) + ((v2 +W ) + (v3 +W )) which shows that(V S2) holds.

Suppose 0 + W ∈ S (we can assume this since 0 ∈ V ). Also, let v + W ∈ S for some v ∈ V . Then,(0 +W ) + (v +W ) = (0 + v) +W = v +W . Thus, (V S3) holds.

Next, let v + W ∈ S where v ∈ V . Let −v + W ∈ S where, since V is closed under scalar multiplica-tion, −v ∈ V . Then, v +W + (−v) +W = (v − v) +W = 0 +W where 0 ∈ S. Therefore, (V S4) holds.

We now verify (V S5). Let v + W ∈ S for some v ∈ V . Then, by the definition in part (c), 1(v + W ) =(1v) +W = v +W as required so (V S5) holds.

Let a, b ∈ F and suppose v+W ∈ S for some v ∈ V . Then, (ab)(v+W ) = (ab)v+W = a(bv)+W = a(bv+W )so (V S6) holds.

For (V S7), let v1+W ∈ S and v2+W ∈ S where v1, v2 ∈ V . Choose a ∈ F . Then, a((v1+W )+(v2+W )) =(a(v1 + v2)) +W = (av1 + av2) +W = (av1 +W ) + (av2 +W ) = a(v1 +W ) + a(v2 +W ) as required. Wehave shown that (V S7) holds.

Finally, let v+W ∈ S where v ∈ V . Pick a, b ∈ F . Then, (a+b)(v+W ) = ((a+b)v)+W = (av+bv)+W =(av +W ) + (b+W ) = a(v +W ) + b(v +W ) so (V S8) holds.

In conclusion, since all of the field axioms hold, S is a vector space.

1.4.13. Show that if S1 and S2 are subsets of a vector space V such that S1 ⊆ S2, then span(S1) ⊆ span(S2).In particular, if S1 ⊆ S2 and span(S1) = V , deduce that span(S2) = V .

Proof. Suppose S1 and S2 are subsets of a vector space V and that S1 ⊆ S2. We wish to show that everyelement of span(S1) is contained in span(S2). Choose z ∈ span(S1). By definition of span, there existsx1, x2, ..., xn ∈ S1 such that z = a1x1 + a2x2 + ... + anxn where a1, a2, ..., an ∈ F . But, since S1 ⊆ S2,

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x1, x2, ..., xn ∈ S2 also which means that z ∈ span(S2) by definition of linear combination. Since z wasarbitrary, we have that span(S1) ⊆ span(S2).

Suppose span(S1) = V . We now show that span(S2) = V . Since span(S1) = V , from above, V ⊆ span(S2).However, we have that S2 ⊆ V which means that span(S2) ⊆ V . That is, x1, x2, ..., xn ∈ S2 is also containedin V so by definition, we know that span(S2) ⊆ V . Since V ⊆ span(S2) and span(S2) ⊆ V , we have thatspan(S2) = V .

1.4.15. Let S1 and S2 be subsets of a vector space V . Prove that span(S1∩S2) ⊆ span(S1)∩span(S2). Givean example in which span(S1 ∩ S2) and span(S1) ∩ span(S2) are equal and one in which they are unequal.

Proof. Suppose S1 and S2 be subsets of a vector space V . Let v ∈ span(S1 ∩ S2), Then, by definition ofspan, there exists x1, x2, ..., xn ∈ S1 ∩ S2 such that v = a1x1 + a2x2 + ... + anxn where a1, a2, ..., an ∈ F .Since x1, x2, ..., xn ∈ S1 ∩S2, by definition of set intersection x1, x2, ..., xn ∈ S1 and x1, x2, ..., xn ∈ S2. Fromthis, we know that v = a1x1 +a2x2 + ...+anxn ∈ span(S1) and v = a1x1 +a2x2 + ...+anxn ∈ span(S2). Bydefinition of set intersection, since v ∈ span(S1) and v ∈ span(S2), we have that v ∈ span(S1) ∩ span(S2).Since v was arbitrary, we have shown that span(S1 ∩ S2) ⊆ span(S1) ∩ span(S2).

Examples. Suppose V = R2, S1 = {(1, 3)} and S2 = {(2, 7)}. Then, S1 ∩ S2 = ∅ so span(S1 ∩ S2) = {0} bydefinition of subspace. Next, we have that span(S1)∩span(S2) = {0} because span(S1) = {(1a, 3a) : a ∈ R}and span(S2) = {(2b, 7b) : b ∈ R} have no elements in common. In this example, span(S1 ∩ S2) =span(S1) ∩ span(S2).

Now consider V = R2, S1 = {(8, 4)} and S2 = {(4, 2)}. Again, we have that S1 ∩ S2 = ∅ so span(S1 ∩S2) = {0} by definition of subspace. However, since span(S1) = {(8a, 4a) : a ∈ R} and span(S2) ={(4b, 2b) : b ∈ R}, we know that span(S1) ∩ span(S2) 6= {0} (since span(S1) = span(S2)). In this example,span(S1 ∩ S2) 6= span(S1) ∩ span(S2)

1.5.9. Let u and v be distinct vectors in a vector space V . Show that {u, v} is linearly dependent if and onlyif u or v is a multiple of the other.

Proof. (⇒) Assume u and v are distinct vectors in a vector space V . Suppose {u, v} is linearly dependent.Assume a, b ∈ F and that they are not all are zero. Then, au+ bv = 0 which can be rewriten as u = − b

av.Since both a, b ∈ F can’t be zero, we can assume a 6= 0. Thus, we see that v is a multiple of u. On the otherhand, if we rewritten the equation to v = −a

bu and assume that b 6= 0, we have that u is a multiple of v. Inconclusion, we have that u or v is a multiple of the other.

(⇐) Now assume that u or v is a multiple of other. Then, we have that u = av or v = bu for a, b ∈ F . Bythe distributive law of logic, we have two cases.

Case I: Suppose u = av. Then, 0 = av − u = av + (−1)u. Since −1 ∈ F , by definition of linear de-pendent, we have {u, v} is linear dependent.

Case II: Suppose v = bu. Then, 0 = bu − v = bu + (−1)v. Since −1 ∈ F , by definition of linear de-pendent, we have {u, v} is linear dependent.

In both cases, we have that {u, v} is linear dependent.

In conclusion, we have shown that {u, v} is linearly dependent if and only if u or v is a multiple of theother

1.5.13. Let V be a vector space over a field of characteristic not equal to two.

(a) Let u and v be distinct vectors in V . Prove that {u, v} is linearly independent if and only if {u+v, u−v}is linearly indepedent.(b) Let u,v, and w be distinct vectors in V . Prove that {u, v, w} is linearly independent if and only if{u+ v, u+ w, v + w} is linearly independent.

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Proof of (a). (⇒) Suppose V is a vector space over a field of characteristic not equal to two. Assume u andv are distinct vectors in V . Suppose {u, v} is linearly independent. We wish to show that {u+ v, u− v} islinearly independent which mean a(u+ v) + b(u− v) = au+ av+ bu− bv = (a+ b)u+ (a− b)v = 0 for somea, b ∈ F such that a = b = 0. Since we know that {u, v} is linearly independent, a + b = 0 and a − b = 0.Since a+ b = 0, we know that b = −a which implies a− b = a− (−a) = a+ a = 2a = 0. Because the fieldcharacteristic is not equal to two, we have a = 0. In a similar argument, since a− b = 0, we know that a = bwhich implies that a+ b = b+ b = 2b = 0. And since the field characteristic is not equal to two, b = 0. Thus,since a = b = 0, we have shown that {u+ v, u− v} is linearly independent.

(⇐) Now assume {u+ v, u− v} is linearly independent. Then, a(u+ v) + b(u− v) = 0 for some a, b ∈ F anda = b = 0. Rearranging terms, au + av + bu + bv = (a + b)u + (a − b)v = 0. We prove by contradiction soassume that {u, v} is linearly dependent. That is, cu+ dv = 0 for some c, d ∈ F and that all cannot be zero.Since (a+ b)u+ (a− b)v = 0 and cu+ dv = 0, c = a+ b and d = a− b. Since {u, v} is linearly dependent,we have that c 6= 0 or d 6= 0. By the distributive law of logic, we have two cases.

Case I: Assume that c 6= 0. Since c = a + b, we have a + b 6= 0 which implies {u + v, u − v} is lin-early dependent because a and b cannot both be 0 since a+ b 6= 0 for a(u+v)+ b(u−v) = 0 (our assumptionthat c 6= 0 would not hold if they are both zero).

Case II: Assume that d 6= 0. Since d = a − b, we have a − b 6= 0 which implies {u + v, u − v} is lin-early dependent because a and b cannot both be 0 since a− b 6= 0 for a(u+v)+ b(u−v) = 0 (our assumptionthat c 6= 0 would not hold if they are both zero).

In either case, we have that {u+ v, u− v} is linearly dependent which is a contradiction. Therefore, {u, v}is linearly independent.

In conclusion, {u, v} is linearly independent if and only if {u+ v, u− v} is linearly indepedent

Proof of (b). (⇒) Let u, v and w be distinct vectors in V . Assume V is a vector space over a field of char-acteristic not equal to two. Suppose Assume that u, v, w is linearly independent which, by definition, meansau+bv+cw = 0 for some a, b, c ∈ F and a = b = c = 0. We wish to prove that {u+v, u+w, v+w} is linearlyindependent. Then, d(u+v)+e(u+w)+f(v+w) = du+dv+eu+ew+fv+fw = (d+e)u+(d+f)v+(e+f)w = 0for some d, e, f ∈ F and d = e = f = 0. Since u, v, w is linearly independent, d + e = 0, d + f = 0 ande + f = 0. We then solve for each variables. Since d + e = 0, d = −e. Then, d + f = −e + f = 0 whichimplies f = e. Since f = e, e+ f = e+ e = 2e = 0. Since the field characteristic is not two, we have e = 0.In a similar argument, since f = e, e + f = f + f = 2f = 0 which implies f = 0 (since char(F ) 6= 2).Lastly, e + f = 0 implies that e = −f . Since e = −f , d + e = d − f = 0 which implies d = f and sod+ f = d+ d = 2d = 0 which mean d = 0 (again, since char(F ) 6= 2). Thus, since d = f = e = 0, we havethat {u+ v, u+ w, v + w} is linearly independent.

(⇐) Now assume that {u+ v, u+w, v+w} is linearly independent. Then, a(u+ v) + b(u+w) + c(v+w) = 0where a, b, c ∈ F and a = b = c = 0. Rearranging terms we have a(u + v) + b(u + w) + c(v + w) =au+ av+ bu+ bw+ cv+ cw = (a+ b)u+ (a+ c)v+ (b+ c)w = 0. We will prove by contradiction so assume{u, v, w} is linearly dependent. Then, du + ev + fw = 0 where d, e, f ∈ F and not all are zero. It followsthat a + b = d, a + c = e and b + c = f because (a + b)u + (a + c)v + (b + c)w = 0 and du + ev + fw = 0.Next, because we assumed {u, v, w} is linearly dependent, we know that d, e, f ∈ F are not all zero so d 6= 0or e 6= 0 or f 6= 0. By the distributive law of logic, we have three cases.

