174012203-Linalg-Friedberg-Solutions.pdf

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   V  ,  , show that  (a  ( a + b)(x )(x + y ) =  ax + ay + bx + by for by  for any  x,  x, y  ∈  V  and  V  and any  a,  a, b ∈  F .  F . Proof. Let x, y ∈ V  and a, b ∈ F . F . No Notte tha thatt (a (a  +  b)  b ) ∈ F   F   (it (it is a scala scalar) r).. By (VS7  (VS7 ), ), we have have that that (a + b  +  b)( )(x x + y  +  y)) = (a + b  +  b))x + (a ( a + b  +  b))y . By (VS8  (VS8 ), ), (a + b  +  b))x  =  ax +  ax  + bx  bx.. Likewi Likewise, se, (a (a + b  +  b))y =  ay  + by  +  by.. Thus, Thus, (a + b)x + (a (a + b)y  = ax  =  ax + bx + ay + ay + by. by. Finally, by (VS1 (VS1 ), we have (a ( a + b)(x )(x + y ) =  ax + ay + ay + bx + by. by. 1.1.9.  Prove Corollaries 1 and 2 of Theorem 1.1 and Theorem 1.2(c). a) Corollary a)  Corollary 1.  The vector  0  described in  VS3 is VS3  is unique. b) Corollary b)  Corollary 2.  The vector  y  described in  VS4 is VS4  is unique. c) Theorem c)  Theorem 1.2 (c). a0 = 0  for each  a ∈  F .  F . Note that 0 denotes the zero vector. Proof of (a).   Suppose ˜0 ∈ V   V   such that x  +  ˜0 = x   for all x ∈ V . V . By (VS3  (VS3 ), ), x  + 0 = x. It follo follows ws that that ˜  ˜ x + 0 =  x + 0. By (VS1 (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   such that x + y˜  = 0. That That is, is, both y   and y˜  are additive inverses of  x (VS4 ). Since x + y  +  y =  = 0 and  x +  x  + y˜ = 0, we have that x + y  +  y =  =  x +  x  + y˜. By (VS1 (VS1 ), we have y  + x  +  x =  = y˜ + x  +  x.. We apply the Cancellation the  Cancellation Law  to Law  to get y get  y  = y˜ as required. Proof of (c). Let a Let  a  ∈  F .  F . By (VS3  (VS3 ), ), a  a00 =  a(0  a (0 + 0). By (VS7  ( VS7 ), ), a  a00 =  a0  a 0 + a +  a0. 0. By (VS1 (VS1 ), 0 + a + a00 =  a0  a 0 + a +  a0. 0. By the Cancellation the  Cancellation Law , 0 =  a0.  a 0. Then, by (VS1 (VS1 ), a ),  a0 0 = 0 as required. 1.2.21. Let  V   V  and  W  be  W  be vector spaces over a field  F  field  F .. Let  Z   Z  = { (v, w) :  v  ∈  V  ∧ w  ∈  W }. Prove that  Z   Z  is a  vector space over  F  over  F  with  with the operations  ( operations  (vv1 , w1 ) + (v (v2 , w2 ) = (v1  + v  +  v2 , w1  + w  +  w2 ) and  c(v1 , w1 ) = (cv1 , cw1 ). Proof.  We will show that  Z  satisfy  Z  satisfy all of the field axioms. We note that  V  and W  and  W  are  are vector spaces. Let (v ( v1 , x1 ), (v2 , x2 ), (v3 , x3 ) ∈  V    V   and (y (y1 , w1 ), (y2 , w2 ), (y3 , w3 ) ∈ W . W . By (VS1 ), (v1 , x1 ) + (v2 , x2 ) = (v1 +  v 2 , x1 +  x 2 ) = (v2 , x2 ) + (v1 , x1 ) = (v2 +  v 1 , x2 +  x 1 ). Thus Thus,, v1 + v2  = v  =  v 2 + v1 . Likewise, (y (y1 , w1 ) + (y2 , w2 ) = (y1 + y2 , w1 + w2 ) = (y2 , w2 ) + (y1 , w1 ) = (y2 + y1 , w2 + w1 ). Thus, w Thus, w 1 + w2  =  w 2 + w1 . Since v Since  v 1 + v2  =  v 2 + v1  and  w 1 + w2  = w  =  w 2 + w1 , by definition, (v (v1 , w1 ) + (v2 , w2 ) = (v1  + v  +  v2 , w1  + w  +  w2 ) = (v2  + v  +  v1 , w2  + w  +  w1 ) = (v2 , w2 ) + (v (v1 , w1 ). We have shown that (VS1 ( VS1 ) holds for Z  for  Z .. Next, by (VS2  (VS2 ), ), ((v ((v1 , x1 ) + (v2 , x2 ))+(v ))+(v3 , x3 ) = (v1 , x1 )+((v )+((v2 , x2 ) + (v3 , x3 )) so (v ( v1 + v2 ) + v3  = v  =  v 1 + (v2 + v3 ). Similarly, ((y ((y1 , w1 ) + (y2 , w2 ))+(y ))+(y3 , w3 ) = (y1 , w1 )+((y )+((y2 , w2 ) + (y3 , w3 )) so (w (w1 + w2 ) + w3  = w  =  w 1 + (w2 + w3 ). It follows that ((v (( v1 + v2 ) + v3 , (w1 + w2 ) + w3 ) = (v1 + (v2 + v3 ), w1 + (w2 + w3 )). Thus, ((v ((v1 , w1 ) + (v2 , w2 )) + (v3 , w3 ) = (v1 , w1 ) + ((v ((v2 , w2 ) + (v ( v3 , w3 )) where (v ( v1 , w1 ), (v2 , w2 ), (v3 , w3 )  ∈  Z .  Z . We have shown shown that (VS2  ( VS2 )) holds for Z  for  Z .. By (VS3 ), VS3 ), there exists (0 v1 , 0v2 ) ∈ V   V   such that (x (xv , yv ) + (0v1 , 0v2 ) + 0 v = (xv , yv ) for all (x ( xv , yv ) ∈ V  . . Similarly Similarly,, there exists (0w1 , 0w2 )  ∈  W  such  W  such that (x (xw , yw ) + (0w1 , 0w2 ) = (xw , yw ) for all (x ( xw , yw )  ∈  W .  W . By definition, (0v1 , 0w1 ) ∈  Z .  Z . Thus, (v, (v, w) + (0v1 , 0w1 ) = (v, w) and so (VS3  ( VS3 )) holds for Z  for  Z .. 1

