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 that ˜ x + 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 apply the 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 a vector 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 . By definition, (0v1 , 0w1 ) ∈ Z. Thus, (v, w) + (0v1 , 0w1 ) = (v, w) and so (VS3 ) holds for Z. 1
By (VS4 ), there exists (−v, −x) ∈ V such that (v, x) + (−v, −x) = (0v1 , 0v2 ) for all (v, x) ∈ V . Likewise, there exists (−y, −w) ∈ V such that (y, w) + (−y, −w) = (0w1 , 0w2 ) for all (y, w) ∈ V . It follows that for all (v, w) ∈ Z, there exists (−v, −w) ∈ Z such that (v, w) + (−v, −w) = (0z1 , 0z2 ) where (0z1 , 0z2 ) ∈ Z so (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) so 1 · 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 that c(v1 , v2 ) = (cv1 , cv2 ). Likewise, c((y1 , w1 ) + (y2 , w2 )) = c(y1 , w1 ) + c(y2 , w2 ) = (cy1 , cw1 ) + (cy2 , cw2 ) which implies 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) which implies (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 of A. 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 X
Aii and tr(B) =
i=0
n X
Bii where i is the row
i=0
and column of the matrix (because the matrix is square). Thus, by definition, tr(aA+bB) =
n X
(aA+bB)ii =
i=0 n X
n n n n X X X X aAii + bBii = a Aii +b (aAii +bBii ). Then, by property of summation, we have i=0
i=0
i=0
i=0
Bii . Finally,
i=0
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 ∈ W whenever a ∈ F and x, y ∈ W . Proof. (⇒) Suppose a subset W of a vector space V is a subspace with addition and scalar multiplication defined 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, we have 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 , we only 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 only if W1 ⊆ W2 or W2 ⊆ W1 . 2
Proof. (⇒) Suppose W1 and W2 are subspaces of a vector space V and assume W1 ∪ W2 is a subspace. We wish to prove by contradiction so assume W1 6⊆ W2 and W2 6⊆ W1 . Since W1 6⊆ W2 , there exists a ∈ W1 but a 6∈ W2 . Also, since W2 6⊆ W1 , there exists b ∈ W2 but b 6∈ W1 . Since W1 ∪ W2 is a subspace, it must be that (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 contradicts our assumption that b 6∈ W1 . Case II: Assume (a + b) ∈ W2 . By the additive inverse property, a + b + (−b) = a ∈ W2 which contradicts our 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 a subspace. 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 that w2 = 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, we have 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 every vector in W1 + W2 has an additive inverse which means that with any arbitrary vector we can always obtain 0 ∈ W1 + W2 . And since, 0 ∈ W1 and 0 ∈ W2 by definition of subspace, it follows that W1 ⊆ W1 + W2 and W2 ⊆ W1 + W2 . Proof of (b). Suppose W is a subspace of V . Assume W1 ⊆ W and W2 ⊆ W . We wish to prove that W1 + W2 ⊆ W . Let u ∈ W1 + W2 . From part (a), we define u = x + y where x ∈ W1 and y ∈ W2 . Since W1 ⊆ W , x ∈ W . Similarly, since W2 ⊆ W , y ∈ W . And since W is a subspace, we know that u = x+y ∈ W where x, y ∈ W . Since W was arbitrary, we have shown that any subspace of V that contains both W1 and W2 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 if and 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 and x2 ∈ 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 knowing that W1 and W2 are disjoint and that x1 − y1 ∈ W1 and y2 − x2 ∈ W2 (W1 and W2 are subspaces of V so both are closed under addition), we have that x1 − y1 = 0 and y2 − x2 = 0. Thus, x1 = y1 and x2 = y2 so x1 + x2 is unique. (⇐) Let W1 and W2 be subspaces of a vector space V . Suppose each vector in V can be uniquely written as x1 + x2 where x1 ∈ W1 and x2 ∈ W2 . Then, it follows that W1 + W2 = V . Next, we let t be in the intersection of W1 and W2 (i.e. t ∈ W1 ∩ W2 ). Since W1 and W2 are subspaces, we know that 0 ∈ W1 and 3
0 ∈ W2 . Then, t = t + 0 where t ∈ W1 and 0 ∈ W2 . Also, t = 0 + t where 0 ∈ W1 and t ∈ W2 . It must follow that t = 0 so W1 ∩ W2 = {0} since t ∈ W1 ∩ W2 . Since W1 + W2 = V and W1 ∩ W2 = {0}, by definition of direct 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 uniquely written 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 = v`1 + W and v2 + W = v`2 + W , then (v1 + W ) + (v2 + W ) = (v`1 + W ) + (v`2 + W ) and a(v1 + W ) = a(v`1 + 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 the quotient 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 subspace of V . Because v + W is a subspace, we have that v ∈ v + W since 0 ∈ v + W . Since v ∈ v + W and v + W is a subspace, it follows that v + v ∈ v + W (since v + W is closed under addition). We wish to prove by contradiction 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 definition 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 + W for 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 and v ∈ W so v + W is closed under addition. Finally, assume f ∈ F . Then, f x = f v + f a which we can rewrite as f x = v + (−v) + f v + f a. Since a ∈ W and W is a subspace, f a ∈ W . Likewise, since v ∈ W , f v ∈ W . And since we have already stated that −v ∈ W , we know that (−v) + cv + ca ∈ W because W is closed under addition. Thus, f x ∈ 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 + b which 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 and v2 + 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 + W and 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 where a ∈ W . Likewise, we have that y = v2 + b where b ∈ W . Since v1 − v2 ∈ W and W is a subspace, we have that v2 − v1 ∈ W (it is clear that v2 + (−1)v1 ∈ W because W is closed under addition and scalar 4
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 = v`1 + W and v2 + W = v`2 + W . We wish to show (v1 + W ) + (v2 + W ) = (v`1 + W ) + (v`2 + W ). However, from part(b) we know this is equivalent to (v1 + v2 ) − (v`1 + v`2 ) ∈ W . Since v1 + W = v`1 + W and v2 + W = v`2 + W , by part (b), we have that v1 − v`1 ∈ W and v2 − v`2 ∈ W . Since W is a subspace, it is closed under addition so (v1 − v`1 ) + (v2 − v`2 ) ∈ W . Rearranging terms, (v1 + v2 ) − (v`1 + v`2 ) ∈ W as required. Now we wish to prove that a(v1 + W ) = a(v`1 + W ) for all a ∈ F . However, we apply part (b) which means showing av1 − av`1 ∈ W is equivalent. From above, we already have v1 − v`1 ∈ W . Since W is a subspace, it is closed under scalar multiplication so we have that a(v1 − v`1 ) ∈ W for all a ∈ F . By the distribution law, we conclude that av1 − av`1 ∈ 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 field axioms. For (V S1), let v1 + W ∈ S and v2 + W ∈ S where v1 , v2 ∈ V . Then, (v1 + W ) + (v2 + W ) = (v1 + v2 ) + W and (v2 + W ) + (v1 + W ) = (v2 + v1 ) + W by the definition in part (c). Since v1 , v2 ∈ V , by (V S1), we know 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 definition in 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 multiplication, −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. We have 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 every element of span(S1 ) is contained in span(S2 ). Choose z ∈ span(S1 ). By definition of span, there exists x1 , x2 , ..., xn ∈ S1 such that z = a1 x1 + a2 x2 + ... + an xn where a1 , a2 , ..., an ∈ F . But, since S1 ⊆ S2 ,