Case I: Suppose d 6= 0. Since a + b = d, a + b 6= 0 which implies {u + v, u + w, v + w} is linearly de-pendent because a and b cannot both be equal to zero (recall that we’d assumed a = b = c = 0).

Case II: Suppose e 6= 0. Since a + c = e, a + c 6= 0 which implies {u + v, u + w, v + w} is linearly de-pendent because a and c cannot both be equal to zero.

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Case III: Suppose f 6= 0. Since b + c = f , b + c 6= 0 which implies {u + v, u + w, v + w} is linearlydependent because b and c cannot both be equal to zero.

Thus, we have that either a, b, or c is not equal to zero which means that {u + v, u + w, v + w} is lin-early dependent so we have a contradiction. Therefore, {u, v, w} must be linearly independent.

In conclusion, {u, v, w} is linearly independent if and only if {u+v, u+w, v+w} is linearly independent.

1.6.11. Let u and v be distinct vectors of a vector space V . Show that if {u, v} is a basis for V and a and bare nonzero scalars, then both {u+ v, au} and {au, bv} are also bases for V .

Proof. Let u and v be distinct vectors of a vector space V . Suppose {u, v} is a basis for V and a and b arenonzero scalars. We wish to show that {u+v, au} and {au, bv} are linearly independent and that they span V .

We begin by proving for {u + v, au}. First, to show that {u + v, au} is linearly independent, we needto show that for x(u + v) + y(au) = 0, x = y = 0 for scalars x, y ∈ F . Rearranging terms, we noticex(u+ v) + y(au) = xu+xv+ yau = (x+ ya)u+ (x)v = 0. Now since {u, v} is linearly independent by defini-tion of basis, it follows that wu+ qv = 0 and w = q = 0 for scalars w, q ∈ F . It follows that x+ ya = w = 0and x = q = 0. Then, because x = 0, ya = 0. Since we know that a is a nonzero scalar, it is clear that y = 0.Thus, the only solution to x(u+ v) + y(au) = 0 is x = y = 0 so {u+ v, au} is linearly independent. Second,to show that {u+ v, au} span V , by definition of basis, we need to show for any p ∈ V , x(u+ v) + y(au) = pwhere x, y ∈ F . Again, rearranging terms, we have (x+ ya)u+ (x)v = p. By definition of basis, {u, v} spanV which implies that for any p ∈ V , xu + yv = p, where scalars x, y ∈ F . It follows that x + ya = x andx = y. Since x = y, we solve for y to obtain y = x−y

a which is defined since a is nonzero. Thus, we have that

{u + v, au} span V (clearly, x(u + v) + y(au) = y(u + v) + x−ya (au) = p). Because {u + v, au} is linearly

independent and {u+ v, au} span V , {u+ v, au} is a basis for V .

Next, we will prove that {au, bv} is a basis for V . First, we show that {au, bv} is linearly independent.That is, m(au) + n(bv) = 0 is satisfy since m = n = 0. Rearranging terms, we obtain (ma)u + (nb)v = 0.Because {u, v} is linearly independent by definition of basis, it follows that hu + jv = 0 and h = j = 0 forscalars w, q ∈ F . This implies that ma = h = 0and nb = j = 0. Since we know that a and b are nonzeroscalars, m = n = 0 so {au, bv} is linearly independent. Second, we will show that {au, bv} span V . That is,by definition of span, w(au)+ q(bv) = f for any f ∈ V . Again, rearranging terms, we have (wa)u+(qb)v = fSince {u, v} span V , xu + yv = f which implies that wa = x and qb = y. Then, it follows that w = x

a and

q = yb so w(au) + q(bv) = x

a (au) + yb (bv) = f . Clearly, {au, bv} span V . Thus, since {au, bv} is linearly

independent and since it span V , by definition of basis, {au, bv} is a basis of V .

In conclusion, we have shown that if {u, v} is a basis for V and a and b are nonzero scalars, then both{u+ v, au} and {au, bv} are also bases for V .

1.6.22. Let W1 and W2 be subspaces of a finite-dimensional vector space V . Determine necessary andsufficient conditions of W1 and W2 so that dim(W1 ∩W2) = dim(W1).

Proof. We believe that dim(W1 ∩W2) = dim(W1) if and only if W1 ⊆W2. We will prove this claim.

(⇒) Let W1 and W2 be subspaces of a finite-dimensional vector space V . Suppose dim(W1∩W2) = dim(W1).Let β be a basis for W1∩W2. By definition of set intersection, we know that W1∩W2 ⊆W1 so we can extendβ to be a basis for W1. It follows that for any x ∈W1, x ∈W1∩W2. This implies that W1 ⊆W1∩W2. SinceW1 ∩W2 ⊆ W1 and W1 ⊆ W1 ∩W2, we have that W1 ∩W2 = W1. Now, by definition of set intersection(specifically, the property of subset intersection), W1 ∩W2 = W1 implies that W1 ⊆W2.

(⇐) Now assume that W1 ⊆ W2. Let β be a basis for W1 ∩W2. Since W1 ⊆ W2, by the definition of setintersection (property of subset intersection), it follows that W1∩W2 = W1. Since W1∩W2 = W1 (which tellsus that W1∩W2 ⊆W1), β can be extended to be a basis for W1. This implies that dim(W1∩W2) = dim(W1).

We have shown that dim(W1 ∩W2) = dim(W1) if and only if W1 ⊆W2.

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1.6.29. (a) Prove that if W1 and W2 are finite-dimensional subspaces of a vector space V , then the subspaceW1 +W2 is finite-dimensional, and dim(W1 +W2) = dim(W1) + dim(W2)− dim(W1 ∩W2).

(b) Let W1 and W2 be finite-dimensional subspaces of a vector space V , and let V = W1 + W2. Deducethat V is the direct sum of W1 and W2 if and only if dim(V ) = dim(W1) + dim(W2).

Proof of (a). Suppose W1 and W2 are finite-dimensional subspaces of a vector space V . Let β be a ba-sis for W1 ∩W2 such that β = {u1, u2, ..., uk} (since we know W1 and W2 are finite-dimensional). SinceW1 ∩ W2 ⊆ W1, we extend β to a basis of W1 so γ = {u1, u2, ..., uk, v1, v2, ..., vm}. Likewise, µ ={u1, u2, ..., uk, w1, w2, ..., wp}. We wish to show that τ = {u1, u2, ..., uk, v1, v2, ..., vm, w1, w2, ..., wp} whichimplies that W1 + W2 is finite-dimensional (as it contains k = m = p number of vectors) and thatdim(W1+W2) = k+m+p = k+m+p+(k−k) = (k+m)+(k+p)−k = dim(W1)+dim(W2)−dim(W1∩W2).By definition of basis, we show that τ span W1 +W2 and that it is linearly independent.

Note that γ span W1 and µ span W2 as they are both bases. That is, for any w1 ∈ W1, a1u1 +... + akuk + b1v1 + ... + bmvm = w1 and, for any w2 ∈ W1, a1u1 + ... + akuk + c1w1 + ... + cpwp =w2 for scalars ak, bm, and cp respectively (Theorem 1.8 ). It is obvious that τ span W1 + W2 becausea1u1 + ...+ akuk + b1v1 + ...+ bmvm + c1w1 + ...+ cpwp = w1 + w2 for any w1 + w2 ∈W1 +W2 (recall ourgiven definition of W1 +W2 in problem 1.3.23 in Homework Set 1 ).

To show that τ is linearly independent, we will prove by contradiction so assume that it is not (τ is linearly de-pendent). Because γ is linearly independent (by definition of basis), we know that for a1u1+...+akuk+b1v1+...+bmvm = 0, ak = bm = 0. Similarly, since µ is linearly indepdenent, for a1u1+...+akuk+c1w1+...+cpwp =0, we know that ak = cp = 0. But, we had assumed that τ is linearly dependent which implies that fora1u1 + ...+ akuk + b1v1 + ...+ bmvm + c1w1 + ...+ cpwp = 0, we have that coefficients ak, bm, and cp are notall zeros. Thus, we have a contradiction so τ must be linearly independent.

By definition of basis, because τ span W1 + W2 and is linearly independent, we have that τ is a basisfor W1 + W2. Thus, it is finite-dimensional and dim(W1 + W2) = (k + m) + (k + p) − k = dim(W1) +dim(W2)− dim(W1 ∩W2) as required.

Proof of (b). Let W1 and W2 be finite-dimensional subspaces of a vector space V , and let V = W1 +W2.

(⇒) Suppose that V = W1 ⊕ W2. We wish to prove that dim(V ) = dim(W1) + dim(W2). BecauseV = W1 ⊕W2, we know that W1 ∩W2 = {0} which implies that dim(W1 ∩W2) = 0 as W1 and W2 have nounique vectors in common (in fact, they don’t have any vectors in common). So, noting that V = W1 +W2,following from from part (a), we have dim(V ) = dim(W1)+dim(W2)−dim(W1∩W2) = dim(W1)+dim(W2)as required.

(⇐) Suppose dim(V ) = dim(W1) + dim(W2). We wish to prove that V = W1 ⊕ W2. Since we knowthat V = W1 + W2, we only need to show that W1 and W2 are disjoint (i.e., W1 ∩W2 = {0}). From part(a), we know that dim(V ) = dim(W1) +dim(W2)−dim(W1 ∩W2). However, from our assumption, we havethat dim(W1 ∩W2) = 0 which means, by definition of dimension, there isn’t any unique number of vectorsin each basis for W1 ∩W2. This implies that W1 and W2 are disjoint. Therefore, since V = W1 + W2 andW1 ∩W2 = {0}, by definition of direct sum, V = W1 ⊕W2 as required.

We have shown that V is the direct sum of W1 and W2 if and only if dim(V ) = dim(W1) + dim(W2)

1.6.33. (a) Let W1 and W2 be subspaces of a vector space V such that V = W1⊕W2. If β1 and β2 are basesfor W1 and W2, respectively, show that β1 ∩ β2 = ∅ and β1 ∪ β2 is a basis for V .

(b) Conversely, let β1 and β2 be disjoint bases for subspaces W1 and W2, respectively, of a vector spaceV . Prove that if β1 ∪ β2 is a basis for V , then V = W1 ⊕W2.

Proof of (a). Let W1 and W2 be subspaces of a vector space V such that V = W1 ⊕ W2. Suppose β1and β2 are bases for W1 and W2 respectively. Define β1 = {u1, u2, ..., un} and β2 = {v1, v2, ..., vm}.

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Then, by definition of basis, β1 and β2 both are linearly independent and span W1 and W2 respectively.Since we know that V = W1 ⊕ W2, we have that W1 ∩ W2 = ∅. By Theorem 1.18, we know thatspan(β1) = W1 and span(β2) = W2. Since span(β1) = W1, span(β2) = W2 and W1 ∩ W2 = ∅, it fol-lows that span(β1) ∩ span(β2) = ∅ which implies that β1 ∩ β2 = ∅.

Now, based on our definition of β1 and β2, we define β1 ∪ β2 = {u1, u2, ..., un, v1, v2, ..., vm}. We firstshow that β1 ∪β2 is linearly independent. However, we will prove by contradiction so assume that β1 ∪β2 islinearly dependent. That is, a1u1 + ...+ anun + b1v1 + ...+ bmvm = 0 where scalars a1, ..., an and b1, ..., bmare not all zero. Because, β1 is linearly independent, we have that a1u1 + ... + anun = 0 where scalarsa1 = a2 = ... = an = 0. Likewise, for β2, b1v1 + ...+ bnvn = 0 where scalars b1 = b2 = ... = bm = 0. Thus, acontradiction so it is obvious that β1 ∪ β2 is linearly independent.