By (VS4 ), there exists ( −v, −x) ∈ V   V    such that (v, (v, x) + (−v, −x) = (0v1 , 0v2 ) for all (v, ( v, x) ∈ V . V . Lik Likewise, there exists ( −y, −w) ∈  V  such  V  such that (y, (y, w) + ( −y, −w) = (0w1 , 0w2 ) for all (y, (y, w)  ∈  V  .   . It follows that for all (v, (v, w)  ∈  Z ,  Z , there exists (−v, −w)  ∈  Z  such  Z  such that (v, ( v, w) + ( −v, −w) = (0z1 , 0z2 ) where (0z1 , 0z2 )  ∈  Z  so (VS4 ) holds for Z  for  Z .. For all (v, (v, x)  ∈  V  , , 1 · (v, x) = (v, x) so 1 · v  =  v  (by (VS5  (VS5 )). )). Also, Also, for all (y, (y, w)  ∈  W ,  W , 1 · (y, w) = (y, w) so 1 · w  = w  =  w.. By definition (letting c (letting  c =  = 1), 1 · (v, w) = (1 · v, 1 · w) = (v, w) so (VS5  (VS5 )) holds for Z  for  Z .. Let a, b ∈ F . F . By (VS6 ), VS6 ), for all (v, (v, x) ∈ V  , , (ab)( ab)(v, v, x) = (a(b(v, x)). )). Lik Likewis ewise, e, for all all (y, w) ∈ W , W , (ab)( ab)(y, y, w) = (a(b(y, w)). Accordingly, we have (ab (ab))v  = a  =  a((bv) bv ) and (ab (ab))w  = a  =  a((bw). bw). By definition, (ab (ab)( )(v, v, w) = ((ab ((ab))v, (ab) ab)w). It follows from above that (( ab) ab)v, (ab) ab)w) = (a(bv) bv ), a(bw)) bw)) so (VS6  (VS6 )) holds for Z  for  Z .. Let c Let  c  ∈  F .  F . By (VS7  (VS7 ), ), c  c(( ((vv1 , x1 ) + (v2 , x2 )) =  c(  c (v1 , x1 ) + c(v2 , x2 ) = (cv1 , cx1 ) + (cv2 , cx2 ) which which implies implies that c(v1 , v2 ) = (cv1 , cv2 ). Likewise, Likewise, c  c(( ((yy1 , w1 ) + (y ( y2 , w2 )) =  c(  c (y1 , w1 ) + c + c((y2 , w2 ) = (cy1 , cw1 ) + (cy ( cy2 , cw2 ) which implies c implies c((w1 , w2 ) = (cw1 , cw2 ). Then, c Then, c(( ((vv1 , w1 )+(v )+(v2 , w2 )) = (c (c(v1 +v2 ), c(w1 +w2 )) = (cv (cv1 +cv2 , cw1 +cw2 ) = c(v1 , w1 ) + c + c((v2 , w2 ) as required. So (VS7  (VS7 )) holds for Z  for  Z .. For all a, b ∈ F , F , by (VS8  (VS8 ), ), (a  + b  +  b)( )(v, v, x) = ((a ((a  + b  +  b))v, (a  + b  +  b))x) = (av + av  +  bv,ax +  bv,ax  + bx  bx)) where (v, (v, x) ∈ V  . . This implies (a (a + b  + b))v  =  av +  av  + bv  bv.. Similarly Similarly,, (a (a + b  + b)( )(y, y, w) = ((a ((a + b  +  b))y, (a + b  + b))w) = (ay + ay  + by,aw  by,aw +  + bw  bw)) which implies (a (a + b  + b))w  = aw  =  aw +  + bw  bw.. It follows that (a (a + b  + b)( )(v, v, w) = ((a ((a + b  + b))v, (a + b  + b))w) = (av + av + bv,aw  bv,aw +  + bw  bw)) where (v, w) ∈  Z .  Z . We have shown that (VS8  ( VS8 )) holds for Z  for  Z .. We have shown that  Z  is  Z  is a vector space. 1.3.5.   Prove that  A that  A +  + A  At is symmetric for any square matrix  A matrix  A.. Proof. Let A Let  A  ∈  M nxn F ). Note that a square matrix  A  is symmetric if  A  A t =  A where  A  where A  A t is the transpose of  nxn(F ). t t t A. That is, we wish to show that (A ( A + A  + A ) =  A +  A  + A  A . By properties of tranpose, (A ( A + A  + At )t =  A t + (A (At )t = At + A =  A  = A  A +  + A  At as required. 1.3.6.   Prove that  tr that  tr (aA + aA + bB  bB)) =  atr(  atr (A) + btr + btr((B )  for any  A, B  ∈  M nxn F ). nxn(F ) Proof. Let A, Let  A, B  ∈  M nxn F ). By definition of trace,  tr(  tr (A) = nxn(F ).

n 

and  tr (B ) = Aii and tr

i=0

n 

 where  i  is the row Bii  where i

i=0

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

n 

i=0

i=0

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

aAii +

n  i=0

bBii  = a  =  a

n  i=0

n 

(aA+ aA + bB) bB )ii  =

i=0 n

Aii + b



Bii . Finally,

i=0

by definition of trace,  atr(  atr (A) + btr + btr((B ) as required. 1.3.18.  Prove that a subset  W  of W  of a vector space  V  space  V    is a subspace of  V  of  V  if  if and only if  0 if  0  ∈  W  and  ax  ax + y  ∈  W  whenever  a ∈  F  and  x, y  ∈  W .  W . Proof. (⇒) Suppose a subset W  of W  of a vector space V  is V  is a subspace with addition and scalar multiplication defined on V  .   . By definitio definition n of subspace subspace,, we know know that that there there exists exists 0 ∈ W . W . We also know know that ax ∈ W  (closed under scalar multiplication) whenever x whenever  x  ∈  W  and a and  a  ∈  F .  F . And since W  since  W  is  is closed under addition, we have that ax that  ax +  + y  y  ∈  W  for  W  for all y all  y  ∈  W .  W . (⇐) Suppose 0  ∈  W  and ax + ax + y  y  ∈  W    W   for any x, any  x, y  ∈  W  and a  ∈  F .  F . Since we already know that 0  ∈  W ,  W , we only need to show that W  that  W  is  is closed under addition and scalar multiplication. Since a Since  a  ∈  F ,  F , setting a setting  a =  = 1 ∈  F ,  F , we have that 1 · x + y  = x  =  x + y  ∈  W  so W  so W  is  is closed under addition. Since y Since  y  ∈  W  and  W  and 0  ∈  W ,  W , we let y let  y =  = 0 ∈  W .  W . This means that ax that  ax +  + 0 =  ax  ∈  W  so W  so  W    is closed under scalar multiplication. 1.3.19. Let  W   W 1 and  W   W 2  be subspaces of a vector space  V  space  V  .   . Prove that  W   W 1 ∪ W 2  is a subspace of  V  of  V  if  if and only  if  W   W 1  ⊆  W 2 or  W 2  ⊆  W 1 . 2

By (VS4 ), there exists ( −v, −x) ∈ V   V    such that (v, (v, x) + (−v, −x) = (0v1 , 0v2 ) for all (v, ( v, x) ∈ V . V . Lik Likewise, there exists ( −y, −w) ∈  V  such  V  such that (y, (y, w) + ( −y, −w) = (0w1 , 0w2 ) for all (y, (y, w)  ∈  V  .   . It follows that for all (v, (v, w)  ∈  Z ,  Z , there exists (−v, −w)  ∈  Z  such  Z  such that (v, ( v, w) + ( −v, −w) = (0z1 , 0z2 ) where (0z1 , 0z2 )  ∈  Z  so (VS4 ) holds for Z  for  Z .. For all (v, (v, x)  ∈  V  , , 1 · (v, x) = (v, x) so 1 · v  =  v  (by (VS5  (VS5 )). )). Also, Also, for all (y, (y, w)  ∈  W ,  W , 1 · (y, w) = (y, w) so 1 · w  = w  =  w.. By definition (letting c (letting  c =  = 1), 1 · (v, w) = (1 · v, 1 · w) = (v, w) so (VS5  (VS5 )) holds for Z  for  Z .. Let a, b ∈ F . F . By (VS6 ), VS6 ), for all (v, (v, x) ∈ V  , , (ab)( ab)(v, v, x) = (a(b(v, x)). )). Lik Likewis ewise, e, for all all (y, w) ∈ W , W , (ab)( ab)(y, y, w) = (a(b(y, w)). Accordingly, we have (ab (ab))v  = a  =  a((bv) bv ) and (ab (ab))w  = a  =  a((bw). bw). By definition, (ab (ab)( )(v, v, w) = ((ab ((ab))v, (ab) ab)w). It follows from above that (( ab) ab)v, (ab) ab)w) = (a(bv) bv ), a(bw)) bw)) so (VS6  (VS6 )) holds for Z  for  Z .. Let c Let  c  ∈  F .  F . By (VS7  (VS7 ), ), c  c(( ((vv1 , x1 ) + (v2 , x2 )) =  c(  c (v1 , x1 ) + c(v2 , x2 ) = (cv1 , cx1 ) + (cv2 , cx2 ) which which implies implies that c(v1 , v2 ) = (cv1 , cv2 ). Likewise, Likewise, c  c(( ((yy1 , w1 ) + (y ( y2 , w2 )) =  c(  c (y1 , w1 ) + c + c((y2 , w2 ) = (cy1 , cw1 ) + (cy ( cy2 , cw2 ) which implies c implies c((w1 , w2 ) = (cw1 , cw2 ). Then, c Then, c(( ((vv1 , w1 )+(v )+(v2 , w2 )) = (c (c(v1 +v2 ), c(w1 +w2 )) = (cv (cv1 +cv2 , cw1 +cw2 ) = c(v1 , w1 ) + c + c((v2 , w2 ) as required. So (VS7  (VS7 )) holds for Z  for  Z .. For all a, b ∈ F , F , by (VS8  (VS8 ), ), (a  + b  +  b)( )(v, v, x) = ((a ((a  + b  +  b))v, (a  + b  +  b))x) = (av + av  +  bv,ax +  bv,ax  + bx  bx)) where (v, (v, x) ∈ V  . . This implies (a (a + b  + b))v  =  av +  av  + bv  bv.. Similarly Similarly,, (a (a + b  + b)( )(y, y, w) = ((a ((a + b  +  b))y, (a + b  + b))w) = (ay + ay  + by,aw  by,aw +  + bw  bw)) which implies (a (a + b  + b))w  = aw  =  aw +  + bw  bw.. It follows that (a (a + b  + b)( )(v, v, w) = ((a ((a + b  + b))v, (a + b  + b))w) = (av + av + bv,aw  bv,aw +  + bw  bw)) where (v, w) ∈  Z .  Z . We have shown that (VS8  ( VS8 )) holds for Z  for  Z .. We have shown that  Z  is  Z  is a vector space. 1.3.5.   Prove that  A that  A +  + A  At is symmetric for any square matrix  A matrix  A.. Proof. Let A Let  A  ∈  M nxn F ). Note that a square matrix  A  is symmetric if  A  A t =  A where  A  where A  A t is the transpose of  nxn(F ). t t t A. That is, we wish to show that (A ( A + A  + A ) =  A +  A  + A  A . By properties of tranpose, (A ( A + A  + At )t =  A t + (A (At )t = At + A =  A  = A  A +  + A  At as required. 1.3.6.   Prove that  tr that  tr (aA + aA + bB  bB)) =  atr(  atr (A) + btr + btr((B )  for any  A, B  ∈  M nxn F ). nxn(F ) Proof. Let A, Let  A, B  ∈  M nxn F ). By definition of trace,  tr(  tr (A) = nxn(F ).