5
x1 , x2 , ..., xn ∈ S2 also which means that z ∈ span(S2 ) by definition of linear combination. Since z was arbitrary, 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 contained in V so by definition, we know that span(S2 ) ⊆ V . Since V ⊆ span(S2 ) and span(S2 ) ⊆ V , we have that span(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 ). Give an 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 of span, there exists x1 , x2 , ..., xn ∈ S1 ∩ S2 such that v = a1 x1 + a2 x2 + ... + an xn 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 . From this, we know that v = a1 x1 + a2 x2 + ... + an xn ∈ span(S1 ) and v = a1 x1 + a2 x2 + ... + an xn ∈ span(S2 ). By definition 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} by definition 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 only if 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 = − ab v. 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 other hand, if we rewritten the equation to v = − ab u and assume that b 6= 0, we have that u is a multiple of v. In conclusion, 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 . By the 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 dependent, 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 dependent, 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 the other 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. 6
Proof of (a). (⇒) Suppose V is a vector space over a field of characteristic not equal to two. Assume u and v are distinct vectors in V . Suppose {u, v} is linearly independent. We wish to show that {u + v, u − v} is linearly independent which mean a(u + v) + b(u − v) = au + av + bu − bv = (a + b)u + (a − b)v = 0 for some a, 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 field characteristic is not equal to two, we have a = 0. In a similar argument, since a − b = 0, we know that a = b which 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 and a = b = 0. Rearranging terms, au + av + bu + bv = (a + b)u + (a − b)v = 0. We prove by contradiction so assume 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 linearly dependent because a and b cannot both be 0 since a + b 6= 0 for a(u + v) + b(u − v) = 0 (our assumption that 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 linearly dependent because a and b cannot both be 0 since a − b 6= 0 for a(u + v) + b(u − v) = 0 (our assumption that 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 characteristic not equal to two. Suppose Assume that u, v, w is linearly independent which, by definition, means au+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 linearly independent. Then, d(u+v)+e(u+w)+f (v+w) = du+dv+eu+ew+f v+f w = (d+e)u+(d+f )v+(e+f )w = 0 for some d, e, f ∈ F and d = e = f = 0. Since u, v, w is linearly independent, d + e = 0, d + f = 0 and e + f = 0. We then solve for each variables. Since d + e = 0, d = −e. Then, d + f = −e + f = 0 which implies 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 so d + f = d + d = 2d = 0 which mean d = 0 (again, since char(F ) 6= 2). Thus, since d = f = e = 0, we have that {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) = 0 where 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 + f w = 0 where d, e, f ∈ F and not all are zero. It follows that a + b = d, a + c = e and b + c = f because (a + b)u + (a + c)v + (b + c)w = 0 and du + ev + f w = 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= 0 or 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 dependent 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 dependent 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 linearly dependent 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 linearly 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 b are 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 are nonzero 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 need to show that for x(u + v) + y(au) = 0, x = y = 0 for scalars x, y ∈ F . Rearranging terms, we notice x(u + v) + y(au) = xu + xv + yau = (x + ya)u + (x)v = 0. Now since {u, v} is linearly independent by definition of basis, it follows that wu + qv = 0 and w = q = 0 for scalars w, q ∈ F . It follows that x + ya = w = 0 and 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) = p where x `, y` ∈ F . Again, rearranging terms, we have (` x + y`a)u + (` x)v = p. By definition of basis, {u, v} span V which implies that for any p ∈ V , x ˜u + y˜v = p, where scalars x ˜, y˜ ∈ F . It follows that x ` + y`a = x ˜ and y which is defined since a is nonzero. Thus, we have that x ` = y˜. Since x ` = y˜, we solve for y` to obtain y` = x˜−˜ a y {u + v, au} span V (clearly, x `(u + v) + y`(au) = y˜(u + v) + x˜−˜ (au) = p). Because {u + v, au} is linearly a 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 for scalars w, q ∈ F . This implies that ma = h = 0and nb = j = 0. Since we know that a and b are nonzero scalars, 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 ˜ + (˜ q b)v = f Since {u, v} span V , x ˜u + y˜v = f which implies that wa ˜ =x ˜ and q˜b = y˜. Then, it follows that w ˜ = xa˜ and q˜ = yb˜ so w(au) ˜ + q˜(bv) = xa˜ (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 and sufficient 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 extend β to be a basis for W1 . It follows that for any x ∈ W1 , x ∈ W1 ∩ W2 . This implies that W1 ⊆ W1 ∩ W2 . Since W1 ∩ 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 set intersection (property of subset intersection), it follows that W1 ∩W2 = W1 . Since W1 ∩W2 = W1 (which tells us 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 . 8