Second, we need to show that β1 ∪ β2 span V . Note the Corollary on page 51 of the text which statesif W is a subspace of a finite-dimensional vector space V , then any basis for W can be extended to a basisfor V . Recall that W1 and W2 are subspaces of a vector space V . Therefore, we extend β1 and β2 to basesfor V . Thus, since β1 and β2 span V (by definition of basis), it follows that β1 ∪ β2 span V .

Thus, since β1 ∪ β2 is linearly independent and span V , by definition of basis, we have that β1 ∪ β2 isa basis for V . In conclusion, we have proved that if β1 and β2 are bases for W1 and W2, respectively, showthat β1 ∩ β2 = ∅ and β1 ∪ β2 is a basis for V .

Proof of (b). Let β1 and β2 be disjoint bases for subspaces W1 and W2, respectively, of a vector space V(i.e., β1 ∩ β2 = ∅). Assume β1 ∪ β2 is a basis for V . We wish to prove that V = W1 ⊕W2. That is, we needto show that W1 ∩W2 = ∅ and V = W1 +W2.

Since β1 ∩ β2 = ∅, we have that span(β1 ∩ β2) = {0}. And since β1 is a basis for W1, β1 span W1.Likewise, we have that β2 span W2. Then, by definition of span, span(β1) = W1 and span(β2) = W2. Notefrom problem 1.4.15 of Homework Set 2 that span(W1 ∩ W2) ⊆ span(W1) + span(W2). It follows thatspan(β1) ∩ span(β2) = W1 ∩W2 = ∅ as β1 ∩ β2 = ∅.

Now we wish to show that V = W1 + W2. We know that span(β1) = W1 and span(β2) = W2. Alsorecall that β1 ∪ β2 is a basis for V . Then, by definition, span(β1 ∪ β2) = V . Suppose each vector inspan(β1 ∪ β2) can be uniquely written as w1 + w2 where w1 ∈ span(β1) and w2 ∈ span(β2). Then, byproblem 1.3.30 of Homework Set 2, it follows that V = W1 +W2.

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Partial Solutions for Linear Algebra by Friedberg et al.

Chapter 2

John K. Nguyen

December 7, 2011

2.1.14. Let V and W be vector spaces and T : V →W be linear.

(a) Prove that T is one-to-one if and only if T carries linearly independent subsets of V onto linearlyindependent subsets of W .

(b) Suppose that T is one-to-one and that S is a subset of V . Prove that S is linearly independent ifand only if T (S) is linearly independent.

(c) Suppose β = {v1, v2, ..., vn} is a basis for V and T is one-to-one and onto. Prove that T (β) ={T (v1), T (v2), ..., T (vn)} is a basis for W .

Proof of (a). Suppose V and W be vector spaces and T : V →W is linear.

(⇒) Suppose T is one-to-one. Let S be a linearly independent subset of V . We wish to prove by contradic-tion so assume that T (S) is linearly dependent. Since S is linearly independent, for a1v1 + ... + anvn = 0,a1 = a2 = ... = an = 0. By definition of linear transformation from V to W , we have that if T is linear,then T (0) = 0. In this case, T (a1v1 + ... + anvn) = T (0) = 0. Note, by Theorem 2.5, this is equivalent toT being one-to-one as we had assumed. But we have that T (S) is linearly dependent which implies that forvectors v1, v2, ..., vn ∈ S, a1T (v1) + ... + anT (vn) = 0 where scalars a1, a2, ..., an are not all zero. Clearly,this contradiction because this would mean a1v1 + ...+ anvn 6= 0 which implies T (a1v1 + ...+ anvn) 6= T (0),contradicting the assumption that T is one-to-one. Therefore, T (S) must be linearly independent. Thus,since S was arbitrary, T carries linearly independent subsets of V onto linearly independent subsets of W .

(⇐) Suppose that T carries linearly independent subsets of V onto linearly independent subsets of W .We prove by contradiction so assume T is not one-to-one. Then, by definition of one-to-one, for somea, b ∈ V such that T (x) 6= T (y). It follows that T (x) − T (y) = 0 which implies that x − y ∈ N(T ) becauseT : V →W is linear.

Proof of (b). Let V and W be vector spaces and T : V → W be linear. Suppose that T is one-to-one andthat S is a subset of V .

(⇒) Suppose that S is linearly independent. Then, by part (a), we have that T (S) is linearly indepen-dent and so we’re done.

(⇐) Now suppose that T (S) is linearly independent. We wish to prove by contradiction so assume that Sis linearly dependent. This implies that for v1, v2, ..., vn ∈ S, a1v1 + ...+ anvn = 0 where scalars a1, a2, ...anare not all zero. Since T : V → W is linear, we have that a1T (v1) + ...+ anT (vn) = 0, again, where scalarsa1, a2, ..., an are not all zero. However, we had assumed that T (S) is linearly independent so we have acontradiction. Thus, S is linearly independent as required.

Proof of (c). Suppose β = {v1, v2, ..., vn} is a basis for V and T is one-to-one and onto. We wish to showthat T (β) is linearly independent and span W . Since T is one-to-one and β is linearly independent (by

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definition of basis), by part (b), T (β) is linearly independent. Next, since β is a basis for V , by Theorem 2.2,R(T ) = span(T (β)). That is, β span R(T ). Now, since T is onto, we know that R(T ) = W . So, since T (β)span R(T ) and R(T ) = W , we have tha β span W . Therefore, by definition of basis, since T (β) is linearlyindependent and span W , T (β) is a basis for W .

2.1.17. Let V and W be finite-dimensional vector spaces and T : V →W be linear.

(a) Prove that if dim(V ) < dim(W ), then T cannot be onto.

(b) Prove that if dim(V ) > dim(W ), then T cannot be one-to-one.

Proof of (a). Let V and W be finite-dimensional vector spaces and T : V →W be linear. Assume dim(V ) <dim(W ). We will prove by contradiction so assume that T is onto. By the Dimension Theorem, sinceV is finite-dimensional then nullity(T ) + rank(T ) = dim(V ) which, by definition of nullity and rank,can be written equivalently as dim(N(T )) + dim(R(T )) = dim(V ). Since T is onto, by Theorem 2.5,rank(T ) = dim(V ) which is equivalent to dim(R(T )) = dim(V ) by definition of rank. This implies thatdim(N(T ))+dim(R(T )) = 0+dim(R(T )) = dim(V ). However, since T is onto, we also know that R(T ) = Waccording to Theorem 2.5. Thus, we have that dim(W ) = dim(V ) which is a contradiction (based ontransitivity). Therefore, T is not onto.

Proof of (b). Let V and W be finite-dimensional vector spaces and T : V →W be linear. Assume dim(V ) >dim(W ). We will prove by contradiction so assume that T is one-to-one. By the Dimension Theorem,since V is finite-dimensional then nullity(T ) + rank(T ) = dim(V ) which, by definition of nullity and rank,can be written equivalently as dim(N(T )) + dim(R(T )) = dim(V ). Since T is one-to-one, by Theorem 2.4,N(T ) = {0} so it follows that dim(N(T )) = 0. This implies that dim(R(T )) = dim(V ). Also, by by Theorem2.5, since T is one-to-one we have that rank(T ) = dim(W ) which is equivalent to dim(R(T )) = dim(W )by definition of rank. Therefore, since dim(R(T )) = dim(V ) and dim(R(T )) = dim(W ), it follows thatdim(W ) = dim(V ) which is a contradiction to our assumption that dim(V ) > dim(W ) so T must not beone-to-one.

2.1.40. Let V be a vector space and W be a subspace of V . Define the mapping η : V → V/W by η(v) = v+Wfor v ∈ V .

(a) Prove that η is a linear transformation from V to V/W and that N(η) = W .

(b) Suppose that V is finite-dimensional. Use (a) and the dimension theorem to derive a formula relat-ing dim(V ), dim(W ), and dim(V/W ).

Proof of (a). Suppose V is a vector space and W is a subspace of V . Define the mapping η : V → V/W byη(v) = v +W for v ∈ V .

We will first prove that η is a linear transformation from V to V/W . That is, by definition of lineartransformation, we need to show that for all x, y ∈ V and c ∈ F , η(x+ y) = η(x) + η(y) and η(cx) = cη(x).Note, in Problem 1.3.31 of Homework Set 2, we have already showed that the coset operations are welldefined. Let a, b ∈ V . By definition of η, η(a+ b) = (a+ b)+W . Then, by Problem 1.3.31 and our definitionof η, (a + b) + W = (a + W ) + (b + W ) = η(a) + η(b) as required. Next, by definition of η, for any c ∈ F ,η(ca) = (ca) + W . According to the definition of coset in Problem 1.3.31, (ca) + W = c(a + W ) which, bydefinition of η is cη(a) as required.

We now prove that N(η) = W . By Theorem 2.1, N(η) is a subspace of V . Recall from Problem 1.3.31that we have proven that the quotient space of V modulo W (V/W) is a vector space so it is defined underall of the field axioms. By definition of null space, N(η) = {x ∈ V : η(x) = 0}. Let n ∈ N(η). Then,η(n) = 0 but we know that 0 ∈ W since W is a subspace. Since n was arbitrary, we have that N(η) ⊆ W .Now let w ∈ W . Since W is closed under addition, w + 0 ∈ W . But, we also have that w + 0 ∈ N(η) sinceη(w + 0) = 0. Thus, we have that W ⊆ N(η). Since N(η) ⊆W and W ⊆ N(η), N(η) = W as required.

2

Proof of (b). We claim that dim(V ) + dim(V/W ) = dim(V ). From the Dimension Theorem, we know thatdim(N(η)) + dim(R(η)) = dim(V ). From part (a), we know that N(η) = W which implies dim(W ) +dim(R(η)) = dim(V ). Also, it follows that R(η) = V/W . Thus, dim(W ) + dim(V/W ) = dim(V )

2.2.13. Let V and W be vector spaces, and let T and U be nonzero linear transformations from V into W .If R(T ) ∩R(U) = {0}, prove that {T,U} is a linearly independent subset of L(V,W ).

Proof. Let V and W be vector spaces, and suppose T : V → W and U : V → W are nonzero. AssumeR(T )∩R(U) = {0}. Since T is nonzero, there exists v1 ∈ V such that T (v1) 6= 0. Likewise for U , there existv2 ∈ V such that U(v2) 6= 0. We wish to show by contradiction so assume aT + bU = T0 (note, T0 denotesthe zero transformation) where scalars a, b ∈ F are not all zero. By the distributive law of logic, we havetwo cases.

Case I: Assume a 6= 0 and b = 0. Note, since a 6= 0, aT + bU = T0 can be rewritten as T = −ba U = U(−ba )

by definition of linear transformation. Then, T (v1) = U(−ba v1) ∈ R(U) by definition of range. This impliesthat T (v1) ∈ R(T ) ∩ R(U). But, we know that R(T ) ∩ R(U) = {0} so, in consideration that T (v1) 6= 0,T (v1) ∈ R(T ) ∩R(U) contradicts the fact that T is nonzero.