n 

and  tr (B ) = Aii and tr

i=0

n 

 where  i  is the row Bii  where i

i=0

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

n 

i=0

i=0

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

aAii +

n  i=0

bBii  = a  =  a

n  i=0

n 

(aA+ aA + bB) bB )ii  =

i=0 n

Aii + b



Bii . Finally,

i=0

by definition of trace,  atr(  atr (A) + btr + btr((B ) as required. 1.3.18.  Prove that a subset  W  of W  of a vector space  V  space  V    is a subspace of  V  of  V  if  if and only if  0 if  0  ∈  W  and  ax  ax + y  ∈  W  whenever  a ∈  F  and  x, y  ∈  W .  W . Proof. (⇒) Suppose a subset W  of W  of a vector space V  is V  is a subspace with addition and scalar multiplication defined on V  .   . By definitio definition n of subspace subspace,, we know know that that there there exists exists 0 ∈ W . W . We also know know that ax ∈ W  (closed under scalar multiplication) whenever x whenever  x  ∈  W  and a and  a  ∈  F .  F . And since W  since  W  is  is closed under addition, we have that ax that  ax +  + y  y  ∈  W  for  W  for all y all  y  ∈  W .  W . (⇐) Suppose 0  ∈  W  and ax + ax + y  y  ∈  W    W   for any x, any  x, y  ∈  W  and a  ∈  F .  F . Since we already know that 0  ∈  W ,  W , we only need to show that W  that  W  is  is closed under addition and scalar multiplication. Since a Since  a  ∈  F ,  F , setting a setting  a =  = 1 ∈  F ,  F , we have that 1 · x + y  = x  =  x + y  ∈  W  so W  so W  is  is closed under addition. Since y Since  y  ∈  W  and  W  and 0  ∈  W ,  W , we let y let  y =  = 0 ∈  W .  W . This means that ax that  ax +  + 0 =  ax  ∈  W  so W  so  W    is closed under scalar multiplication. 1.3.19. Let  W   W 1 and  W   W 2  be subspaces of a vector space  V  space  V  .   . Prove that  W   W 1 ∪ W 2  is a subspace of  V  of  V  if  if and only  if  W   W 1  ⊆  W 2 or  W 2  ⊆  W 1 . 2

Proof. (⇒) Suppose W  Suppose  W 1 and W  and  W 2  are subspaces of a vector space  V    V   and assume W  assume  W 1  ∪ W 2  is a subspace. We ⊆  W 2 and W  ⊆  W 1 . Since W  ⊆ W 2 , there exists a wish to prove by contradiction so assume  W 1   and  W 2   Since  W 1   exists  a  ∈  W 1 but ∈  W 2 . Also,   W 1 , there exists b  ∈  W 2 but b   ∈  W 1 . Sinc a   Also, since since W 2 ⊆ Sincee W 1  ∪ W 2  is a subspace, it must be that (a (a + b  + b)) ∈  W 1 or (a + b  + b)) ∈  W 2 . We have two cases. Case I:   Assume (a (a  + b  +  b)) ∈ W 1 . By the additive additive inverse inverse property, property, a  + b  +  b +  + (−a) = b ∈ W 1   which contra∈  W 1 . dicts our assumption that b that  b   Case II:  Assume (a (a + b  +  b))  ∈  W 2 . By the additive additive inverse inverse property property, a + b  + b +  + ( −b) =  a  ∈  W 2  which contradicts our assumption that  a   ∈  W 2 . We have shown that if  W  if  W 1  ∪ W 2  is a subspace, then  W 1  ⊆  W 2 or W  or  W 2  ⊆  W 1 . (⇐) Suppose W  Suppose  W 1  ⊆  W 2 or W  or  W 2  ⊆  W 1 . We have two cases. Case I: Assume I:  Assume W   W 1  ⊆  W 2 . Then, by definition of union, W  union,  W 1  ∪ W 2  = W   =  W 2  which is a subspace of  V  of  V  . . Case II: Assume II:  Assume W   W 2  ⊆  W 1 . Then, by definition of union, W  union,  W 1  ∪ W 2  = W   =  W 1  which is a subspace of  V  of  V  . . In either cases, we have that  W 1 ∪ W 2  is a subspace if  W   W 1  ⊆  W 2 or W  or  W 2  ⊆  W 1 . 1.3.23. Let  W 1 and  W 2  be subspaces of a vector space  V  space  V  .  . a) Prove a)  Prove that  W 1  + W   +  W 2  is a subspace of  V   V   that contains both  W 1 and  W 2 . b) Prove b)  Prove that any subspace of  V  of  V    that contains both  W   W 1 and  W   W 2  must also contain  W 1  + W   +  W 2 . Proof of (a).   We define W 1  +  W 2 = {w1  +  w 2 |w1 ∈ W 1  ∧  w 2 ∈ W 2 }. We first show show that that W 1  +  W 2 is a subspace. First, since 0  ∈  W 1  and 0  ∈  W 2 , we have that 0 = 0 + 0  ∈  W 1  + W   +  W 2 . Next, if  w  w 1 , w2  ∈  W 1  + W   +  W 2 , there exists x exists  x  ∈  W 1 and y and  y  ∈  W 2  such that w that  w 1  =  x +  x  + y  y.. Similarly, there exists z exists  z  ∈  W 1 and t and  t  ∈  W 2  such that w2  = z  =  z + t. Then, w Then,  w 1 + w2  = (x + y) + (z + t) = (x + z ) + (y + t) ∈  W 1 + W 2  (since W   (since  W 1  and W   and  W 2  are subspaces, x + z  ∈  W 1  and y  and  y + t  ∈  W 2 ) so W  so  W 1 + W 2  is closed under addition. Lastly, since W  since  W 1  and W   and  W 2  are subspaces, we have that q that  qw and  qw any  q  ∈  F   Thus,  aw 1  + aw  +  aw2  =  a(  +  w2 ) ∈  W 1  + W   +  W 2 w1  ∈  W 1 and q w2  ∈  W 2  for any q   F   by definition. Thus, aw  a (w1  + w so W  so  W 1 + W   +  W 2  is closed closed under scalar multiplicati multiplication. on. Thus, Thus, W   W 1  + W   +  W 2  is a subspace of  V   V  . . We now show W 1 ⊆ W 1 +  W 2 and W 2 ⊆ W 1 +  W 2 . By defin definit itio ion n of subspa subspace ce,, we know know that that every every vector in W  in  W 1 + W 2  has an additive inverse which means that with any arbitrary vector we can always obtain 0  ∈  W 1  + W   +  W 2 . And since, 0  ∈  W 1  and 0  ∈  W 2  by definition of subspace, it follows that  W 1  ⊆  W 1  + W   +  W 2 and W 2  ⊆  W 1 + W   +  W 2 . Proof of (b).   Suppose W   W   is a subspace of  V . V . Assu Assume me W 1 ⊆ W  and W 2 ⊆ W . W . We wish wish to prove prove that that W 1  + W   +  W 2 ⊆  W .  W . Let u  ∈  W 1  + W   +  W 2 . From part (a), (a), we define u  =  x +  x  + y  y  where x  ∈  W 1 and y ∈  W 2 . Since Since W 1  ⊆  W ,  W ,  x  ∈  W .  W . Similarly, since W  since W 2  ⊆  W ,  W ,  y  ∈  W .  W . And since W  since  W  is  is a subspace, we know that u that  u =  = x  x + y  ∈  W  where x, where  x, y  ∈  W .  W . Since W  Since  W  was  was arbitrary, we have shown that any subspace of  V  of  V    that contains both  W 1 and W 2  must also contain W  contain  W 1  + W   +  W 2 . 1.3.30. Let  W 1 and  W 2  be subspaces of a vector space  V . V . Prove Prove that  V  is V  is the direct sum of  W 1 and  W 2 if  and only if each vector in  V  in  V  can  can be uniquely written as  x1  + x  +  x2  where  x1  ∈  W 1 and  x2  ∈  W 2 . Proof. (⇒) Suppose W  Suppose  W 1 and W  and  W 2  are subspaces of a vector space  V  .   . Assume Assume that that V  is V  is a direct sum of  W 1 and W 2  (that is, V  = W 1  ⊕ W 2 ). Then, Then, each vecto vectorr in V  can V  can be written as x1  + x  +  x2   where x1 ∈ W 1 and x2  ∈  W 2 . We wish to show that  x 1 + x2  is unique so assume  x 1 + x2  = y  =  y1 + y2  where y  where  y 1  ∈  W 1 and y and  y 2  ∈  W 2 . It follows that  x 1  − y1  =  y2  − x2 . Because V  Because  V  =  W 1 ⊕ W 2 , we know that W  that  W 1 ∩ W 2  =  { 0}. Then, in knowing  − ∈  − ∈ that W 1 and W  and  W 2  are disjoint and that x1 y1  W 1 and y and  y 2 x2  W 2 (W 1 and W  and  W 2  are subspaces of  V  so both are closed under addition), we have that  x 1  − y1  = 0 and y and  y 2  − x2  = 0. Thus, Thus, x1 =  y 1 and x2  =  y 2 so x1  + x  +  x2  is unique. (⇐) Let W 1 and W 2  be subspaces of a vector space  V .  V  . Suppose each each vector in V  in  V  can  can be uniquely written as x1  + x  +  x 2   where x1 ∈ W 1 and x2 ∈ W 2 . Then, Then, it follows follows that that W 1  + W   +  W 2 = V . V . Next Next,, we let let t  be in the intersection of  W 1 and W 2  (i.e. t  ∈  W 1  ∩ W 2 ). Since W  Since  W 1 and W  and  W 2  are subspaces, we know that 0  ∈  W 1 and 3