1.6.29. (a) Prove that if W1 and W2 are finite-dimensional subspaces of a vector space V , then the subspace W1 + 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 . Deduce that 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 basis for W1 ∩ W2 such that β = {u1 , u2 , ..., uk } (since we know W1 and W2 are finite-dimensional). Since W1 ∩ 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 } which implies that W1 + W2 is finite-dimensional (as it contains k = m = p number of vectors) and that dim(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 , a1 u1 + ... + ak uk + b1 v1 + ... + bm vm = w1 and, for any w2 ∈ W1 , a1 u1 + ... + ak uk + c1 w1 + ... + cp wp = w2 for scalars ak , bm , and cp respectively (Theorem 1.8 ). It is obvious that τ span W1 + W2 because a1 u1 + ... + ak uk + b1 v1 + ... + bm vm + c1 w1 + ... + cp wp = w1 + w2 for any w1 + w2 ∈ W1 + W2 (recall our given 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 dependent). Because γ is linearly independent (by definition of basis), we know that for a1 u1 +...+ak uk +b1 v1 + ...+bm vm = 0, ak = bm = 0. Similarly, since µ is linearly indepdenent, for a1 u1 +...+ak uk +c1 w1 +...+cp wp = 0, we know that ak = cp = 0. But, we had assumed that τ is linearly dependent which implies that for a1 u1 + ... + ak uk + b1 v1 + ... + bm vm + c1 w1 + ... + cp wp = 0, we have that coefficients ak , bm , and cp are not all 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 basis for 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 ). Because V = W1 ⊕ W2 , we know that W1 ∩ W2 = {0} which implies that dim(W1 ∩ W2 ) = 0 as W1 and W2 have no unique 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 know that 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 have that dim(W1 ∩ W2 ) = 0 which means, by definition of dimension, there isn’t any unique number of vectors in each basis for W1 ∩ W2 . This implies that W1 and W2 are disjoint. Therefore, since V = W1 + W2 and W1 ∩ 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 bases for 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 space V . 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 β1 and β2 are bases for W1 and W2 respectively. Define β1 = {u1 , u2 , ..., un } and β2 = {v1 , v2 , ..., vm }. 9
Then, by Since we span(β1 ) lows that
definition of basis, β1 and β2 both are linearly independent and span W1 and W2 respectively. know that V = W1 ⊕ W2 , we have that W1 ∩ W2 = ∅. By Theorem 1.18, we know that = W1 and span(β2 ) = W2 . Since span(β1 ) = W1 , span(β2 ) = W2 and W1 ∩ W2 = ∅, it folspan(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 first show that β1 ∪ β2 is linearly independent. However, we will prove by contradiction so assume that β1 ∪ β2 is linearly dependent. That is, a1 u1 + ... + an un + b1 v1 + ... + bm vm = 0 where scalars a1 , ..., an and b1 , ..., bm are not all zero. Because, β1 is linearly independent, we have that a1 u1 + ... + an un = 0 where scalars a1 = a2 = ... = an = 0. Likewise, for β2 , b1 v1 + ... + bn vn = 0 where scalars b1 = b2 = ... = bm = 0. Thus, a contradiction 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 states if W is a subspace of a finite-dimensional vector space V , then any basis for W can be extended to a basis for V . Recall that W1 and W2 are subspaces of a vector space V . Therefore, we extend β1 and β2 to bases for 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 is a basis for V . In conclusion, we have proved that if β1 and β2 are bases for W1 and W2 , respectively, show that 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 need to 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 . Note from problem 1.4.15 of Homework Set 2 that span(W1 ∩ W2 ) ⊆ span(W1 ) + span(W2 ). It follows that span(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 . Also recall that β1 ∪ β2 is a basis for V . Then, by definition, span(β1 ∪ β2 ) = V . Suppose each vector in span(β1 ∪ β2 ) can be uniquely written as w1 + w2 where w1 ∈ span(β1 ) and w2 ∈ span(β2 ). Then, by problem 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 linearly independent 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 if and 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. {T (v1 ), T (v2 ), ..., T (vn )} is a basis for W .
Prove that T (β) =
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 contradiction so assume that T (S) is linearly dependent. Since S is linearly independent, for a1 v1 + ... + an vn = 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 (a1 v1 + ... + an vn ) = T (0) = 0. Note, by Theorem 2.5, this is equivalent to T being one-to-one as we had assumed. But we have that T (S) is linearly dependent which implies that for vectors v1 , v2 , ..., vn ∈ S, a1 T (v1 ) + ... + an T (vn ) = 0 where scalars a1 , a2 , ..., an are not all zero. Clearly, this contradiction because this would mean a1 v1 + ... + an vn 6= 0 which implies T (a1 v1 + ... + an vn ) 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 some a, b ∈ V such that T (x) 6= T (y). It follows that T (x) − T (y) = 0 which implies that x − y ∈ N (T ) because T : 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 and that S is a subset of V . (⇒) Suppose that S is linearly independent. Then, by part (a), we have that T (S) is linearly independent and so we’re done. (⇐) Now suppose that T (S) is linearly independent. We wish to prove by contradiction so assume that S is linearly dependent. This implies that for v1 , v2 , ..., vn ∈ S, a1 v1 + ... + an vn = 0 where scalars a1 , a2 , ...an are not all zero. Since T : V → W is linear, we have that a1 T (v1 ) + ... + an T (vn ) = 0, again, where scalars a1 , a2 , ..., an are not all zero. However, we had assumed that T (S) is linearly independent so we have a contradiction. 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 show that T (β) is linearly independent and span W . Since T is one-to-one and β is linearly independent (by
1
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 linearly independent 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, 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 onto, by Theorem 2.5, rank(T ) = dim(V ) which is equivalent to dim(R(T )) = dim(V ) by definition of rank. This implies that dim(N (T ))+dim(R(T )) = 0+dim(R(T )) = dim(V ). However, since T is onto, we also know that R(T ) = W according to Theorem 2.5. Thus, we have that dim(W ) = dim(V ) which is a contradiction (based on transitivity). 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 Theorem 2.5, 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 that dim(W ) = dim(V ) which is a contradiction to our assumption that dim(V ) > dim(W ) so T must not be one-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+W for 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 relating 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 by η(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 linear transformation, 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 well defined. Let a, b ∈ V . By definition of η, η(a + b) = (a + b) + W . Then, by Problem 1.3.31 and our definition of 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, by definition 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 1.3.31 that we have proven that the quotient space of V modulo W (V/W) is a vector space so it is defined under all 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) since η(w + 0) = 0. Thus, we have that W ⊆ N (η). Since N (η) ⊆ W and W ⊆ N (η), N (η) = W as required.