Case II: Assume a = 0 and b 6= 0. Note, since b 6= 0, aT + bU = T0 can be rewritten as U = −ab T = T (−ab ).

Then, U(v2) = T (−ab v2) ∈ R(T ) by definition of range. This implies that U(v2) ∈ R(T ) ∩ R(U). But, weknow that R(T ) ∩ R(U) = {0} so, in consideration that U(v2) 6= 0, U(v2) ∈ R(T ) ∩ R(U) contradicts thefact that U is nonzero.

In either case, we have a contradiction. Therefore, for aT + bU = T0, we have that a = b = 0 which,by definition of linear independence, means that {T,U} is linearly independent.

2.2.15. Let V and W be vector spaces, and let S be a subset of V . Define S0 = {T ∈ L(V,W ) : T (x) = 0for all x ∈ S}. Prove the following statement.

(a) S0 is a subspace of L(V,W ).(b) If S1 and S2 are subsets of V and S1 ⊆ S2, then S0

2 ⊆ S01 .

(c) If V1 and V2 are subspaces of V then (V1 + V2)0 = V 01 ∩ V 0

2 .

Proof of (a). Let T,U ∈ S0 and choose a ∈ F . By Theorem 1.3, we wish to show that T0 ∈ S0 and additionand scalar multiplication is closed for S0. First, it is clear that for any x ∈ S, T0(x) = 0 since T0 is the zerotransformation. Next, by definition, (T +U)(x) = T (x) +U(x) = 0 + 0 = 0 so T +U ∈ S0 which means thatS0 is closed under addition. Finally, we have that aT (x) = T (ax) = 0 by definition of linear transformationso aT (x) ∈ S0. Thus, we have that S0 is a subspace of L(V,W ).

Proof of (b). Suppose S1 and S2 are subsets of V . Also assume S1 ⊆ S2. Let T ∈ S02 . By definition, we

have that for all x ∈ S2, T (x) = 0. However, since S1 ⊆ S2, we know that x ∈ S1 also. This implies that forall x ∈ S1, T (x) = 0 so we have T ∈ S0

1 . By defiition of subset, we can conclude that S02 ⊆ S0

1 .

Proof of (c). Suppose V1 and V2 are subspaces of V .

(⊆) Let T ∈ (V1 + V2)0. Then, by definition, for all a ∈ V1 + V2, T (a) = 0. By the definition on page22 of set addition, we have that a = v1 + v2 for all v1 ∈ V1 and v2 ∈ V2 so T (v1 + v2) = 0. Since V1 is asubspace, we know that 0 ∈ V1. So, setting v1 = 0, we have that T (0 + v2) = T (v2) = 0 for all v2 ∈ V2which implies that T ∈ V 0

2 . With similar consideration, since V2 is a subspace, 0 ∈ V2 so we have thatT (v1 + 0) = T (v1) = 0 for all v1 ∈ V1. This implies that T ∈ V 0

1 . Since T ∈ V 02 and T ∈ V 0

1 , it follows thatT ∈ V 0

1 ∩ V 02 . Therefore, we have that (V1 + V2)0 ⊆ V 0

1 ∩ V 02 .

(⊇) Now let T ∈ V 01 ∩ V 0

2 . By definition, we have that for all v1 ∈ V1, T (v1) = 0. Likewise, for all v2 ∈ V2,T (v2) = 0. Note that by definition of linear transformation, T (v1) + T (v2) = T (v1 + v2). So, T (v1 + v2) = 0for all v1 + v2 ∈ V1 + V2 (by definition of subspace addition) which implies that T ∈ (V1 + V2)0. Thus, itfollows that (V1 + V2)0 ⊇ V 0

1 ∩ V 02 .

3

Since (V1 + V2)0 ⊆ V 01 ∩ V 0

2 and (V1 + V2)0 ⊇ V 01 ∩ V 0

2 , we have that (V1 + V2)0 = V 01 ∩ V 0

2 as required.

2.3.11. Let V be a vector space, and let T : V → V be linear. Prove that T 2 = T0 if and only if R(T ) ⊆ N(T ).

Proof. Suppose V be a vector space and let T : V → V be linear.

(⇒) Suppose T 2 = T0. Pick a ∈ R(T ). By definition of range, there exists v ∈ V such that T (v) = a.Because T 2 = T0 and T (v) = a, it follows that T 2(v) = T (T (v)) = T (a) = 0 which means a ∈ N(T ). Thus,we have that R(T ) ⊆ N(T ).

(⇐) Suppose R(T ) ⊆ N(T ). Choose a ∈ R(T ). By definition of range, there exists v ∈ V such that T (v) = a.However, since R(T ) ⊆ N(T ), a ∈ N(T ). So, by definition of nullspace, we have that T (v) = a = 0 whichimplies that T (a) = 0. Then, T 2(v) = T (T (v)) = T (a) = 0. Thus, we have that T 2 = T 0.

In conclusion, we have shown that T 2 = T0 if and only if R(T ) ⊆ N(T ).

2.3.12. Let V , W , and Z be vector spaces, and let T : V →W and U : W → Z be linear.

(a) Prove that if UT is one-to-one, then T is one-to-one. Must U also be one-to-one?(b) Prove that if UT is onto, then U is onto. Must T also be onto?(c) Prove that if U and T are one-to-one and onto, then UT is also.

Proof of (a). Suppose V , W , and Z are vector spaces, and suppose that T : V →W and U : W → Z is linear.Assume UT is one-to-one. We wish to prove by contradiction so assume that T is not one-to-one. Then,by definition of linear transformation, there exists a, b ∈ V such that a 6= b and T (a) = T (b) which can berewritten as T (a)−T (b) = 0. By Theorem 2.10, it follows that UT (a)−UT (b) = U(T (a)−T (b)) = U(0) = 0.Since UT (a) − UT (b) = 0, we have that UT (a) = UT (b) which is a contradiction since a 6= b and UT isassumed to be one-to-one. Therefore, T must be one-to-one.

Proposition (a). We claim that U does not need to be one-to-one.

Proof of Proposition (a). It is sufficient to provide a counterexample. Let V = R, W = R2 and Z = R.Define T : R→ R2 by T (r1) = (r1, 0) and define U : R2 → R by U(r1, r2) = r1. Then, UT (r1) = U(T (r1)) =U(r1, 0) = r1 (since UT : R→ R). Thus, we have that T and UT is one-to-one. However, U is not one-to-onesince an element of Z is mapped to by multiple elements of W . That is, we might have different r2 but r1will always result from the transformation. For example, U(1, 1) = U(1, 2) = U(1, 3) = 1.

Proof of (b). Suppose V , W , and Z are vector spaces, and suppose that T : V → W and U : W → Z islinear. Assume UT is onto. By Theorem 2.9, UT : V → Z is linear. Since UT is onto, for all v ∈ V ,UT (v) = z where z ∈ Z. We wish to prove by contradiction so assume that U is not onto. Then, we knowthat not all vector in Z are mapped to (they do not have a corresponding vector from W ). That is, thereexists z ∈ Z such that U(w) 6= z for all w ∈ W . Clearly this contradicts the assumption that UT is ontosince all vectors in Z need to be mapped to by definition of onto. Thus, U is onto.

Proposition (b). We claim that T does not need to be onto.

Proof of Proposition (b). It is sufficient to show a counterexample. Let V = R, W = R2 and Z = R. DefineT : R→ R2 by T (r1) = (r1, 0) and define U : R2 → R by U(r1, r2) = r1. Clearly UT is onto as every elementof Z will be mapped to (U is also onto as every element of Z is mapped to by how we define U). However,T is not onto since it cannot correspond to the second coordinate of W . That is, the second coordinate willalways be zero by how we defined T so there will be elements of W that does not have anything mapped to.For example, (0, 3) ∈W does not get mapped to.

Proof of (c). Suppose V , W , and Z are vector spaces, and suppose that T : V → W and U : W → Z islinear. Assume U and T are one-to-one and onto.

We first show that UT is one-to-one. Suppose UT (x) = UT (y). We wish to show that x = y by the

4

definition of one-to-one. Then, U(T (x)) = U(T (y)). Since U is one-to-one, we have that T (x) = T (y). SinceT is one-to-one, it follows that x = y as required.

We now show that UT is onto. Let v ∈ V . Since T is onto, we have that T (v) = w for some w ∈ W .Also, let z ∈ Z. Since U is onto, we have that U(w) = z. It follows that UT (v) = U(T (v)) = U(w) = zwhich implies that UT is onto.

In conclusion, if U and T are one-to-one and onto, then UT is also.

2.3.16. Let V be a finite-dimensional vector space, and let T : V → V be linear.

(a) If rank(T ) = rank(T 2), prove that R(T ) ∩N(T ) = {0}. Deduce that V = R(T )⊕N(T ).(b) Prove that V = R(T k)⊕N(T k) for some positive integer k.

Proof of (a). Let V be a finite-dimensional vector space, and let T : V → V be linear. Suppose rank(T ) =rank(T 2). Since V is finite-dimensional, by the Dimension Theorem, we have that rank(T ) + nullity(T ) =dim(V ) and rank(T 2) + nullity(T 2) = dim(V ) which implies that rank(T ) + nullity(T ) = rank(T 2) +nullity(T 2). Because rank(T ) = rank(T 2), it follows that nullity(T ) = nullity(T 2) which is equivalent todim(N(T )) = dim(N(T 2)). Clearly, we have that N(T ) is a subspace of N(T 2) because for all t ∈ N(T ),T 2(t) = T (T (t)) = T (0) = 0 which implies t ∈ N(T 2) so N(T ) ⊆ N(T 2). So, since dim(N(T )) =dim(N(T 2)) and N(T ) is a subspace of N(T 2), we have that N(T ) = N(T 2). Choose x ∈ R(T ) andalso let x ∈ N(T ). Since x ∈ R(T ), by definition of range, there exists a ∈ V such that T (a) = x. Also,since x ∈ N(T ), by definition of nullspace, T (x) = 0. Then, with consideration that T (a) = x, we havethat T (x) = T (T (a)) = T 2(a) = 0 so it follows that a ∈ N(T 2). But, since we have that N(T ) = N(T 2),a ∈ N(T ) also. And because a ∈ N(T ), by definition of nullspace, T (a) = 0 which, since we have thatT (a) = x, means T (a) = x = 0. Since we had arbitrarily that x ∈ N(T ) and x ∈ R(T ), we can concludethat R(T ) ∩N(T ) = {0}.

Proposition (a). V = R(T )⊕N(T )

Proof of Proposition (a). By definition of direct sum, we wish to prove that R(T ) ∩ N(T ) = {0} and V =R(T ) + N(T ). From part (a), we already know that R(T ) ∩ N(T ) = {0}. Recall problem 1.6.29 fromHomework Set 3. Clearly, R(T ) and N(T ) are finite-dimensional subspaces of vector space V so by part(a) of the exercise, we have that dim(R(T ) + N(T )) = dim(R(T )) + dim(N(T )) − dim(R(T ) ∩N(T )) andR(T ) + N(T ) is finite-dimensional. Since we have determined that R(T ) ∩ N(T ) = {0}, we have thatdim(R(T ) + N(T )) = dim(R(T )) + dim(N(T )). By the Dimension Theorem, we know that dim(V ) =dim(R(T )) + dim(N(T )) so it follows that dim(V ) = dim(R(T ) +N(T )). Thus, since R(T ) +N(T ) and Vis finite-dimensional and dim(V ) = dim(R(T ) + N(T )), it follows that V = R(T ) + N(T ) as required. Inconclusion, since we have shown that R(T )∩N(T ) = {0} and V = R(T ) +N(T ), by the definition of directsum, V = R(T )⊕N(T ).