0 ∈  W 2 . Then, t = t +0 where t  ∈  W 1  and 0  ∈  W 2 . Also, t = 0 + t where 0  ∈  W 1 and t  ∈  W 2 . It must follow that t  = 0 so W 1  ∩ W 2  =  { 0}  since t  ∈  W 1 ∩ W 2 . Since W 1  + W 2  =  V  and W 1  ∩ W 2  =  { 0}, by definition of  direct sum, we have  V  = W 1  ⊕ W 2  as required. We conclude that V   is the direct sum of  W 1 and W 2   if and only if each vector in V   can be uniquely written as x 1  + 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  = v 2  + 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   W . Since v ∈   W , we have by definition that v + v ∈   v + W , a contradiction. contradiction so suppose v ∈ 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 v 1  + W  = v 2  + W . Suppose x  ∈  v 1 + W  and y  ∈  v 1 + W . By definition of coset, we have that x = v 1 + a where a  ∈  W . Likewise, we have that y  = v 2 + b where b  ∈  W . Since v 1 + W  = v 2 + W , we have that x = y. That is, v 1 + a = v 2 + b which can be rewritten as  v 1 − v2  = b − a. Since a  ∈  W , b  ∈  W  and W  is a subspace, we have that  b − a  ∈  W . Thus, since v 1  − v2  = b − a, it follows that v 1  − v2  ∈  W   as required. (⇐) Now suppose v 1  − v2 ∈  W . We wish to show that v 1  + W  = v 2  + W . That is, v1  + W  ⊆  v 2  + W  and v2  + W  ⊆  v 1  + 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  −  v 2 , a ∈ W , we set b = (v1  −  v 2 ) + a. Then, y = v 2  + (v1  − v2 ) + a = v 1  + a. Since x  = v 1  + a, we have that x = y so x  ∈  v 2  + W . Because x  ∈  v 1  + W  and x  ∈  v 2  + W , v 1  + W  ⊆  v 2  + 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  −  v 2 ∈ W  and W  is a subspace, we have that v2  −  v 1 ∈ W  (it is clear that v2  + (−1)v1 ∈ W   because W   is closed under addition and scalar 4

multiplication). Now, since v 2  − v1 , b ∈  W , we set a  = (v2 − v1 ) + b. Then, x  = v 1  + (v2  − v1 ) + b = v 2  + b. Since y  = v 2  + b, x  = y so y  ∈  v 1  + W . Since y  ∈  v 2  + W  and y  ∈  v 1  + W , v 2  + W  ⊆  v 1 + W . Because v 1  + W  ⊆  v 2  + W  and v 2  + W  ⊆  v 1  + W , we conclude that v 1 + W  = v 2  + 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 v 1  + W  = v`1  + W  and v 2  + W  = v`2  + W , by part (b), we have that  v 1  − v`1  ∈  W  and v 2 − 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  −  a ` v1 ∈ 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  av 1 − a ` v1  ∈  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 S 1), let v 1  + W  ∈ S  and v 2 + W  ∈  S  where v 1 , v2  ∈  V   . Then, (v1  + W ) + (v2 + W ) = (v1 + v2 ) + W  and (v2  + W ) + (v1  + W ) = (v2  + v 1 ) + W  by the definition in part (c). Since v1 , v2 ∈ V   , by (V S 1), we know that v1 +v2  ∈  V  and v2 +v1  ∈  V  so v1 +v2  =  v 2 +v1 . Thus, (v1 +v2 )+W  = (v2 +v1 )+W  so (V S 1) 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  + v 2 ) + W ) + (v3  + W ) = ((v1  + v2 ) + v 3 ) + W  = (v1  + (v2  + v3 )) + W  = (v1  + W ) + ((v2  + v3 ) + W ) = (v1  + W ) + ((v2  + W ) + (v3  + W )) which shows that (V S 2) 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 S 3) 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 S 4) holds. We now verify (V S 5 ). 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 S 5) 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 S 6) holds. For (V S 7), let v 1 + W  ∈  S  and v 2 + W  ∈ S  where v 1 , 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 S 7) 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 S 8) holds. In conclusion, since all of the field axioms hold,  S  is a vector space. 1.4.13.  Show that if  S 1  and  S 2  are subsets of a vector space  V  such that  S 1  ⊆  S 2 , then  span(S 1 )  ⊆  span(S 2 ). In particular, if  S 1  ⊆  S 2 and   span(S 1 ) =  V   , deduce that  span(S 2 ) =  V  . Proof.   Suppose S 1 and S 2  are subsets of a vector space V   and that S 1 ⊆  S 2 . We wish to show that every element of  span(S 1 ) is contained in  span(S 2 ). Choose z ∈  span(S 1 ). By definition of span, there exists x1 , x2 ,...,xn ∈ S 1  such that z = a1 x1  +  a 2 x2  +  ... + a n xn   where a1 , a2 ,...,an ∈ F . But, since S 1 ⊆ S 2 ,