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Proof of (b). We claim that dim(V ) + dim(V /W ) = dim(V ). From the Dimension Theorem, we know that dim(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. Assume R(T ) ∩ R(U ) = {0}. Since T is nonzero, there exists v1 ∈ V such that T (v1 ) 6= 0. Likewise for U , there exist v2 ∈ V such that U (v2 ) 6= 0. We wish to show by contradiction so assume aT + bU = T0 (note, T0 denotes the zero transformation) where scalars a, b ∈ F are not all zero. By the distributive law of logic, we have two cases. −b Case I: Assume a 6= 0 and b = 0. Note, since a 6= 0, aT + bU = T0 can be rewritten as T = −b a U = U( a ) −b by definition of linear transformation. Then, T (v1 ) = U ( a v1 ) ∈ R(U ) by definition of range. This implies that 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. −a Case II: Assume a = 0 and b 6= 0. Note, since b 6= 0, aT + bU = T0 can be rewritten as U = −a b T = T ( b ). −a Then, U (v2 ) = T ( b v2 ) ∈ R(T ) by definition of range. This implies that U (v2 ) ∈ R(T ) ∩ R(U ). But, we know that R(T ) ∩ R(U ) = {0} so, in consideration that U (v2 ) 6= 0, U (v2 ) ∈ R(T ) ∩ R(U ) contradicts the fact 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 S 0 = {T ∈ L(V, W ) : T (x) = 0 for all x ∈ S}. Prove the following statement. (a) S 0 is a subspace of L(V, W ). (b) If S1 and S2 are subsets of V and S1 ⊆ S2 , then S20 ⊆ S10 . (c) If V1 and V2 are subspaces of V then (V1 + V2 )0 = V10 ∩ V20 . Proof of (a). Let T, U ∈ S 0 and choose a ∈ F . By Theorem 1.3, we wish to show that T0 ∈ S 0 and addition and scalar multiplication is closed for S 0 . First, it is clear that for any x ∈ S, T0 (x) = 0 since T0 is the zero transformation. Next, by definition, (T + U )(x) = T (x) + U (x) = 0 + 0 = 0 so T + U ∈ S 0 which means that S 0 is closed under addition. Finally, we have that aT (x) = T (ax) = 0 by definition of linear transformation so aT (x) ∈ S 0 . Thus, we have that S 0 is a subspace of L(V, W ). Proof of (b). Suppose S1 and S2 are subsets of V . Also assume S1 ⊆ S2 . Let T ∈ S20 . 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 for all x ∈ S1 , T (x) = 0 so we have T ∈ S10 . By defiition of subset, we can conclude that S20 ⊆ S10 . 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 page 22 of set addition, we have that a = v1 + v2 for all v1 ∈ V1 and v2 ∈ V2 so T (v1 + v2 ) = 0. Since V1 is a subspace, we know that 0 ∈ V1 . So, setting v1 = 0, we have that T (0 + v2 ) = T (v2 ) = 0 for all v2 ∈ V2 which implies that T ∈ V20 . With similar consideration, since V2 is a subspace, 0 ∈ V2 so we have that T (v1 + 0) = T (v1 ) = 0 for all v1 ∈ V1 . This implies that T ∈ V10 . Since T ∈ V20 and T ∈ V10 , it follows that T ∈ V10 ∩ V20 . Therefore, we have that (V1 + V2 )0 ⊆ V10 ∩ V20 . (⊇) Now let T ∈ V10 ∩ V20 . 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 ) = 0 for all v1 + v2 ∈ V1 + V2 (by definition of subspace addition) which implies that T ∈ (V1 + V2 )0 . Thus, it follows that (V1 + V2 )0 ⊇ V10 ∩ V20 . 3
Since (V1 + V2 )0 ⊆ V10 ∩ V20 and (V1 + V2 )0 ⊇ V10 ∩ V20 , we have that (V1 + V2 )0 = V10 ∩ V20 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 which implies 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 U T is one-to-one, then T is one-to-one. Must U also be one-to-one? (b) Prove that if U T 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 U T 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 U T 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 be rewritten as T (a)−T (b) = 0. By Theorem 2.10, it follows that U T (a)−U T (b) = U (T (a)−T (b)) = U (0) = 0. Since U T (a) − U T (b) = 0, we have that U T (a) = U T (b) which is a contradiction since a 6= b and U T is assumed 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, U T (r1 ) = U (T (r1 )) = U (r1 , 0) = r1 (since U T : R → R). Thus, we have that T and U T is one-to-one. However, U is not one-to-one since an element of Z is mapped to by multiple elements of W . That is, we might have different r2 but r1 will 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 is linear. Assume U T is onto. By Theorem 2.9, U T : V → Z is linear. Since U T is onto, for all v ∈ V , U T (v) = z where z ∈ Z. We wish to prove by contradiction so assume that U is not onto. Then, we know that not all vector in Z are mapped to (they do not have a corresponding vector from W ). That is, there exists z ∈ Z such that U (w) 6= z for all w ∈ W . Clearly this contradicts the assumption that U T is onto since 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. Define T : R → R2 by T (r1 ) = (r1 , 0) and define U : R2 → R by U (r1 , r2 ) = r1 . Clearly U T is onto as every element of 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 will always 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 is linear. Assume U and T are one-to-one and onto. We first show that U T is one-to-one. Suppose U T (x) = U T (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). Since T is one-to-one, it follows that x = y as required. We now show that U T 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 U T (v) = U (T (v)) = U (w) = z which implies that U T is onto. In conclusion, if U and T are one-to-one and onto, then U T 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 to dim(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 ) and also 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 have that 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 that T (a) = x, means T (a) = x = 0. Since we had arbitrarily that x ∈ N (T ) and x ∈ R(T ), we can conclude that 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 from Homework 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 )) and R(T ) + N (T ) is finite-dimensional. Since we have determined that R(T ) ∩ N (T ) = {0}, we have that dim(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 V is finite-dimensional and dim(V ) = dim(R(T ) + N (T )), it follows that V = R(T ) + N (T ) as required. In conclusion, since we have shown that R(T ) ∩ N (T ) = {0} and V = R(T ) + N (T ), by the definition of direct sum, 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 prove that V = R(T k ) ⊕ N (T k ) for some k ∈ Z+ . By definition of direct sum, this is sufficient to showing that R(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, in the same manner as the proof to Proposition (a), we have that R(T k ) + N (T k ) and V is finite-dimensional and 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 so suppose that A is invertible. Then, by definition of invertibility, there exists an n × n matrix B such that AB = BA = I. We can rewrite such expression as B = A−1 (note, equivalently, A = B −1 ). Then, since A2 = O, we have that A2 A−1 = OA−1 = O. However, since A is invertible and because A2 A−1 = O, it 5