Proof of (b). Let V be a finite-dimensional vector space, and let T : V → V be linear. We wish to provethat V = R(T k) ⊕ N(T k) for some k ∈ Z+. By definition of direct sum, this is sufficient to showing thatR(T k) ∩ N(T k) = {0} and V = R(T k) + N(T k). In consideration of part (a), if we assume rank(T k) =rank(T 2k) then clearly R(T k) ∩N(T k) = {0} following the same argument as we had shown. Similarly, inthe same manner as the proof to Proposition (a), we have that R(T k) +N(T k) and V is finite-dimensionaland dim(V ) = dim(R(T k) +N(T k)) which means that V = R(T k) +N(T k) as required. So, since R(T k) ∩N(T k) = {0} and dim(V ) = dim(R(T k) +N(T k)), by definition of direct sum, V = R(T k)⊕N(T k).

2.4.7. Let A be an n× n matrix.(a) Suppose that A2 = O. Prove that A is not invertible.(b) Suppose that AB = O for some nonzero n× n matrix B. Could A be invertible? Explain.

Proof of (a). Let A be an n × n matrix and suppose that A2 = O. We will prove by contradiction sosuppose that A is invertible. Then, by definition of invertibility, there exists an n × n matrix B such thatAB = BA = I. We can rewrite such expression as B = A−1 (note, equivalently, A = B−1). Then, sinceA2 = O, we have that A2A−1 = OA−1 = O. However, since A is invertible and because A2A−1 = O, it

5

follows that A2A−1 = AAA−1 = A(AA−1) = AI = O which implies that A = O. This is a contradiction tothe assumption that A is invertible (i.e., A = B−1). Thus, we conclude that A is not invertible.

Proof of (b). We claim that A is not invertible. For the sake of contradiction, assume that A is invertible.Suppose that AB = O for some nonzero n × n matrix B. Then, A−1AB = A−1O = O. But, becauseA−1AB = O and since A is invertible, A−1AB = (A−1A)B = IB = O which implies that B = O. However,this is a contradiction since we had that B is nonzero. Thus, A is not invertible.

2.4.9. Let A and B be n× n matrices such that AB is invertible. Prove that A and B are invertible. Givean example to show that arbitrary matrices A and B need not be invertible if AB is invertible.

Proof. Let A and B be n × n matrices such that AB is invertible. Since AB is invertible, by definition ofinvertibility, there exists an n × n matrix C such that ABC = CAB = In. Since ABC = A(BC) = In,we have that A is invertible as BC is the multiplicative inverse of A by definition of invertibility (i.e.,BC = A−1). Similarly, since CAB = (CA)B = In, we have that B is invertible as CA is the multiplicativeinverse of B by definition of invertibility (i.e., AB = B−1). Thus, we have that A and B are invertible.

Example: Let A =

(1 0 00 0 1

)and B =

1 00 00 1

respectively. Clearly, A and B are not invertible as

they are not n×n matrices (by definition of invertibility). However, their product is AB =

(1 00 1

)which

is invertible as the identity is always is its own inverse.

2.4.10. Let A and B be n× n matrices such that AB = In.(a) Use Exercise 9 to conclude that A and B are invertible.(b) Prove A = B−1 (and hence B = A−1).(c) State and prove analogous results for linear transformations defined on finite-dimensional vector spaces.

Proof of (a). Let A and B be n × n matrices such that AB = In. Then, AB is invertible as the identitymatrix is invertible. So since AB is invertible, by 2.4.9, we have that A is invertible and B is invertible.

Proof of (b). From (a), we know that AB = In, A and B is invertible. Since A is invertible and AB = In,by definition of invertibility, B is the multiplicative inverse of A. That is, B = A−1. Then, it follows thatB−1 = (A−1)−1 = A by property of inverse.

Proposition 2.4.10 (c). Let V and W be finite-dimensional vector spaces and let T : V → W and U :W → V be linear. If UT = IV then TU = IW .

Proof of Proposition 2.4.10 (c). Let V and W be finite-dimensional vector spaces and let T : V → Wand U : W → V be linear. Assume UT = IV . We wish to prove that TU = IW . For any v ∈ V ,TU(T (v)) = T (UT (v)) = T (IV (v)) = T (v). Since T (v) ∈ W and v was arbitrary, it follows that TU = IWas required.

Note that this sufficiently shows that T is invertible (U is the inverse of T ).

2.4.24. Let T : V → Z be a linear transformation of a vector space V on to vector space Z. Define themapping

T : V/N(T )→ Z by T (v +N(T )) = T (v)

for any coset v +N(T ) in V/N(T ).

(a) Prove that T is well-defined; that is, prove that if v +N(T ) = v′ +N(T ), then T (v) = T (v′).(b) Prove that T is linear.(c) Prove that T is an isomorphism.(d) Prove that the diagram shown in Figure 2.3 commutes; that is, prove that T = T η.

6

Proof of (a). Let T : V → Z be a linear transformation of a vector space V on to vector space Z. Supposev + N(T ) = v′ + N(T ). By definition, T (v + N(T )) = T (v) and T (v′ + N(T )) = T (v′). Since v + N(T ) =v′ +N(T ), it follows that T (v) = T (v′) as required.

Proof of (b). We wish to prove that T is linear which, by definition of linear transformation, is equivalentto showing that for all x, y ∈ V/N(T ) and c ∈ F , T (x + y) = T (x) + T (y) and T (cx) = cT (x). Choosea, b ∈ V/N(T ) and c ∈ F . By definition of the mapping and the well-definedness of T from part (a), T (a+b) =(a+ b) +N(T ) = (a+N(T )) + (b+N(T )) = T (a) + T (b) and T (ca) = (ca) +N(T ) = c(a+N(T )) = cT (a).Thus, we can conclude that T is linear.

Proof of (c). By definition of isomorphism, we must show that T is linear and invertible. From part (b), wehave that T is linear. By Theorem 2.5, to show that T is invertible is equivalent to showing that dim(T ) =dim(V ). By the Dimension Theorem, we know that dim(N(T )) + dim(R(T )) = dim((V )/N(T )).

Proof of (d). We wish to prove that T = T η.

2.5.10. Prove that if A and B are similar n × n matrices, then tr(A) = tr(B). Hint: Use Exercise 13 ofSection 2.3.

Proof. Suppose A and B are similar n × n matrices. By definition, there exists an invertible matrix Qsuch that A = Q−1BQ. From Exercise 2.3.13, we had that tr(AB) = tr(BA). It follows that tr(A) =tr(Q−1BQ) = ((Q−1B)Q) = tr((BQ−1)Q) = (B(Q−1Q)) = tr(B) as required.

2.5.13. Let V be a finite-dimensional vector space over a field F , and let β = {x1, x2, ..., xn} be an orderedbasis for V . Let Q be an n× n invertible matrix with entries from F . Define

x′j =

n∑n=1

Qijxi 1 ≤ j ≤ n

and set β′ = {x′1, x′2, ..., x′n}. Prove that β′ is a basis for V and hence that Q is the change of coordinatematrix changing β′-coordinates into β-coordinates.

2.6.13. Let V be a finite-dimensional vector space of F . For every subset S of V , define the annilator S0

of S asS0 = {f ∈ V ∗ : f(x) = 0 ∀x ∈ S}.

(a) Prove that S0 is a subspace of V ∗.(b) If W is a subspace of V and x 6∈W , prove that there exists f ∈W 0 such that f(x) 6= 0.(c) Prove (S0)0 = span(ψ(S)), where ψ is defined as in Theorem 2.26.(d) For subspaces W1 and W2, prove that W1 = W2 if and only if W 0

1 = W 02 .

(e) For subspaces W1 and W2, show that (W1 +W2)0 = W 01 ∩W 0

2 .

Proof of (a). By Theorem 1.3, we wish to show that 0 ∈ S0 and addition and scalar multiplication is closed forS0. Let f, g ∈ S0 and a ∈ F . Clearly, f(0) = 0 ∈ S0. Next, by definition, (f+g)(x) = f(x)+g(x) = 0+0 = 0so S0 is closed under addition. Lastly, by definition, af(x) = f(ax) = 0 so S0 is closed under scalarmultiplication. Thus, we have shown that S0 is a subspace of V ∗.

Proof of (b). Suppose W is a subspace of V and x 6∈W . Let β = {x1, x2, ..., xk} be a basis for W . Since Wis a subspace of V , we can extend β to V so β1 = {x1, ..., xk, xk+1, ..., xn}. Then, let (β1)∗ be the dual basisto β1 and define (β1)∗ = {f1, ..., fk, fk+1, ..., fn}. Since we have that fi(xi) 6= 0, W 0 ∩ {f1, ..., fk}. Sincexk+1, ..., xn 6∈W , it follows that {fk+1, ..., fn} ∈W 0. However, x 6∈W implies that for x = w+ (a1x1 + ...+aixi) where i > k, there exists an ao 6= 0. That is, fo(x) = ao 6= 0 as required.

Proof of (c). (⊆) Pick s ∈ S such that ψ(s) = s1. Then, s1(f) = f(s) = 0 so ψ(s) ∈ (S0)0. Since ψ and fis linear, it follows that

7

Proof of (d). Let W1 and W2 be subspaces.

(⇒) Suppose W 01 = W 0

2 . We wish to prove by contradiction so assume W1 6= W2. Then there existsx 6∈ W1 and x ∈ W2. From part (b), we know that there exists f ∈ W 0

1 such that f(x) 6= 0 which is acontradiction since W 0

1 = W 02 and f(x) 6∈W 0

2 . Thus, we have that W1 = W2.

(⇐) Now suppose that W1 = W2. Clearly by definition, since W1 = W2, W 01 = W 0

2 .

Thus, we conclude that W1 = W2 if and only if W 01 = W 0

2 .

Proof of (e). (⊆) Let f ∈ (W1 +W2)0. Then, f(w1) = f(w2) = 0 for w1 ∈W1 and w2 ∈W2. Then, we havethat f ∈W 0

1 and f ∈W 02 so (W1 +W2)0 ⊆W 0

1 ∩W 02 .

(⊇) Let f ∈ W 01 ∩W 0

2 . Then, we have that f(W1) = 0 and f(W2) = 0 which implies that for w1 ∈ W1 andw2 ∈W2, f(w1 + w2) = f(w1) + f(w2) = 0 + 0 = 0. That is, f ∈ (W1 +W2)0 so W 0

1 ∩W 02 ⊆ (W1 +W2)0.

Since we have that (W1+W2)0 ⊆W 01 ∩W 0

2 and W 01 ∩W 0

2 ⊆ (W1+W2)0, we can conclude that (W1+W2)0 =W 0

1 ∩W 02 .