5

x1 , x2 ,...,xn ∈ S 2   also which means that z ∈   span(S 2 ) by definition of linear combination. Since z was arbitrary, we have that span(S 1 )  ⊆  span(S 2 ). Suppose span(S 1 ) =  V   . We now show that  span(S 2 ) =  V   . Since span(S 1 ) = V   , from above, V  ⊆  span(S 2 ). However, we have that S 2  ⊆  V  which means that span(S 2 )  ⊆  V   . That is, x 1 , x2 ,...,xn  ∈  S 2  is also contained in V  so by definition, we know that  span(S 2 )  ⊆  V . Since V  ⊆  span(S 2 ) and span(S 2 )  ⊆  V   , we have that span(S 2 ) =  V  . 1.4.15. Let  S 1 and  S 2  be subsets of a vector space  V   . Prove that  span(S 1 ∩ S 2 )  ⊆  span(S 1 ) ∩ span(S 2 ). Give  an example in which  span(S 1  ∩ S 2 ) and  span(S 1 ) ∩ span(S 2 ) are equal and one in which they are unequal. Proof.   Suppose S 1 and S 2   be subsets of a vector space V . Let v ∈  span(S 1  ∩ S 2 ), Then, by definition of  span, there exists x1 , x2 ,...,xn ∈ S 1  ∩ S 2   such that v = a1 x1  + a 2 x2  + ... + an xn   where a1 , a2 ,...,an ∈ F . Since x 1 , x2 ,...,xn  ∈  S 1 ∩ S 2 , by definition of set intersection  x 1 , x2 ,...,xn  ∈  S 1  and  x 1 , x2 ,...,xn  ∈  S 2 . From this, we know that v  = a 1 x1 + a2 x2 + ... + an xn  ∈  span(S 1 ) and v  = a 1 x1 + a2 x2 + ... + an xn  ∈  span(S 2 ). By definition of set intersection, since v  ∈  span(S 1 ) and v ∈  span(S 2 ), we have that v  ∈  span(S 1 ) ∩ span(S 2 ). Since v  was arbitrary, we have shown that  span(S 1  ∩ S 2 )  ⊆  span(S 1 ) ∩ span(S 2 ). Examples.   Suppose V  = R 2 , S 1  =  {(1, 3)}  and  S 2  =  { (2, 7)}. Then, S 1  ∩ S 2  =  ∅  so  span(S 1  ∩ S 2 ) =  {0}  by definition of subspace. Next, we have that  span(S 1 ) ∩ span(S 2 ) =  {0}  because span(S 1 ) =  { (1a, 3a) :  a  ∈ R} and   span(S 2 ) = {(2b, 7b) : b ∈ R}   have no elements in common. In this example,   span(S 1 ∩  S 2 ) = span(S 1 ) ∩ span(S 2 ). Now consider V  = R2 , S 1 = {(8, 4)} and S 2 = {(4, 2)}. Again, we have that S 1  ∩  S 2 = ∅ so span(S 1  ∩ S 2 ) = {0}   by definition of subspace. However, since   span(S 1 ) = {(8a, 4a) : a ∈ R} and   span(S 2 ) = {(4b, 2b) : b  ∈ R}, we know that  span(S 1 ) ∩ span(S 2 )   =  { 0}  (since span(S 1 ) = span(S 2 )). In this example, span(S 1  ∩ S 2 )   =  span(S 1 ) ∩ span(S 2 ) 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   = 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   = 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   = 0 or d   = 0. By the distributive law of logic, we have two cases.

 0. Since c = a +  b, we have a +  b =  0 which implies {u +  v, u  −  v }   is linCase I:   Assume that c = early dependent because a and b cannot both be 0 since a + b   = 0 for a(u + v) + b(u − v) = 0 (our assumption that c   = 0 would not hold if they are both zero). Case II:   Assume that d =  0. Since d = a  −  b, we have a  −  b =  0 which implies {u + v, u  −  v }   is linearly dependent because a and b cannot both be 0 since a − b   = 0 for a(u + v) + b(u − v) = 0 (our assumption that c   = 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.  2). In a similar argument, since f  = e, e + f  = f  + f  = 2f  = 0 which implies f   = 0 (since   char(F ) = 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 )   = 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   =0  0. By the distributive law of logic, we have three cases. or e   = 0 or f  =

 0. Since a + b = d, a + b =  0 which implies {u + v, u + w, v +  w }   is linearly deCase I:   Suppose d = pendent because a  and  b  cannot both be equal to zero (recall that we’d assumed  a = b = c = 0). Case II:   Suppose e =  0. Since a + c = e, a + c =  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  =  0. Since b +  c = f , b +  c =  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 ˜ x x `  = y˜. Since x `  = y˜, we solve for y` to obtain y` = a which is defined since a is nonzero. Thus, we have that ˜ 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 ˜ + (˜ qb)v = f  ˜ x Since  { u, v}  span V  , x ˜u + y˜v =  f   which implies that wa = ˜ x ˜  and q˜b = y˜. Then, it follows that w = ˜ and a ˜ ˜ y y ˜ x q˜  = b so w(au) ˜ + q˜ (bv) = a (au) + b (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  W 1 and  W 2   be subspaces of a finite-dimensional vector space  V . Determine necessary and  sufficient conditions of  W 1 and  W 2  so that  dim(W 1  ∩ W 2 ) = dim(W 1 ). Proof.  We believe that dim(W 1  ∩ W 2 ) =  dim(W 1 ) if and only if  W 1  ⊆  W 2 . We will prove this claim. (⇒) Let W 1  and W 2  be subspaces of a finite-dimensional vector space  V   . Suppose dim(W 1 ∩ W 2 ) =  dim(W 1 ). Let β  be a basis for W 1 ∩ W 2 . By definition of set intersection, we know that  W 1 ∩ W 2  ⊆  W 1  so we can extend β   to be a basis for W 1 . It follows that for any  x  ∈  W 1 , x  ∈  W 1 ∩ W 2 . This implies that W 1  ⊆  W 1 ∩ W 2 . Since W 1  ∩ W 2 ⊆  W 1 and W 1 ⊆  W 1  ∩ W 2 , we have that W 1  ∩ W 2 = W 1 . Now, by definition of set intersection (specifically, the property of subset intersection),  W 1  ∩ W 2  = W 1  implies that W 1  ⊆  W 2 . (⇐) Now assume that W 1 ⊆ W 2 . Let β  be a basis for W 1  ∩ W 2 . Since W 1 ⊆ W 2 , by the definition of set intersection (property of subset intersection), it follows that  W 1 ∩ W 2  =  W 1 . Since W 1 ∩ W 2  =  W 1  (which tells us that W 1 ∩ W 2  ⊆  W 1 ), β   can be extended to be a basis for  W 1 . This implies that dim(W 1 ∩ W 2 ) =  dim(W 1 ). We have shown that dim(W 1  ∩ W 2 ) =  dim(W 1 ) if and only if  W 1  ⊆  W 2 . 8