follows that A2 A−1 = AAA−1 = A(AA−1 ) = AI = O which implies that A = O. This is a contradiction to the 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−1 AB = A−1 O = O. But, because A−1 AB = O and since A is invertible, A−1 AB = (A−1 A)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. Give an 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 of invertibility, 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 multiplicative inverse of B by definition of invertibility (i.e., AB = B −1 ). Thus, we have that A and B are invertible. 1 0 1 0 0 Example: Let A = and B = 0 0 respectively. Clearly, A and B are not invertible as 0 0 1 0 1 1 0 they are not n × n matrices (by definition of invertibility). However, their product is AB = which 0 1 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 identity matrix 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 that B −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 U T = IV then T U = IW . Proof of 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. Assume U T = IV . We wish to prove that T U = IW . For any v ∈ V , T U (T (v)) = T (U T (v)) = T (IV (v)) = T (v). Since T (v) ∈ W and v was arbitrary, it follows that T U = IW as 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 the mapping T¯ : V /N (T ) → Z by T¯(v + N (T )) = T (v) for any coset v + N (T ) in V /N (T ). (a) (b) (c) (d)
Prove Prove Prove Prove
that that that that
T¯ is well-defined; that is, prove that if v + N (T ) = v 0 + N (T ), then T (v) = T (v 0 ). T¯ is linear. T¯ is an isomorphism. 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. Suppose v + N (T ) = v 0 + N (T ). By definition, T¯(v + N (T )) = T (v) and T¯(v 0 + N (T )) = T (v 0 ). Since v + N (T ) = v 0 + N (T ), it follows that T (v) = T (v 0 ) as required. Proof of (b). We wish to prove that T¯ is linear which, by definition of linear transformation, is equivalent to showing that for all x, y ∈ V /N (T ) and c ∈ F , T¯(x + y) = T (x) + T (y) and T¯(cx) = cT (x). Choose a, 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), we have 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 of Section 2.3. Proof. Suppose A and B are similar n × n matrices. By definition, there exists an invertible matrix Q such that A = Q−1 BQ. From Exercise 2.3.13, we had that tr(AB) = tr(BA). It follows that tr(A) = tr(Q−1 BQ) = ((Q−1 B)Q) = tr((BQ−1 )Q) = (B(Q−1 Q)) = 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 ordered basis for V . Let Q be an n × n invertible matrix with entries from F . Define x0j =
n X
Qij xi
1≤j≤n
n=1
and set β 0 = {x01 , x02 , ..., x0n }. Prove that β 0 is a basis for V and hence that Q is the change of coordinate matrix changing β 0 -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 S 0 of S as S 0 = {f ∈ V ∗ : f (x) = 0 ∀x ∈ S}. (a) (b) (c) (d) (e)
Prove that S 0 is a subspace of V ∗ . If W is a subspace of V and x 6∈ W , prove that there exists f ∈ W 0 such that f (x) 6= 0. Prove (S 0 )0 = span(ψ(S)), where ψ is defined as in Theorem 2.26. For subspaces W1 and W2 , prove that W1 = W2 if and only if W10 = W20 . For subspaces W1 and W2 , show that (W1 + W2 )0 = W10 ∩ W20 .
Proof of (a). By Theorem 1.3, we wish to show that 0 ∈ S 0 and addition and scalar multiplication is closed for S 0 . Let f, g ∈ S 0 and a ∈ F . Clearly, f (0) = 0 ∈ S 0 . Next, by definition, (f +g)(x) = f (x)+g(x) = 0+0 = 0 so S 0 is closed under addition. Lastly, by definition, af (x) = f (ax) = 0 so S 0 is closed under scalar multiplication. Thus, we have shown that S 0 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 W is a subspace of V , we can extend β to V so β1 = {x1 , ..., xk , xk+1 , ..., xn }. Then, let (β1 )∗ be the dual basis to β1 and define (β1 )∗ = {f1 , ..., fk , fk+1 , ..., fn }. Since we have that fi (xi ) 6= 0, W 0 ∩ {f1 , ..., fk }. Since xk+1 , ..., xn 6∈ W , it follows that {fk+1 , ..., fn } ∈ W 0 . However, x 6∈ W implies that for x = w + (a1 x1 + ... + ai xi ) 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) ∈ (S 0 )0 . Since ψ and f is linear, it follows that
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Proof of (d). Let W1 and W2 be subspaces. (⇒) Suppose W10 = W20 . We wish to prove by contradiction so assume W1 6= W2 . Then there exists x 6∈ W1 and x ∈ W2 . From part (b), we know that there exists f ∈ W10 such that f (x) 6= 0 which is a contradiction since W10 = W20 and f (x) 6∈ W20 . Thus, we have that W1 = W2 . (⇐) Now suppose that W1 = W2 . Clearly by definition, since W1 = W2 , W10 = W20 . Thus, we conclude that W1 = W2 if and only if W10 = W20 . Proof of (e). (⊆) Let f ∈ (W1 + W2 )0 . Then, f (w1 ) = f (w2 ) = 0 for w1 ∈ W1 and w2 ∈ W2 . Then, we have that f ∈ W10 and f ∈ W20 so (W1 + W2 )0 ⊆ W10 ∩ W20 . (⊇) Let f ∈ W10 ∩ W20 . Then, we have that f (W1 ) = 0 and f (W2 ) = 0 which implies that for w1 ∈ W1 and w2 ∈ W2 , f (w1 + w2 ) = f (w1 ) + f (w2 ) = 0 + 0 = 0. That is, f ∈ (W1 + W2 )0 so W10 ∩ W20 ⊆ (W1 + W2 )0 . Since we have that (W1 + W2 )0 ⊆ W10 ∩ W20 and W10 ∩ W20 ⊆ (W1 + W2 )0 , we can conclude that (W1 + W2 )0 = W10 ∩ W20 . 2.6.14. Prove that if W is a subspace of V , then dim(W ) + dim(W 0 ) = dim(V ). Hint: Extend an ordered basis {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 extend it 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, we need to show that γ span W 0 . We will prove by contradiction so assume that γ does not span W 0 . Since W 0 ⊆ V ∗ , we have that a1 f1 + ... + an fn ∈ W 0 where a1 , ..., an are not all zero. That is, there exists an ao such that ao 6= 0. This implies that f (vo ) = ao 6= 0 for vo ∈ W , a contradiction. Thus, γ span W 0 . Since γ is 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 that N (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 so f ∈ 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.