2.6.14. Prove that if W is a subspace of V , then dim(W ) + dim(W 0) = dim(V ). Hint: Extend an orderedbasis {x1, x2, ..., xk} of W to an ordered basis β = {x1, x2, ..., xn} of V . Let β∗ = {f1, f2, ..., fn}. Prove that{fk+1, fk+2, ..., fn} is a basis for W 0.

Proof. Suppose W is a subspace of V . Let βW = {x1, x2, ..., xk} be an ordered basis of W and extendit onto an ordered basis β = {x1, x2, ..., xn} of V . Let β∗ = {f1, f2, ..., fn}. We wish to show that γ ={fk+1, fk+2, ..., fn} is a basis for W 0 which is equivalent to showing that dim(W ) + dim(W 0) = dim(V ).Clearly, γ ∈W 0 and it must be linearly independent since it is a subset of a basis. By definition of basis, weneed to show that γ span W 0. We will prove by contradiction so assume that γ does not span W 0. SinceW 0 ⊆ V ∗, we have that a1f1 + ...+ anfn ∈ W 0 where a1, ..., an are not all zero. That is, there exists an aosuch that ao 6= 0. This implies that f(vo) = ao 6= 0 for vo ∈ W , a contradiction. Thus, γ span W 0. Since γis linearly independent and since it generates W 0, we have that γ is a basis for W 0 as required.

2.6.15. Suppose that W is a finite-dimensional vector space and that T : V → W is linear. Prove thatN(T t) = (R(T ))0.

Proof. Suppose that W is a finite-dimensional vector space and that T : V →W is linear.

(⊆) Let f ∈ N(T t). By definition of nullspace, we have that T t(f) = 0. Then, for all v ∈ V , f(T (v)) = 0.This implies that f(R(T )) = 0 which, by definition of annilator, means that f ∈ (R(T ))0. Thus, N(T t) ⊆(R(T ))0.

(⊇) Now let f ∈ (R(T ))0. By definition of annilator, we have that f(R(T )) = 0. By definition of range,this means that for all v ∈ V , we have that f(T (v)) = 0. Since f(T (v)) = 0, we have that T t(f) = 0 sof ∈ N(T t) by definition of nullspace. Thus, (R(T ))0 ⊆ N(T t).

Since N(T t) ⊆ (R(T ))0 and (R(T ))0 ⊆ N(T t), we have that N(T t) = (R(T ))0.

2.6.16. Use Exercises 14 and 15 to deduce that rank(LAt) = rank(LA) for any A ∈Mm×n(F ).

Proof. We wish to show that rank(LAt) = rank(LA) for any A ∈Mm×n(F ). Note, LA : V → W and LAt :W ∗ → V ∗. Also, by definition of dual space, we know that dim(V ∗) = dim(V ) and dim(W ∗) = dim(W ).So, from definition of rank and by 2.6.14 (Homework Set 5 ) and 2.6.15 we have that

8

rank(LAt) = dim(R(LAt))

= dim(W )− dim(W ∗)− dim(R(LAt))

= dim(W )− dim(N(LAt))

= dim(W )− dim((R(LA))0)

= dim(R(LA))

= rank(LA)

as required.

9

Partial Solutions for Linear Algebra by Friedberg et al.

Chapter 3

John K. Nguyen

December 7, 2011

3.2.14. Let T,U : V →W be linear transformations.(a) Prove that R(T + U) ⊆ R(T ) +R(U)(b) Prove that if W is finite-dimensional, then rank(T + U) ≤ rank(T ) + rank(U).(c) Deduce from (b) that rank(A+B) ≤ rank(A) + rank(B) for any m× n matrices A and B.

Proof of (a). Let T,U : V →W be linear transformations. We wish to prove that R(T +U) ⊆ R(T ) +R(U)so assume that x ∈ R(T + U). Then, by definition of range, there exists v ∈ V such that (T + U)(x) = v.By definition of linear transformation, we have that T (x) + U(x) = v which implies that x ∈ R(T ) + R(U)according to the definition of range. Thus, R(T + U) ⊆ R(T ) +R(U).

Proof of (b). Assume W is finite-dimensional. By Theorem 2.1, we know that R(T ) and R(U) are subspacesof W and, as such, they are both finite-dimensional. From (a), we know that R(T + U) ⊆ R(T ) + R(U) sowe have dim(R(T + U)) ≤ dim(R(T ) + R(U)) ≤ dim(R(T )) + dim(R(U)). By definition of rank, we havethat rank(T + U) ≤ rank(T ) + rank(U).

Proof of (c). Suppose A and B are both m×n matrices. By definition of left-multiplication transformation,we have that LA : Fm → Fn and LB : Fm → Fn. In consideration of (b) and Theorem 2.15, we have thatrank(LA+B) = rank(LA + LB) ≤ rank(LA) + rank(LB) which implies that rank(A + B) ≤ rank(A) +rank(B) as required.

3.2.17. Prove that if B is a 3 × 1 matrix and C is a 1 × 3 matrix, then the 3 × 3 matrix BC has rank atmost 1. Conversely, show that if A is any 3 × 3 matrix having rank 1, then there exists a 3 × 1 matrix Band a 1× 3 matrix C such that A = BC.

Proof. Suppose B is an 3 × 1 matrix and C is an 1 × 3 matrix. By Theorem 3.5, we know that the rankof any matrix equals the maximum number of its linearly independent columns. Thus, clearly rank(B) ≤ 1since B is 3× 1. By definition of matrix multiplication, we have that BC is an 3× 3 matrix so it is defined.Then, from Theorem 3.7, we have that rank(BC) ≤ rank(B). Since we know that rank(B) ≤ 1, it followsthat rank(BC) ≤ rank(B) ≤ 1 which implies that rank(BC) ≤ 1.

To show the converse, suppose A is any 3×3 matrix having rank 1. By definition, we have that LA : F 3 → F 3.Let β be the standard ordered basis for F 3 and assume f : β → F 1. Then, by the Freeness Theorem, thereexists a unique linear transformation LB : F 3 → F 1 extending f . Now suppose γ is a standard ordered basisfor F 1 and assume g : γ → F 3. Also by the Freeness Theorem, there exists a unique linear transformationLC : F 1 → F 3 extending g. So, since LB : F 3 → F 1, we have that B is a 3× 1 matrix. Similarly, we knowthat C is a 1 × 3 matrix. It follows that LBLC : F 3 → F 3 which implies that BC is 3 × 3 (we also knowthis by definition of matrix multiplication). Thus, A = BC as required.

3.2.19. Let A be an m× n matrix with rank m and B be an n× p matrix with rank n. Determine the rankof AB. Justify your answer.

Proof. We claim that rank(AB) = m.

1

Let A be an m × n matrix with rank m and B be an n × p matrix with rank n. By definition of ma-trix multiplication, we know that AB is an m × p matrix. By definition, we have that LA : Fm → Fn,LB : Fn → F p and LAB : Fm → F p respectively. Observe that since LB : F p → Fn is onto (recallTheorem 2.15 ), we have R(AB) = R(LALB) = LALB(F p) = LA(LB(F p)) = LA(Fn) = R(LA). There-fore, rank(AB) = dim(R(LALB)) = dim(R(LA)) = rank(A). And since rank(A) = m, we have thatrank(AB) = m.

2

Partial Solutions for Linear Algebra by Friedberg et al.

Chapter 4

John K. Nguyen

December 7, 2011

4.1.11. Let δ : M2×2(F )→ F be a function with the following three properties.(i) δ is a linear function of each row of the matrix when the other row is held fixed.(ii) If the two rows of A ∈M2(F ) are identical, then δ(A) = 0.(iii) If I is the 2× 2 identity matrix, then δ(I) = 1.

Prove that δ(A) = det(A) for all A ∈M2×2(F ).

Proof. Let A ∈ M2×2(F ) and defined it as

(a bc d

)where a, b, c, d are scalars. First, by the properties

above, notice that δ

(1 11 1

)= δ

(1 00 1

)+ δ

(0 11 0

)= 1 + δ

(0 11 0

)= 0 which implies that

δ

(0 11 0

)= −1. Now, in knowing this and from the properties above, we have the following

δ(A) = δ

(a bc d

)= aδ

(1 0c d

)+ bδ

(0 1c d

)= acδ

(1 01 0

)+ adδ

(1 00 1

)+ bcδ

(0 11 0

)+ bdδ

(0 10 1

)= ac(0) + ad(1) + bc(−1) + bd(0)

= ad− bc

= det

((a bc d

))= det(A).

Since A was arbitrary, we have shown that δ(A) = det(A).

4.2.25. Prove that det(kA) = kndet(A) for any A ∈Mn×n(F ).

Proof. Choose A ∈Mn×n(F ). Then, by Theorem 4.8,

det(kA) = det(kInA)

= det(kIn)det(A)

= kndet(In)det(A)

= kn · 1det(A)

= kndet(A).

Since A was arbitrary, we have that det(kA) = kndet(A) for any A ∈Mn×n(F ) as required.

1

4.3.10. A matrix M ∈ Mn×n(F ) is called nilpotent if, for some positive integer k, Mk = 0, where 0 is then× n zero matrix. Prove that if M is nilpotent, then det(M) = 0.

Proof. Suppose M ∈ Mn×n(F ) such that it is nilpotent. Then, by definition, there exists some k ∈ Z suchthat Mk = 0, where 0 is the n×n zero matrix. This implies that det(Mk) = 0. By Theorem 4.7, we know thatdet(AB) = det(A)det(B) for any A,B ∈ Mn×n(F ). So, by induction, we know that det(Mk) = (det(M))k

for all k ∈ Z. Since the determinant of the zero matrix is zero, (det(M))k = 0 which implies that det(M) = 0as required.

4.3.11. A matrix M ∈Mn×n(F ) is called skew-symmetric if M t = −M . Prove that if M is a skew-symmetricand n is odd, then M is not invertible. What happens if n is even?

Proof. Choose M ∈Mn×n(F ) such that it is skew-symmetric. Then, by Theorem 4.8 and since M t = M ,

det(M) = det(M t)

= det(−M)

= (−1)ndet(M).

Since we know that n is odd, we have that det(M) = −det(M). Rearranging terms, we have that 2det(M) = 0which implies that det(M) = 0 so, by the corollary on page 223, M is not invertible.

If n is even, then we would have det(M) = det(M) which would not imply anything.

4.3.13. For M ∈Mn×n(C), let M be the matrix such that (M)ij = Mij for all i, j, where Mij is the complexconjugate of Mij.

(a) Prove that det(M) = det(M).(b) A matrix Q ∈ Mn×n(C) is called unitary if QQ∗ = I, where Q∗ = Qt. Prove that if Q is a unitarymatrix, then |det(Q)| = 1.

Proof of (a). Let M ∈ Mn×n(C) and suppose M be the matrix such that (M)ij = Mij for all i, j, whereMij is the complex conjugate of Mij .

2

Proof of (b). Let Q ∈ Mn×n(C) be a unitary matrix. Then, by definition, QQ∗ = QQt = I. Then, fromTheorem 4.8 and part (a), we have that

det(I) = det(QQt)

= det(Q)det(Qt)

= det(Q)det(Qt)

= det(Q)det(Q)

= |det(Q)|2.

Since det(I) = 1, we have that |det(Q)|2 = 1 which can further be reduced to |det(Q)| = 1 as required.