1.6.29.  (a) Prove that if  W 1 and  W 2  are finite-dimensional subspaces of a vector space  V   , then the subspace  W 1  + W 2   is finite-dimensional, and  dim(W 1  + W 2 ) =  dim(W 1 ) + dim(W 2 ) − dim(W 1  ∩ W 2 ). (b) Let  W 1 and  W 2  be finite-dimensional subspaces of a vector space  V   , and let  V  = W 1  +  W 2 . Deduce  that  V  is the direct sum of  W 1 and  W 2  if and only if  dim(V  ) =  dim(W 1 ) + dim(W 2 ). Proof of (a).   Suppose W 1 and W 2   are finite-dimensional subspaces of a vector space V . Let β   be a basis for W 1  ∩  W 2  such that β  = {u1 , u2 ,...,uk }  (since we know W 1 and W 2   are finite-dimensional). Since W 1 ∩ W 2 ⊆ W 1 , we extend β   to a basis of  W 1 so γ  = {u1 , u2 ,...,uk , v1 , v2 ,...,vm }. Likewise, µ = {u1 , u2 ,...,uk , w1 , w2 ,...,w p }. We wish to show that τ  = {u1 , u2 ,...,uk , v1 , v2 ,...,vm , w1 , w2 ,...,w p }  which implies that W 1 + W 2   is finite-dimensional (as it contains k = m = p   number of vectors) and that dim(W 1 +W 2 ) =  k +m+ p = k +m+ p+(k − k) = (k+m)+ (k+ p) − k = dim(W 1 )+ dim(W 2 ) − dim(W 1 ∩ W 2 ). By definition of basis, we show that  τ   span W 1  + W 2  and that it is linearly independent. Note that γ   span W 1 and µ   span W 2   as they are both bases. That is, for any w1 ∈ W 1 , a1 u1 + ... +  a k uk + b 1 v1 + ...  +  b m vm = w1   and, for any w2 ∈ W 1 , a1 u1 + ...  +  a k uk + c 1 w1 + ...  +  c p w p = w2   for scalars ak , bm , and c p   respectively (Theorem 1.8 ). It is obvious that τ   span W 1  +  W 2   because a1 u1  + ... + ak uk  + b1 v1  + ... + bm vm + c1 w1  + ... + c pw p  = w 1  + w2  for any w 1 + w2  ∈  W 1  + W 2  (recall our given definition of  W 1  + W 2  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 a 1 u1 +...+ak uk +b1 v1 + ...+bm vm  = 0, a k  =  b m  = 0. Similarly, since µ is linearly indepdenent, for a1 u1 +...+ak uk +c1 w1 +...+c p w p  = 0, we know that ak = c p   = 0. But, we had assumed that τ   is linearly dependent which implies that for a1 u1 + ... + ak uk + b1 v1 + ... + bm vm + c1 w1 + ... + c p w p  = 0, we have that coefficients a k , b m , and c p  are not all zeros. Thus, we have a contradiction so  τ   must be linearly independent. By definition of basis, b ecause τ   span W 1 + W 2   and is linearly independent, we have that τ   is a basis for W 1  +  W 2 . Thus, it is finite-dimensional and dim(W 1  +  W 2 ) = (k + m) + (k + p)  −  k = dim(W 1 ) + dim(W 2 ) − dim(W 1  ∩ W 2 ) as required. Proof of (b). Let W 1 and W 2  be finite-dimensional subspaces of a vector space  V  , and let V  = W 1  + W 2 . (⇒) Suppose that V  = W 1 ⊕  W 2 . We wish to prove that dim(V  ) = dim(W 1 ) + dim(W 2 ). Because V  = W 1 ⊕ W 2 , we know that W 1 ∩ W 2  =  { 0}  which implies that dim(W 1 ∩ W 2 ) = 0 as W 1 and W 2  have no unique vectors in common (in fact, they don’t have any vectors in common). So, noting that  V  = W 1 + W 2 , following from from part (a), we have dim(V  ) =  dim(W 1 )+dim(W 2 ) − dim(W 1 ∩ W 2 ) =  dim(W 1 )+dim(W 2 ) as required. (⇐) Suppose dim(V  ) = dim(W 1 ) + dim(W 2 ). We wish to prove that V  = W 1 ⊕  W 2 . Since we know that V  = W 1  + W 2 , we only need to show that W 1 and W 2  are disjoint (i.e., W 1  ∩ W 2 =  { 0}). From part (a), we know that dim(V  ) = dim(W 1 ) + dim(W 2 ) − dim(W 1 ∩ W 2 ). However, from our assumption, we have that dim(W 1  ∩ W 2 ) = 0 which means, by definition of dimension, there isn’t any unique number of vectors in each basis for W 1  ∩ W 2 . This implies that W 1 and W 2  are disjoint. Therefore, since V  = W 1  + W 2 and W 1  ∩ W 2  =  { 0}, by definition of direct sum,  V  = W 1  ⊕ W 2  as required. We have shown that V  is the direct sum of  W 1 and W 2  if and only if  dim(V  ) =  dim(W 1 ) + dim(W 2 ) 1.6.33.  (a) Let  W 1 and  W 2  be subspaces of a vector space  V   such that  V  = W 1 ⊕ W 2 . If  β 1 and  β 2  are bases   for  W 1 and  W 2 , respectively, show that  β 1  ∩ β 2  =  ∅  and  β 1  ∪ β 2  is a basis for  V  . (b) Conversely, let  β 1 and  β 2  be disjoint bases for subspaces  W 1 and  W 2 , respectively, of a vector space  V   . Prove that if  β 1 ∪ β 2  is a basis for  V  , then  V  = W 1 ⊕ W 2 . Proof of (a). Let W 1 and W 2   be subspaces of a vector space V   such that V  = W 1 ⊕  W 2 . Suppose β 1 and β 2   are bases for W 1 and W 2 respectively. Define β 1 = {u1 , u2 ,...,un } and β 2 = {v1 , v2 ,...,vm }. 9

Then, by definition of basis, β 1 and β 2   both are linearly independent and span W 1 and W 2   respectively. Since we know that V  = W 1 ⊕  W 2 , we have that W 1 ∩  W 2 = ∅. By   Theorem 1.18 , we know that span(β 1 ) = W 1 and  span(β 2 ) = W 2 . Since   span(β 1 ) = W 1 ,   span(β 2 ) = W 2 and W 1  ∩  W 2 = ∅, it follows that span(β 1 ) ∩ span(β 2 ) =  ∅  which implies that β 1 ∩ β 2  =  ∅. Now, based on our definition of  β 1 and β 2 , we define β 1 ∪  β 2 = {u1 , u2 ,...,un , v1 , v2 ,...,vm}. We first show that β 1 ∪ β 2  is linearly independent. However, we will prove by contradiction so assume that  β 1 ∪ β 2 is linearly dependent. That is, a 1 u1  + ... + an un + b1 v1  + ... + bm vm  = 0 where scalars  a 1 ,...,an and b 1 ,...,bm are not all zero. Because, β 1   is linearly independent, we have that a1 u1  +  ... + a n un   = 0 where scalars a1  = a 2  = ... = a n  = 0. Likewise, for β 2 , b 1 v1  + ... + bn vn  = 0 where scalars  b 1  = b 2  =  ... = b m  = 0. Thus, a contradiction so it is obvious that β 1  ∪ β 2   is linearly independent. Second, we need to show that β 1  ∪  β 2   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 W 1 and W 2  are subspaces of a vector space  V   . Therefore, we extend β 1 and β 2  to bases for V   . Thus, since β 1 and β 2  span V  (by definition of basis), it follows that β 1 ∪ β 2  span V  . Thus, since β 1  ∪  β 2   is linearly independent and span V   , by definition of basis, we have that β 1  ∪  β 2 is a basis for V   . In conclusion, we have proved that if  β 1 and β 2  are bases for  W 1 and W 2 , respectively, show that β 1 ∩ β 2  =  ∅  and  β 1  ∪ β 2  is a basis for  V  . Proof of (b). Let β 1 and β 2  be disjoint bases for subspaces W 1 and W 2 , respectively, of a vector space V  (i.e., β 1 ∩ β 2  =  ∅ ). Assume β 1 ∪ β 2  is a basis for  V   . We wish to prove that V  = W 1 ⊕ W 2 . That is, we need to show that W 1  ∩ W 2  =  ∅  and  V  = W 1  + W 2 . Since β 1 ∩  β 2 = ∅, we have that   span(β 1 ∩  β 2 ) = {0}. And since β 1   is a basis for W 1 , β 1   span W 1 . Likewise, we have that β 2  span W 2 . Then, by definition of span, span(β 1 ) =  W 1 and span(β 2 ) =  W 2 . Note from problem   1.4.15  of  Homework Set 2   that  span(W 1  ∩  W 2 ) ⊆   span(W 1 ) + span(W 2 ). It follows that span(β 1 ) ∩ span(β 2 ) =  W 1  ∩ W 2  =  ∅  as  β 1  ∩ β 2  =  ∅ . Now we wish to show that V  = W 1  +  W 2 . We know that   span(β 1 ) = W 1 and   span(β 2 ) = W 2 . Also recall that β 1 ∪  β 2   is a basis for V . Then, by definition,   span(β 1 ∪  β 2 ) = V . Suppose each vector in span(β 1  ∪  β 2 ) can be uniquely written as w1  +  w 2   where w1 ∈   span(β 1 ) and w2 ∈  span(β 2 ). Then, by problem 1.3.30  of   Homework Set 2 , it follows that V  = W 1  + W 2 .