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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 subspaces of W and, as such, they are both finite-dimensional. From (a), we know that R(T + U ) ⊆ R(T ) + R(U ) so we have dim(R(T + U )) ≤ dim(R(T ) + R(U )) ≤ dim(R(T )) + dim(R(U )). By definition of rank, we have that 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 : F m → F n and LB : F m → F n . In consideration of (b) and Theorem 2.15, we have that rank(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 at most 1. Conversely, show that if A is any 3 × 3 matrix having rank 1, then there exists a 3 × 1 matrix B and 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 rank of any matrix equals the maximum number of its linearly independent columns. Thus, clearly rank(B) ≤ 1 since 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 follows that 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, there exists a unique linear transformation LB : F 3 → F 1 extending f . Now suppose γ is a standard ordered basis for F 1 and assume g : γ → F 3 . Also by the Freeness Theorem, there exists a unique linear transformation LC : F 1 → F 3 extending g. So, since LB : F 3 → F 1 , we have that B is a 3 × 1 matrix. Similarly, we know that C is a 1 × 3 matrix. It follows that LB LC : F 3 → F 3 which implies that BC is 3 × 3 (we also know this 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 rank of 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 matrix multiplication, we know that AB is an m × p matrix. By definition, we have that LA : F m → F n , LB : F n → F p and LAB : F m → F p respectively. Observe that since LB : F p → F n is onto (recall Theorem 2.15 ), we have R(AB) = R(LA LB ) = LA LB (F p ) = LA (LB (F p )) = LA (F n ) = R(LA ). Therefore, rank(AB) = dim(R(LA LB )) = dim(R(LA )) = rank(A). And since rank(A) = m, we have that rank(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 ). a b Proof. Let A ∈ M2×2 (F ) and defined it as where a, b, c, d are scalars. First, by the properties c d 1 1 1 0 0 1 0 1 above, notice that δ = δ +δ = 1+δ = 0 which implies that 1 1 0 1 1 0 1 0 0 1 δ = −1. Now, in knowing this and from the properties above, we have the following 1 0
a b δ(A) = δ c d 1 0 0 = aδ + bδ c d c 1 0 = acδ + adδ 1 0
1 d 1 0
0 1
+ bcδ
0 1
1 0
+ bdδ
0 0
1 1
= ac(0) + ad(1) + bc(−1) + bd(0) = ad − bc a = det c
b d
= det(A). Since A was arbitrary, we have shown that δ(A) = det(A). 4.2.25. Prove that det(kA) = k n det(A) for any A ∈ Mn×n (F ). Proof. Choose A ∈ Mn×n (F ). Then, by Theorem 4.8, det(kA) = det(kIn A) = det(kIn )det(A) = k n det(In )det(A) = k n · 1det(A) = k n det(A). Since A was arbitrary, we have that det(kA) = k n det(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, M k = 0, where 0 is the n × 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 such that M k = 0, where 0 is the n×n zero matrix. This implies that det(M k ) = 0. By Theorem 4.7, we know that det(AB) = det(A)det(B) for any A, B ∈ Mn×n (F ). So, by induction, we know that det(M k ) = (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 ) = 0 as 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-symmetric and 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)n det(M ). Since we know that n is odd, we have that det(M ) = −det(M ). Rearranging terms, we have that 2det(M ) = 0 which 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 complex conjugate 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 unitary matrix, 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, where Mij 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, from Theorem 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 A B M= O 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(M¯11 ) + 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 × t matrix and that A B M= . O C Now, taking the cofactor expansion along the first column, det(M ) = a11 det(M¯11 ) − ... ± ak1 det(M¯k1 ) + 0 + ... + 0 A¯11 B¯1 A¯k1 B¯k = a11 det − ... ± det O C O C ¯ ¯ = a11 det(A11 )det(C) − ... ± ak1 det(Ak1 )det(C) = (a11 det(A¯11 )) − ... ± ak1 det(A¯k1 )det(C) = det(A)det(C).
4.3.22. Let T : Pn (F ) → F n+1 be the linear transformation defined in Exercise 22 of Section 2.4 by T (f ) = (f (c0 ), f (c1 ), ..., f (cn )), where c0 , c1 , ..., cn are distinct scalars in an infinite field F. Let β be the standard ordered basis for Pn (F ) and γ be the standard ordered basis for F n+1 . (a) Show that M = [T ]γβ has the form
(c) Prove that det(M ) =
Y
(cj − ci ),
0≤i
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 ) → F n+1 is the linear transformation defined in Exercise 22 of Section 2.4 by T (f ) = (f (c0 ), f (c1 ), ..., f (cn )), where c0 , c1 , ..., cn are distinct scalars in an infinite field F. Let β be the standard ordered basis for Pn (F ) and γ be the standard ordered basis for F n+1 .
Proof of (c). We will prove that det(M ) =
Y
(cj − ci ),
0≤i
the product of all the terms of the of the form cj − ci for 0 ≤ i < j ≤ n.