4.3.21. Prove that if M ∈Mn×n(F ) can be written in the form

M =

(A BO C

)where A and C are square matrices, then det(M) = det(A) · det(C).

Proof. Let A be an k×k matrix, B be a k×t matrix and C be a t ×t matrix. We wish to prove by induction.

For our base case, suppose k = 1. Then, expanding down the first column,

det(M) = a · det(M11) + 0 + ...+ 0 = a · det(C) = det(A)det(C).

So, this holds true for k = 1.

Now suppose that A is an (k − 1) × (k − 1) matrix and B is an (k − 1) × t matrix and C is a t × tmatrix and that

M =

(A BO C

).

Now, taking the cofactor expansion along the first column,

det(M) = a11det(M11)− ...± ak1det(Mk1) + 0 + ...+ 0

= a11det

(A11 B1

O C

)− ...± det

(Ak1 BkO C

)= a11det(A11)det(C)− ...± ak1det(Ak1)det(C)

= (a11det(A11))− ...± ak1det(Ak1)det(C)

= det(A)det(C).

4.3.22. Let T : Pn(F ) → Fn+1 be the linear transformation defined in Exercise 22 of Section 2.4 byT (f) = (f(c0), f(c1), ..., f(cn)), where c0, c1, ..., cn are distinct scalars in an infinite field F. Let β be thestandard ordered basis for Pn(F ) and γ be the standard ordered basis for Fn+1.(a) Show that M = [T ]γβ has the form

(c) Prove that

det(M) =∏

0≤i<j≤n

(cj − ci),

the product of all the terms of the of the form cj − ci for 0 ≤ i < j ≤ n.

3

Proof of (a). Suppose T : Pn(F ) → Fn+1 is the linear transformation defined in Exercise 22 of Section 2.4by T (f) = (f(c0), f(c1), ..., f(cn)), where c0, c1, ..., cn are distinct scalars in an infinite field F. Let β be thestandard ordered basis for Pn(F ) and γ be the standard ordered basis for Fn+1.

Proof of (c). We will prove that

det(M) =∏

0≤i<j≤n

(cj − ci),

the product of all the terms of the of the form cj − ci for 0 ≤ i < j ≤ n.

4

Partial Solutions for Linear Algebra by Friedberg et al.

Chapter 5

John K. Nguyen

December 7, 2011

5.2.11. Let A be an n × n matrix that is similar to an upper triangular matrix and has the distinct eigen-values λ1, λ2, ..., λk with corresponding multiplicities m1,m2, ...,mk. Prove the following statements.

(a) tra(A) =

k∑i=1

miλi.

(b) det(A) = (λ1)m1(λ2)m2 ...(λk)mk .

Proof of (a). Since A be an n× n matrix that is similar to an upper triangular matrix, say B, that has thedistinct eigenvalues λ1, λ2, ..., λk with corresponding multiplicities m1,m2, ...,mk. By definition of similar,there exists an invertible matrix Q such that A = Q−1BQ. Then, in consideration of Exercise 2.3.13,

tr(A) = tr(Q−1BQ) = tr(Q−1QB) = tr(B) =

k∑i=1

miλi.

Proof of (b). Since A be an n× n matrix that is similar to an upper triangular matrix, say B, that has thedistinct eigenvalues λ1, λ2, ..., λk with corresponding multiplicities m1,m2, ...,mk. By definition of similar,there exists an invertible matrix Q such that A = Q−1BQ. Recall Theorem 4.7 which states that forany A,B ∈ M2×2(F ), det(AB) = det(A)det(B). In consideration of this theorem and the corollary onpage 223, we have that det(A) = det(Q−1BQ) = det(Q−1)det(B)det(Q) = 1

det(Q)det(B)det(Q) = det(B) =

(λ1)m1(λ2)m2 ...(λk)mk as required.

5.2.12. Let T be an invertible linear operator on a finite-dimensional vector space V .

(a) Recall that for any eigenvalue λ of T , λ−1 is an eigenvalue of T−1. Prove that the eigenspace of Tcorresponding to λ is the same as the eigenspace of T−1 corresponding to λ−1.(b) Prove that if T is diagonalizable, then T−1 is diagonalizable.

Proof of (a). Pick v ∈ Eλ. Then, by definition, T (v) = λv. Taking the inverse of both sides, we getT−1T (v) = T−1(λv) which means v = λT−1(v). Then, by definition, v ∈ Eλ−1 so we have that theeigenspace of T corresponding to λ is the same as the eigenspace of T−1 corresponding to λ−1.

Proof of (b). Suppose T is diagonalizable. Then T has n linearly independent eigenvectors. From part (a)we know that T−1 also has the same n eigenvectors. Thus, T−1 is diagonalizable.

5.2.13. Let A ∈Mn×n(F ). Recall from Exercise 14 of Section 5.1 that A and At have the same character-istic polynomial and hence share the same eigenvalues with the same multiplicities. For any eigenvalue λ ofA and At, let Eλ and E′λ denote the corresponding eigenspaces for A and At, respectively.

(a) Show by way of example that for a given common eigenvalue, these two eigenspaces need not be thesame.(b) Prove that for any eigenvalue λ, dim(Eλ) = dim(E′λ).(c) Prove that if A is diagonalizable, then At is also diagonalizable.

1

Example for (a). Define A =

(1 23 4

). Now, notice that A(1, 0) = (1, 3) but At(1, 0) = (1, 2). Thus,

EA 6= EtA.

Proof of (b). Suppose dim(F ) = n. Then, by definition, we know that dim(Eλ) = n− rank(A− λI). Then,taking the transpose (recall that rank(A) = rank(At) by a previous exercise), we have that dim(Eλ) =n− rank((A− λI)t) = n− rank(At − λI) = dim(E′λ) as required.

Proof of (c). Suppose A is diagonalizable. Then, there exists an invertible matrix Q such that B = Q−1AQis a diagonal matrix (recall that a square matrix is diagonalizable if it is similar to a diagonal matrix accordingto Section 5.1 ). Now, taking the transpose of both sides yields Bt = (Q−1AQ)t = QtAt(Q−1)t. Clear,y Bt

is diagonal so by definition, we have that At is diagonalizable.

5.2.18a. Prove that if T and U are simultaneously diagonalizable operators, then T and U commute (i.e.,UT = TU).

Proof. Suppose T and U are simultaneously diagonalizable operators. Then, by definition, there exists anordered basis β = {v1, v2, ..., vn} such that T (vi) = λvi and U(vi) = αvi where i = 1, 2, ..., n. It follows thatTU(vi) = T (U(vi)) = T (αvi) = αT (vi) = αλvi = λαvi = λU(vi) = U(λvi) = U(T (vi)) = UT (vi). So, wecan conclude that UT = TU as required.

5.4.13. Let T be a linear operator on a vector space V , let v be a nonzero vector in V , and let W be theT-cyclic subspace of V generated by v. For any w ∈ V , prove that w ∈ W if and only if there exists apolynomial g(t) such that w = g(T )(v).

Proof. (⇒) Suppose w ∈ W and assume dim(W ) = n. Let β = {v, Tv, ..., Tn−1v} be an ordered basis forW . Then, by definition, w = a0v + a1Tv + ... + an−1T

n−1v for some scalars a0, a1, ...an−1 (that is, w is alinear combination of the elements of β). Let g(t) = a0 + a1t+ ...+ an−1t

n−1. Then, g(t) is a polynomial oft of degree less than or equal to n− 1. This also means that w = g(T )v.

(⇐) Suppose there exists a polynomial g(t) such that w = g(T )v. Then, g(t) = a0 + a1t + ... + antn

and w = a0v + a1Tv + ...+ anTnv for some scalars a0, a1, ..., an. Since v, Tv, ..., Tnv ∈W , w ∈W .

5.4.16. Let T be a linear operator on a finite-dimensional vector space V .

(a) Prove that if the characteristic polynomial of T splits, then so does the characteristic polynomial ofthe restriction of T to any T-invariant subspace of V .(b) Deduce that if the characteristic polynomial of T splits, then any nontrivial T-invariant subspace of Vcontains an eigenvector of T .

Proof of (a). Suppose the characteristic polynomial of T , say f , splits. Let W be a T − invariant subspaceof V and g is the characteristic polynomial of TW . Then, g divides f so there exists a polynomial r suchthat f = gr. Suppose that f has n degrees. Note that the amount of zeros that g have is less than or equalto degree of g and analogously for r. But, it follows that n = deg(g) = deg(f). So, g has deg(g) zeros, h hasdeg(h) zeros. Thus, we can factor g to deg(g) factors which means that the characteristic polynomial of therestriction of T to a T-invariant subspace splits.

Proof of (b). Suppose W is a nontrivial T-invariant subspace of V . Let f be the characteristic polynomialof TW . By part (a), we know that f splits. Pick λ to be the root of f . Then, f(λ) = det(TW − λI) = 0 .But this means that TW − λI is not invertible so there exists a nonzero w ∈ W such that (T − λI)(w) = 0which implies that w is an eigenvector of T . Since w is arbitrary, we have shown that if the characteristicpolynomial of T splits, then any nontrivial T-invariant subspace of V contains an eigenvector of T .

5.4.20. Let T be a linear operator on a vector space V , and suppose that V is a T-cyclic subspace of itself.Prove that if U is a linear operator on V , then UT = TU if and only if U = g(T ) for some polynomial g(t).Hint: Suppose that V is generated by v. Choose g(t) according to Exercise 13 so that g(T )(v) = U(v).

2

Proof. (⇒) Suppose UT = TU . Then, since V = span(v, T (v), T 2(v), ...), U(v) = a0 +a1T (v)+ ...+anTn(v)

for some scalars a0, a1, ..., an. So, U(v) = g(T )(v) where g(v) = a0 +a1x+ ...+anxn. Suppose x ∈ V . Then,

x = b0 + b1T (v) + ...+ bmTm(v) for some scalars b0, b1, ..., bm. It follows that

U(x) = U(b0 + b1T (v) + ...+ bmTm(v)) = b0U(T 0(v)) + b1U(T (v)) + ...+ bmU(Tm(v))

= b0T0(U(v)) + b1T (U(v)) + ...+ bmT

m(U(v))

= b0T0(g(T )(v)) + b1T (g(T )(v)) + ...+ bmT

m(g(T )(v))

= b0g(T )(T 0(v)) + b1g(T )(T (v)) + ...+ bmg(T )(Tm(v))

= g(T )(b0T0(v) + b1T

1(v) + ...+ bmTm(v))

= g(T )(x).

Thus, U = g(T ) for some polynomial g.

(⇐) Suppose U = g(T ) for some polynomial g. Define g(x) = a0 + a1x+ ...+ anxn. Then,

UT (x) = a0T0(T (x)) + a1T (T (x)) + ...+ anT

n(T (x))

= a0T (T 0(x)) + a1T (T (x)) + ...+ anT (Tn(x))

= T (a0T0(x) + a1T (x) + ...+ anT

n(x))

= TU(x).

Therefore, we have that UT = TU .

We have shown that if U is a linear operator on V , then UT = TU if and only if U = g(T ) for somepolynomial g(t).

3

Partial Solutions for Linear Algebra by Friedberg et al.