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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. 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 contradiction so assume that T (S ) is linearly dependent. Since S  is linearly independent, for  a 1 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  a 1 v1 + ... + an vn   = 0 which implies T (a1 v1 + ... + an vn )  =  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)  =  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  v 1 , v2 ,...,vn  ∈ S ,  a 1 v1  + ... + an vn  = 0 where scalars a 1 , a2 ,...an are not all zero. Since T  : V  → W  is linear, we have that  a 1 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

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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 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 = 0. Likewise for U , there exist R(T ) ∩ R(U ) = {0}. Since T  is nonzero, there exists  v 1  ∈ V  such that T (v1 )  v2 ∈ V   such that U (v2 )  = 0. We wish to show by contradiction so assume aT  + bU  = T 0  (note, T 0  denotes the zero transformation) where scalars a, b ∈ F   are not all zero. By the distributive law of logic, we have two cases. Case I:  Assume a  = 0 and b  = 0. Note, since a  = 0, aT  + bU  = T 0  can be rewritten as T  = −ab U  = U ( −ab ) by definition of linear transformation. Then, T (v1 ) = U ( −ab 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 )  = 0, T (v1 ) ∈ R(T ) ∩ R(U ) contradicts the fact that  T   is nonzero.

 0. Note, since b =  0, aT  + bU  = T 0  can be rewritten as  U  = −ba T  = T ( −ba ). Case II: Assume a  = 0 and 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 )  = 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  = T 0 , 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  S 1 and  S 2  are subsets of  V  and  S 1  ⊆ S 2 , then  S 20 ⊆ S 10 . (c) If  V 1 and  V 2  are subspaces of  V   then  (V 1 + V 2 )0 = V 10 ∩ V 20 . Proof of (a). Let T , U  ∈ S 0 and choose a ∈ F . By Theorem 1.3 , we wish to show that T 0  ∈ S 0 and addition and scalar multiplication is closed for S 0 . First, it is clear that for any  x ∈ S ,  T 0 (x) = 0 since T 0  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 S 1 and S 2  are subsets of  V . Also assume S 1 ⊆ S 2 . Let T  ∈ S 20 . By definition, we have that for all x ∈ S 2 , T (x) = 0. However, since S 1  ⊆ S 2 , we know that x ∈ S 1  also. This implies that for all x ∈ S 1 , T (x) = 0 so we have  T  ∈ S 10 . By defiition of subset, we can conclude that S 20 ⊆ S 10 . Proof of (c).   Suppose V 1 and V 2  are subspaces of  V . (⊆) Let T  ∈ (V 1  +  V 2 )0 . Then, by definition, for all a ∈ V 1  +  V 2 , T (a) = 0. By the definition on  page  22  of set addition, we have that a = v1  + v2  for all v1 ∈  V 1 and v2 ∈  V 2 so T (v1  + v 2 ) = 0. Since V 1 is a subspace, we know that 0 ∈ V 1 . So, setting v1  = 0, we have that T (0 + v 2 ) = T (v2 ) = 0 for all v2 ∈ V 2 which implies that T  ∈ V 20 . With similar consideration, since V 2   is a subspace, 0 ∈ V 2   so we have that T (v1  + 0) = T (v1 ) = 0 for all  v 1  ∈ V 1 . This implies that T  ∈ V 10 . Since T  ∈ V 20 and T  ∈ V 10 , it follows that T  ∈ V 10 ∩ V 20 . Therefore, we have that (V 1  + V 2 )0 ⊆ V 10 ∩ V 20 . (⊇) Now let T  ∈ V 10 ∩ V 20 . By definition, we have that for all  v 1 ∈ V 1 , T (v1 ) = 0. Likewise, for all v 2  ∈ V 2 , 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 ∈  V 1  + V 2  (by definition of subspace addition) which implies that T  ∈ (V 1  + V 2 )0 . Thus, it follows that (V 1 + V 2 )0 ⊇ V 10 ∩ V 20 . 3

Since (V 1  + V 2 )0 ⊆ V 10 ∩ V 20  and (V 1 + V 2 )0 ⊇ V 10 ∩ V 20 , we have that (V 1  + V 2 )0 = V 10 ∩ V 20  as required. 2.3.11. Let  V  be a vector space, and let  T  : V  → V  be linear. Prove that  T 2 = T 0  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 = T 0 . Pick a ∈ R(T ). By definition of range, there exists v ∈ V   such that T (v) = a. Because T 2 = T 0 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 = T 0  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   =  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  = 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  : R 2 → R by U (r1 , r2 ) =  r 1 . Then, U T (r1 ) =  U (T (r1 )) = U (r1 , 0) =  r 1  (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)  = 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  = R 2 and Z  = R. Define T  : R → R2 by T (r1 ) = (r1 , 0) and define U  : R 2 → R by U (r1 , r2 ) = r 1 . 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 A 2 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 = I B = 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 = I n . Since ABC  = A(BC ) = I n , 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 C AB = (CA)B =  I n , we have that B  is invertible as C A is the multiplicative inverse 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

0 0 0 1

1 0 and B =  0 0    respectively. Clearly, A and B  are not invertible as 0 1 1 0

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

0 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  = I n . (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 = I n . 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 =  I n , A and B is invertible. Since A  is invertible and AB = I n , 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  = I V    then  T U  = I W . 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  = I V  . We wish to prove that T U  = I W . For any v ∈ V , T U (T (v)) = T (U T (v)) = T (I V   (v)) = T (v). Since T (v)  ∈  W  and v  was arbitrary, it follows that  T U  = I W  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  ¯ : V /N (T ) → Z by T (v + ¯ T  N (T )) =  T (v)  for any coset  v  + N (T ) in  V /N (T ). ¯ well-defined; that is, prove that if  v + N (T ) =  v  + N (T ), then  T (v) =  T (v  ). (a) Prove that  T  is ¯ is linear. (b) Prove that  T   ¯ is an isomorphism. (c) Prove that  T   (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 . Suppose ¯ ¯  + N (T )) = T (v  ). Since v + N (T ) = v + N (T ) = v  + N (T ). By definition, T (v + N (T )) = T (v) and T (v   v + N (T ), it follows that  T (v) =  T (v ) as required. ¯ linear which, by definition of linear transformation, is equivalent Proof of (b).  We wish to prove that T  is ¯ ¯ 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). ¯ linear. Thus, we can conclude that T  is ¯ linear and invertible. From part (b), we Proof of (c).  By definition of isomorphism, we must show that T  is ¯ ¯ ¯ = have that T   is linear. By Theorem 2.5 , to show that T  is invertible is equivalent to showing that  dim(T ) ¯ + dim(R(T )) ¯ =  dim((V )/N (T )). dim(V ). By the Dimension Theorem , we know that dim(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  n

Q x x = 

ij i

j

1 ≤  j  ≤ n

n=1

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 coordinate  matrix 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  S 0 of  S  as  S 0 = {f  ∈ V ∗ : f (x) = 0 ∀x ∈ S }. (a) Prove that  S 0 is a subspace of  V  ∗ . (b) If  W  is a subspace of  V  and  x  = 0. ∈ W , prove that there exists  f  ∈ W 0 such that  f (x)  0 0 (c) Prove  (S  ) = span(ψ(S )), where  ψ is defined as in Theorem 2.26. (d) For subspaces  W 1 and  W 2 , prove that  W 1  = W 2  if and only if  W 10 = W 20 . (e) For subspaces  W 1 and  W 2 , show that  (W 1  + W 2 )0 = W 10 ∩ W 20 . 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 s o S 0 is closed under scalar multiplication. Thus, we have shown that  S 0 is a subspace of  V  ∗ .