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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 eigenvalues λ1 , λ2 , ..., λk with corresponding multiplicities m1 , m2 , ..., mk . Prove the following statements. (a) tra(A) =
k X
mi λi .
i=1
(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 the distinct eigenvalues λ1 , λ2 , ..., λk with corresponding multiplicities m1 , m2 , ..., mk . By definition of similar, there exists an invertible matrix Q such that A = Q−1 BQ. Then, in consideration of Exercise 2.3.13, k X −1 −1 tr(A) = tr(Q BQ) = tr(Q QB) = tr(B) = mi λi . i=1
Proof of (b). Since A be an n × n matrix that is similar to an upper triangular matrix, say B, that has the distinct eigenvalues λ1 , λ2 , ..., λk with corresponding multiplicities m1 , m2 , ..., mk . By definition of similar, there exists an invertible matrix Q such that A = Q−1 BQ. Recall Theorem 4.7 which states that for any A, B ∈ M2×2 (F ), det(AB) = det(A)det(B). In consideration of this theorem and the corollary on 1 page 223, we have that det(A) = det(Q−1 BQ) = det(Q−1 )det(B)det(Q) = det(Q) det(B)det(Q) = det(B) = m1 m2 mk (λ1 ) (λ2 ) ...(λk ) 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 T corresponding 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 get T −1 T (v) = T −1 (λv) which means v = λT −1 (v). Then, by definition, v ∈ Eλ−1 so we have that the eigenspace 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 characteristic polynomial and hence share the same eigenvalues with the same multiplicities. For any eigenvalue λ of A and At , let Eλ and Eλ0 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 the same. (b) Prove that for any eigenvalue λ, dim(Eλ ) = dim(Eλ0 ). (c) Prove that if A is diagonalizable, then At is also diagonalizable.
1
Example for (a). Define A =
1 3
2 4
. Now, notice that A(1, 0) = (1, 3) but At (1, 0) = (1, 2). Thus,
t EA 6= EA .
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(Eλ0 ) as required. Proof of (c). Suppose A is diagonalizable. Then, there exists an invertible matrix Q such that B = Q−1 AQ is a diagonal matrix (recall that a square matrix is diagonalizable if it is similar to a diagonal matrix according to Section 5.1 ). Now, taking the transpose of both sides yields B t = (Q−1 AQ)t = Qt At (Q−1 )t . Clear,y B t 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., U T = T U ). Proof. Suppose T and U are simultaneously diagonalizable operators. Then, by definition, there exists an ordered basis β = {v1 , v2 , ..., vn } such that T (vi ) = λvi and U (vi ) = αvi where i = 1, 2, ..., n. It follows that T U (vi ) = T (U (vi )) = T (αvi ) = αT (vi ) = αλvi = λαvi = λU (vi ) = U (λvi ) = U (T (vi )) = U T (vi ). So, we can conclude that U T = T U 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 the T-cyclic subspace of V generated by v. For any w ∈ V , prove that w ∈ W if and only if there exists a polynomial g(t) such that w = g(T )(v). Proof. (⇒) Suppose w ∈ W and assume dim(W ) = n. Let β = {v, T v, ..., T n−1 v} be an ordered basis for W . Then, by definition, w = a0 v + a1 T v + ... + an−1 T n−1 v for some scalars a0 , a1 , ...an−1 (that is, w is a linear combination of the elements of β). Let g(t) = a0 + a1 t + ... + an−1 tn−1 . Then, g(t) is a polynomial of t 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 + a1 t + ... + an tn and w = a0 v + a1 T v + ... + an T n v for some scalars a0 , a1 , ..., an . Since v, T v, ..., T n v ∈ 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 of the 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 V contains an eigenvector of T . Proof of (a). Suppose the characteristic polynomial of T , say f , splits. Let W be a T − invariant subspace of V and g is the characteristic polynomial of TW . Then, g divides f so there exists a polynomial r such that f = gr. Suppose that f has n degrees. Note that the amount of zeros that g have is less than or equal to degree of g and analogously for r. But, it follows that n = deg(g) = deg(f ). So, g has deg(g) zeros, h has deg(h) zeros. Thus, we can factor g to deg(g) factors which means that the characteristic polynomial of the restriction 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 polynomial of 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) = 0 which implies that w is an eigenvector of T . Since w is arbitrary, we have shown that if the characteristic polynomial 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 U T = T U 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).
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Proof. (⇒) Suppose U T = T U . Then, since V = span(v, T (v), T 2 (v), ...), U (v) = a0 + a1 T (v) + ... + an T n (v) for some scalars a0 , a1 , ..., an . So, U (v) = g(T )(v) where g(v) = a0 + a1 x + ... + an xn . Suppose x ∈ V . Then, x = b0 + b1 T (v) + ... + bm T m (v) for some scalars b0 , b1 , ..., bm . It follows that U (x) = U (b0 + b1 T (v) + ... + bm T m (v)) = b0 U (T 0 (v)) + b1 U (T (v)) + ... + bm U (T m (v)) = b0 T 0 (U (v)) + b1 T (U (v)) + ... + bm T m (U (v)) = b0 T 0 (g(T )(v)) + b1 T (g(T )(v)) + ... + bm T m (g(T )(v)) = b0 g(T )(T 0 (v)) + b1 g(T )(T (v)) + ... + bm g(T )(T m (v)) = g(T )(b0 T 0 (v) + b1 T 1 (v) + ... + bm T m (v)) = g(T )(x). Thus, U = g(T ) for some polynomial g. (⇐) Suppose U = g(T ) for some polynomial g. Define g(x) = a0 + a1 x + ... + an xn . Then, U T (x) = a0 T 0 (T (x)) + a1 T (T (x)) + ... + an T n (T (x)) = a0 T (T 0 (x)) + a1 T (T (x)) + ... + an T (T n (x)) = T (a0 T 0 (x) + a1 T (x) + ... + an T n (x)) = T U (x). Therefore, we have that U T = T U . We have shown that if U is a linear operator on V , then U T = T U if and only if U = g(T ) for some polynomial g(t).