Chapter 6

John K. Nguyen

December 7, 2011

6.1.18. Let V be a vector space over F , where F = R or F = C, and let W be an inner product space overF with product 〈·, ·〉. If T : V → W is linear, prove that 〈x, y〉′ = 〈T (x), T (y)〉 defines an inner product onV if and only if T is one-to-one.

Proof. (⇒) Suppose 〈x, y〉′ = 〈T (x), T (y)〉 defines an inner product on V . Let T (x) = T (y) which, since Tis linear, implies T (x) − T (y) = T (x − y) = 0. From our definition above, we have that 〈x − y, x − y〉′ =〈T (x− y), T (x− y)〉 = 〈T (x), T (x− y)〉 − 〈T (y), T (x− y)〉 = 0. By definition (particularly part (d) on page330) it clearly follows that x− y = 0 which means x = y so T is one-to-one.

(⇐) Assume that T is one-to-one. That is, T (x) = T (y) implies x = y. Then, 〈x+z, y〉′ = 〈T (x+z), T (y)〉 =〈T (x), T (y)〉+ 〈T (z), T (y)〉. Since W is an inner product space, we have that 〈x+ z, y〉′ = 〈x, y〉′ + 〈z, y〉′ soit follows immediately that 〈T (x), T (y)〉 = 〈x, y〉′.

We have proven that 〈x, y〉′ = 〈T (x), T (y)〉 defines an inner product on V if and only if T is one-to-one.

6.1.12. Let {v1, v2, ..., vk} be an orthogonal set in V , and let a1, a2, ..., ak be scalars. Prove that∣∣∣∣∣∣∣∣∣∣

k∑i=1

aivi

∣∣∣∣∣∣∣∣∣∣2

=

k∑i=1

|ai|2||vi||2.

Proof. We apply the definition of inner product space and Theorem 6.2 to get the following∣∣∣∣∣∣∣∣∣∣

k∑i=1

aivi

∣∣∣∣∣∣∣∣∣∣2

=

⟨k∑

j=1

ajvj ,

k∑i=1

aivi

⟩=

k∑j=1

k∑i=1

aj ai < vj , vi >=

k∑i=1

|ai|2||vi||2.

as required.

6.2.13. Let V be an inner product space, S and S0 be subsets of V , and W be a finite-dimensional subspaceof V . Prove the following results.(a) S0 ⊆ S implies that S⊥ ⊆ S⊥0 .(b) S ⊆ (S⊥)⊥; so span(S) ⊆ (S⊥)⊥.(c) W = (W⊥)⊥. Hint: Use Exercise 6.(d) V = W ⊕W⊥. (See the exercises of Section 1.3)

Proof of (a). Suppose S0 ⊆ S. Let x ∈ S⊥. Then, by definition, for all y ∈ S, < x, y >= 0 which impliesthat y ∈ S0. But S0 ⊆ S, so x ∈ S⊥0 . Thus, S⊥ ⊆ S⊥0 as required.

Proof of (b). Let x ∈ S. Then, < x, y >= 0 for all y ∈ S⊥. But this also means that x ∈ (S⊥)⊥. So,S ⊆ (S⊥)⊥.

Proof of (c). (⊆) By part (b), we have that W ⊆ (W⊥)⊥.

(⊇) We will prove the contrapositive so assume that x 6∈ W . Then, by Exercise 6.2.6, there exists y ∈ W⊥

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such that < x, y > 6= 0. This implies that x 6∈ (W⊥)⊥.

Thus, we have shown that W = (W⊥)⊥.

Proof of (d). By definition of direct sum, we wish to show that W ∩W⊥ = {0} and V = W + W⊥. Sincethe only vector that is orthogonal to itself is the zero vector, we have that W ∩W⊥ = {0}. Let v ∈ V . ByTheorem 6.6, there exist unique vector u ∈W and z ∈W⊥ such that v = u+z so we have that V = W +W⊥.We have shown that V = W ⊕W⊥.

6.2.14. Let W1 and W2 be subspaces of a finite-dimensional inner product space. Prove that (W1 +W2)⊥ =W⊥1 ∩W⊥2 and (W1 ∩W2)⊥ = W⊥1 + W⊥2 . Hint for the second equation: Apply Exercise 13(c) to the firstequation.

Proof of (a). (⊆) Let x ∈ (W1 + W2)⊥. By definition, it follows immediately that x ∈ W⊥1 ∩ W⊥2 . So,(W1 + W2)⊥ ⊆W⊥1 ∩W⊥2 .

(⊇) Let x ∈ W⊥1 ∩ W⊥2 . Then, x ∈ W⊥1 and x ∈ W⊥2 . By linearity, it follows that x ∈ (W1 + W2)⊥.So, (W1 + W2)⊥ ⊇W⊥1 ∩W⊥2 .

Thus, since (W1 + W2)⊥ ⊆W⊥1 ∩W⊥2 and (W1 + W2)⊥ ⊇W⊥1 ∩W⊥2 , (W1 + W2)⊥ = W⊥1 ∩W⊥2 .

Proof of (b). (⊆) Let x ∈ (W1 ∩W2)⊥. Then, since W⊥1 ⊆W⊥1 + W⊥2 and W⊥2 ⊆W⊥1 + W⊥2 by a previousexercise, it follows that x ∈W⊥1 +W⊥2 so (W1 ∩W2)⊥ ⊆W⊥1 +W⊥2 (since x is perpendicular to vectors thatis in both W1 and W2).

(⊇) Let x ∈ W⊥1 + W⊥2 . Then, by definition, x = w1 + w2 where w1 ∈ W⊥1 and w2 ∈ W⊥2 . But,W⊥1 ⊆W⊥1 +W⊥2 and W⊥2 ⊆W⊥1 +W⊥2 so, in consideration of Exercise 6.2.13c, (W1 ∩W2)⊥ ⊆W⊥1 +W⊥2which means that x ∈ (W1 ∩W2)⊥ (i.e., w1 and w2 is perpendicular to the intersection of W1 and W2).Therefore, (W1 ∩W2)⊥ ⊇W⊥1 + W⊥2 .

In conclusion, since (W1 ∩W2)⊥ ⊆W⊥1 + W⊥2 and (W1 ∩W2)⊥ ⊇W⊥1 + W⊥2 , we have that (W1 ∩W2)⊥ =W⊥1 + W⊥2 .

6.2.22. Let V = C([0, 1]) with the inner product < f, g >=∫ 1

0f(t)g(t)dt. Let W be the subspace spanned by

the linearly independent set {t,√t}.

(a) Find an orthonormal basis for W .

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(b) Let h(t) = t2. Use the orthonormal basis obtained in (a) to obtain the ”best” (closest) approximation ofh in W .

6.3.6. Let T be a linear operator on an inner product space V . Let U1 = T + T ∗ and U2 = TT ∗. Prove thatU1 = U∗1 and U2 = U∗2 .

Proof. Let T be a linear operator on an inner product space V and suppose that U1 = T +T ∗ and U2 = TT ∗.We first prove that U1 = U∗1 . By our assumption and Theorem 6.11,

U∗1 = (T + T ∗)∗ = T ∗ + (T ∗)∗ = T ∗ + T = U1.

In a similar argument, we will show that U2 = U∗2 .

U∗2 = (TT ∗)∗ = T ∗(T ∗)∗ = T ∗T = U2.

Thus, we have that U1 = U∗1 and U2 = U∗2 as required.

6.3.8. Let V be a finite-dimensional inner product space, and let T be a linear operator on V . Prove that ifT is invertible, then T ∗ is invertible and (T ∗)−1 = (T−1)∗.

Proof. Let V be a finite-dimensional inner product space, and let T be a linear operator on V . Choosex, y ∈ V . Suppose T is invertible. Then,

〈T ∗(T−1)∗(x), y〉 = 〈(T−1)∗(x), T (y)〉= 〈x, T−1T (y)〉= 〈x, y〉

which implies that T ∗(T−1)∗ = I. Therefore, (T ∗)−1 = (T−1)∗.

6.3.9. Prove that if V = W ⊕W⊥ and T is the projection on W along W⊥, then T = T ∗. Hint: Recall thatN(T ) = W⊥.

Proof. Suppose that V = W ⊕W⊥ and T is the projection on W along W⊥. Since V = W ⊕W⊥, we havethat V = W +W⊥. Let v1, v2 ∈ V . Then, there exists w1, w2 ∈W and w1, w2 ∈W⊥ such that v1 = w1 + w1

and v2 = w2 + w2. Now, in consideration that T is the projection on W along W⊥,

〈v1, T (v2)〉 = 〈w1 + w1, T (w2 + w2)〉= 〈w1 + w1, w2〉= 〈w1, w2〉.

Similarly,

〈T (v1), v2〉 = 〈T (w1 + w1), w2 + w2〉= 〈w1, w2 + w2〉= 〈w1, w2〉.

So, it follows that 〈v1, T (v2)〉 = 〈T (v1), v2〉 which means 〈v1, T (v2)〉 = 〈v1, T ∗(v2)〉. This implies thatT = T ∗

6.3.11. For a linear operator T on an inner product space V , prove that T ∗T = T0 implies T = T0. Is thesame result true if we assume that TT ∗ = T0.

Proof. Let T be a linear operator on an inner product space V . Suppose T ∗T = T0. Pick x ∈ V . Then,〈T ∗T (x), x〉 = 〈T0(x), x〉 = 〈0, x〉 = 0 by Theorem 6.1(c). But, we also have that 〈T ∗T (x), x〉 = 〈T (x), T (x)〉so 〈T (x), T (x)〉 = ||T (x)||2 = 0. It follows that T (x) = 0 which implies T = T0.

We claim that the same result is true if we assume that TT ∗ = T0.

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Proof. Let T be a linear operator on an inner product space V . Suppose TT ∗ = T0. Pick x ∈ V . Then,〈TT ∗(x), x〉 = 〈T0(x), x〉 = 〈0, x〉 = 0. But 〈TT ∗(x), x〉 = 〈T ∗(x), T ∗(x)〉 which, analogously to the argumentabove, we have that T ∗(x) = 0 so T ∗ = T0. Then, (T ∗)∗ = T ∗0 so T = T0 (note that the adjoint of the zerooperator is the zero operator).

6.3.12. Let V be an inner product space, and let T be a linear operator on V . Prove the following results:

(a) R(T ∗)⊥ = N(T )(b) If V is finite-dimensional, then R(T ∗) = N(T )⊥. Hint: Use Exercise 13(c) of Section 6.2.

Proof of (a). By definition, x ∈ R(T ∗)⊥ if and only if 〈x, T ∗(y)〉 = 〈T (x), y〉 = 0 for all y ∈ V . By Theorem6.1, this is true if and only if T (x) = 0 which means x ∈ N(T ). Thus, x ∈ R(T ∗)⊥ if and only if x ∈ N(T )so we have that R(T ∗)⊥ = N(T ).

Proof of (b). Suppose V is finite-dimmensional. From part (a), we know that R(T ∗)⊥ = N(T ). It followsthat (R(T ∗)⊥) = N(T )⊥. In consideration of 6.2.13c, we have that (R(T ∗)⊥) = R(T ∗). Thus, since(R(T ∗)⊥) = N(T )⊥ and (R(T ∗)⊥) = R(T ∗), we have that R(T ∗) = N(T )⊥.

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