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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 over F with product h·, ·i. If T : V → W is linear, prove that hx, yi0 = hT (x), T (y)i defines an inner product on V if and only if T is one-to-one. Proof. (⇒) Suppose hx, yi0 = hT (x), T (y)i defines an inner product on V . Let T (x) = T (y) which, since T is linear, implies T (x) − T (y) = T (x − y) = 0. From our definition above, we have that hx − y, x − yi0 = hT (x − y), T (x − y)i = hT (x), T (x − y)i − hT (y), T (x − y)i = 0. By definition (particularly part (d) on page 330) 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, hx + z, yi0 = hT (x + z), T (y)i = hT (x), T (y)i + hT (z), T (y)i. Since W is an inner product space, we have that hx + z, yi0 = hx, yi0 + hz, yi0 so it follows immediately that hT (x), T (y)i = hx, yi0 . We have proven that hx, yi0 = hT (x), T (y)i 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 2 k X X a i vi = |ai |2 ||vi ||2 . i=1
i=1
Proof. We apply the definition of inner product space and Theorem 6.2 to get the following 2 * k k + k k k X k X X X X X |ai |2 ||vi ||2 . aj a ¯i < vj , vi >= ai v i = aj vj , ai vi = i=1
j=1
i=1
j=1 i=1
i=1
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 subspace of V . Prove the following results. (a) S0 ⊆ S implies that S ⊥ ⊆ S0⊥ . (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 implies that y ∈ S0 . But S0 ⊆ S, so x ∈ S0⊥ . Thus, S ⊥ ⊆ S0⊥ 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 ⊥ 1
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 ⊥ . Since the only vector that is orthogonal to itself is the zero vector, we have that W ∩ W ⊥ = {0}. Let v ∈ V . By Theorem 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 )⊥ = W1⊥ ∩ W2⊥ and (W1 ∩ W2 )⊥ = W1⊥ + W2⊥ . Hint for the second equation: Apply Exercise 13(c) to the first equation. Proof of (a). (⊆) Let x ∈ (W1 + W2 )⊥ . By definition, it follows immediately that x ∈ W1⊥ ∩ W2⊥ . So, (W1 + W2 )⊥ ⊆ W1⊥ ∩ W2⊥ . (⊇) Let x ∈ W1⊥ ∩ W2⊥ . Then, x ∈ W1⊥ and x ∈ W2⊥ . By linearity, it follows that x ∈ (W1 + W2 )⊥ . So, (W1 + W2 )⊥ ⊇ W1⊥ ∩ W2⊥ . Thus, since (W1 + W2 )⊥ ⊆ W1⊥ ∩ W2⊥ and (W1 + W2 )⊥ ⊇ W1⊥ ∩ W2⊥ , (W1 + W2 )⊥ = W1⊥ ∩ W2⊥ . Proof of (b). (⊆) Let x ∈ (W1 ∩ W2 )⊥ . Then, since W1⊥ ⊆ W1⊥ + W2⊥ and W2⊥ ⊆ W1⊥ + W2⊥ by a previous exercise, it follows that x ∈ W1⊥ + W2⊥ so (W1 ∩ W2 )⊥ ⊆ W1⊥ + W2⊥ (since x is perpendicular to vectors that is in both W1 and W2 ). (⊇) Let x ∈ W1⊥ + W2⊥ . Then, by definition, x = w1 + w2 where w1 ∈ W1⊥ and w2 ∈ W2⊥ . But, W1⊥ ⊆ W1⊥ + W2⊥ and W2⊥ ⊆ W1⊥ + W2⊥ so, in consideration of Exercise 6.2.13c, (W1 ∩ W2 )⊥ ⊆ W1⊥ + W2⊥ which means that x ∈ (W1 ∩ W2 )⊥ (i.e., w1 and w2 is perpendicular to the intersection of W1 and W2 ). Therefore, (W1 ∩ W2 )⊥ ⊇ W1⊥ + W2⊥ . In conclusion, since (W1 ∩ W2 )⊥ ⊆ W1⊥ + W2⊥ and (W1 ∩ W2 )⊥ ⊇ W1⊥ + W2⊥ , we have that (W1 ∩ W2 )⊥ = W1⊥ + W2⊥ . R1 6.2.22. Let V = C([0, 1]) with the inner product < f, g >= 0 f (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 of h in W . 6.3.6. Let T be a linear operator on an inner product space V . Let U1 = T + T ∗ and U2 = T T ∗ . Prove that U1 = U1∗ and U2 = U2∗ . Proof. Let T be a linear operator on an inner product space V and suppose that U1 = T +T ∗ and U2 = T T ∗ . We first prove that U1 = U1∗ . By our assumption and Theorem 6.11, U1∗ = (T + T ∗ )∗ = T ∗ + (T ∗ )∗ = T ∗ + T = U1 . In a similar argument, we will show that U2 = U2∗ . U2∗ = (T T ∗ )∗ = T ∗ (T ∗ )∗ = T ∗ T = U2 . Thus, we have that U1 = U1∗ and U2 = U2∗ 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 if T 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 . Choose x, y ∈ V . Suppose T is invertible. Then, hT ∗ (T −1 )∗ (x), yi = h(T −1 )∗ (x), T (y)i = hx, T −1 T (y)i = hx, yi 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 that N (T ) = W ⊥ . Proof. Suppose that V = W ⊕ W ⊥ and T is the projection on W along W ⊥ . Since V = W ⊕ W ⊥ , we have that V = W + W ⊥ . Let v1 , v2 ∈ V . Then, there exists w1 , w2 ∈ W and w˜1 , w˜2 ∈ W ⊥ such that v1 = w1 + w˜1 and v2 = w2 + w˜2 . Now, in consideration that T is the projection on W along W ⊥ , hv1 , T (v2 )i = hw1 + w˜1 , T (w2 + w˜2 )i = hw1 + w˜1 , w2 i = hw1 , w2 i. Similarly, hT (v1 ), v2 i = hT (w1 + w˜1 ), w2 + w˜2 i = hw1 , w2 + w˜2 i = hw1 , w2 i. So, it follows that hv1 , T (v2 )i = hT (v1 ), v2 i which means hv1 , T (v2 )i = hv1 , T ∗ (v2 )i. This implies that T = T∗ 6.3.11. For a linear operator T on an inner product space V , prove that T ∗ T = T0 implies T = T0 . Is the same result true if we assume that T T ∗ = T0 . Proof. Let T be a linear operator on an inner product space V . Suppose T ∗ T = T0 . Pick x ∈ V . Then, hT ∗ T (x), xi = hT0 (x), xi = h0, xi = 0 by Theorem 6.1(c). But, we also have that hT ∗ T (x), xi = hT (x), T (x)i so hT (x), T (x)i = ||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 T T ∗ = T0 .
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Proof. Let T be a linear operator on an inner product space V . Suppose T T ∗ = T0 . Pick x ∈ V . Then, hT T ∗ (x), xi = hT0 (x), xi = h0, xi = 0. But hT T ∗ (x), xi = hT ∗ (x), T ∗ (x)i which, analogously to the argument above, we have that T ∗ (x) = 0 so T ∗ = T0 . Then, (T ∗ )∗ = T0∗ so T = T0 (note that the adjoint of the zero operator 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 hx, T ∗ (y)i = hT (x), yi = 0 for all y ∈ V . By Theorem 6.1, 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 follows that (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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