∈ W . Let β  = {x1 , x2 ,...,xk }  be a basis for W . Since W  Proof of (b).   Suppose W  is a subspace of  V  and x  is a subspace of  V , we can extend β  to V  so β 1  = {x1 ,...,xk , xk+1 ,...,xn }. Then, let (β 1 )∗ be the dual basis to β 1  and define (β 1 )∗ = {f 1 ,...,f k , f k+1 ,...,f n }. Since we have that f i (xi )  = 0, W 0 ∩ {f 1 ,...,f k }. Since xk+1 ,...,xn   ∈ W , it follows that {f k+1 ,...,f n } ∈ W 0 . However, x  ∈ W  implies that for x = w + (a1 x1 + ... + ai xi ) where i > k , there exists an  a o   = 0. That is, f o (x) =  a o   = 0 as required. Proof of (c). (⊆) Pick s  ∈  S  such that ψ (s) = s 1 . Then, s 1 (f ) = f (s) = 0 so  ψ(s)  ∈  (S 0 )0 . Since ψ and f  is linear, it follows that

7

Proof of (d). Let W 1 and W 2  be subspaces. (⇒) Suppose W 10 = W 20 . We wish to prove by contradiction so assume W 1  = W 2 . Then there exists 0 ∈ W 1 and x ∈ W 2 . From part (b), we know that there exists f  ∈ W 1   such that f (x)  x  = 0 which is a 0 0 0 ∈ W 2 . Thus, we have that W 1  = W 2 . contradiction since W 1 = W 2 and f (x)  (⇐) Now suppose that  W 1  = W 2 . Clearly by definition, since W 1  = W 2 , W 10 = W 20 . Thus, we conclude that W 1  = W 2  if and only if  W 10 = W 20 . Proof of (e). (⊆) Let f  ∈ (W 1 + W 2 )0 . Then, f (w1 ) =  f (w2 ) = 0 for w 1  ∈ W 1  and  w 2  ∈ W 2 . Then, we have that f  ∈ W 10 and f  ∈ W 20 so (W 1  + W 2 )0 ⊆ W 10 ∩ W 20 . (⊇) Let f  ∈ W 10 ∩ W 20 . Then, we have that  f (W 1 ) = 0 and  f (W 2 ) = 0 which implies that for  w 1  ∈ W 1 and w2  ∈ W 2 , f (w1  + w2 ) = f (w1 ) + f (w2 ) = 0 + 0 = 0. That is, f  ∈ (W 1 + W 2 )0 so W 10 ∩ W 20 ⊆ (W 1  + W 2 )0 . Since we have that (W 1 + W 2 )0 ⊆ W 10 ∩ W 20 and W 10 ∩ W 20 ⊆ (W 1 +W 2 )0 , we can conclude that (W 1 + W 2 )0 = W 10 ∩ W 20 . 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 ∗ = {f 1 , f 2 ,...,f n }. Prove that  {f k+1 , f k+2 ,...,f n }  is a basis for  W 0 . Proof.   Suppose W   is a subspace of  V . Let β W  = {x1 , x2 ,...,xk }   be an ordered basis of  W   and extend it onto an ordered basis β  = {x1 , x2 ,...,xn } of  V . Let β ∗ = {f 1 , f 2 ,...,f n }. We wish to show that γ  = {f k+1 , f k+2 ,...,f n }  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  a 1 f 1  + ... + an f n  ∈ W 0 where a 1 ,...,an  are not all zero. That is, there exists an  a o such that a o   = 0. This implies that  f (vo ) =  a o   = 0 for v o  ∈ 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(LA ) =  rank(LA) for any  A ∈ M m×n (F ). t

Proof.  We wish to show that  rank(LA ) =  rank(LA ) for any  A ∈ M m×n (F ). Note, L A  :  V  → W  and L A : 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 t

t

8

rank(LA ) =  dim(R(LA )) = dim(W ) − dim(W ∗ ) − dim(R(LA )) = dim(W ) − dim(N (LA )) t

t

t

t

= 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 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 . 3.2.14.

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  +  L B ) ≤   rank (LA ) +  rank (LB ) which implies that   rank(A  +  B ) ≤   rank(A) + rank(B ) as required.  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 . 3.2.17.

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  B C  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  L B  :  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 =  B C  as required. 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. 3.2.19.

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 Let  δ  :  M 2×2 (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 ∈ M 2 (F ) are identical, then  δ (A) = 0. (iii) If  I  is the  2 × 2 identity matrix, then  δ (I ) = 1. 4.1.11.

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

  ( ) and defined it as  where 1 1 1 0 0 1 a c

Proof. Let A ∈ M 2×2 F  above, notice that δ  δ 

0 1 1 0

= δ 

1 1

0 1

b d

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

 +  δ 

1 0

= 1 +  δ 

0 1 1 0

  = 0 which implies that

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

  1 0 0 1 + 1 0 1 0 a c

=  aδ 

b d

c

=  acδ 

 bδ 

d

+ adδ 

1 0

c

d

0 1

+ bcδ 

0 1 1 0

+ bdδ 

0 1 0 1

=  ac (0) + ad(1) + bc(−1) +  bd(0) =  ad − bc =  det



a 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 ∈ M n×n (F ).

Proof.   Choose  A ∈ M n×n (F ). Then, by Theorem 4.8 , det(kA ) =  det (kI n A)

=  det (kI n )det(A) =  k n det(I n )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 ∈ M n×n (F ) as required.

1

  A matrix  M  ∈ M n×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. 4.3.10.

Proof.   Suppose M  ∈  M n×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 ∈  M n×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. matrix  M  ∈ M n×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?  4.3.11.  A

Proof.   Choose  M  ∈ M n×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 2 det(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. For  M  ∈ M n×n (C ), let  M  be the matrix such that  ( M )ij =  M ij  for all  i, j , where  M ij  is the complex  conjugate of  M ij . (a) Prove that  det(M ) =  det (M ). (b) A matrix  Q ∈ M n×n (C )   is called unitary if  QQ∗ = I , where  Q∗ = Qt . Prove that if  Q  is a unitary  matrix, then  |det(Q)| = 1. 4.3.13.

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

2

Proof of (b). Let Q ∈ M n×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  ∈ M n×n (F ) can be written in the form  M  =



A O

B 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, ¯11 ) + 0 +  ...  + 0 =  a · det(C ) =  det (A)det(C ). det(M ) =  a · det(M  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 M  =



A O

B C 



.

Now, taking the cofactor expansion along the first column, ¯11 ) − ... ± ak1 det(M ¯k1 ) + 0 +  ...  + 0 det(M ) =  a 11 det(M  =  a 11 det



A¯11 O

B¯1 C 



− ... ± det



A¯k1 O

B¯k C 



=  a 11 det(A¯11 )det(C ) − ... ± ak1 det(A¯k1 )det(C ) = ( a11 det(A¯11 )) − ... ± ak1 det(A¯k1 )det(C ) =  det (A)det(C ).

Let  T  : P n (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  P n (F ) and  γ  be the standard ordered basis for  F n+1 . (a) Show that  M  = [T ]γ  β  has the form  4.3.22.

(c) Prove that  det(M ) =



(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  :  P n (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  P n (F ) and γ  be the standard ordered basis for  F  n+1 .

Proof of (c).  We will prove that det(M ) =



(cj  − ci ),

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

4

Partial Solutions for Linear Algebra by Friedberg et al. Chapter 5 John K. Nguyen December 7, 2011 Let  A  be an  n × n  matrix that is similar to an upper triangular matrix and has the distinct eigenvalues  λ1 , λ2 ,...,λ k  with corresponding multiplicities  m 1 , m2 ,...,m k . Prove the following statements. 5.2.11.

k

 )=

(a) tra(A

mi λi .

i=1

(b) det(A) = (λ1 )m (λ2 )m ...(λk )m . 1

k

2

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 ,...,λ k   with corresponding multiplicities m1 , m2 ,...,m k . 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

tr(A) =  tr (Q

1

−

BQ) =  tr (Q

1

−

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 ,...,λ k   with corresponding multiplicities m1 , m2 ,...,m k . 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 ∈ M 2×2 (F ), det(AB) = det(A)det(B ). In consideration of this theorem and the corollary on page 223, we have that det(A) =  det (Q−1 BQ ) =  det (Q−1 )det(B )det(Q) = det1(Q) det(B )det(Q) =  det (B ) = (λ1 )m (λ2 )m ...(λk )m as required. 1

5.2.12.

2

k

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 ∈ E λ . 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 ∈ E λ−   so we have that the eigenspace of  T   corresponding to  λ  is the same as the eigenspace of  T −1 corresponding to  λ −1 . 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. Let  A  ∈  M n×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  E λ and  E λ  denote the corresponding eigenspaces for  A  and  At , respectively. 5.2.13.

(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(E λ ) =  dim(E λ ). (c) Prove that if  A  is diagonalizable, then  At is also diagonalizable.

1

Example for (a).   Define A = t =  E A . E A  

1 2 3 4

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

Proof of (b).   Suppose  dim (F ) =  n . Then, by definition, we know that  dim (E λ ) =  n − rank(A − λI ). Then, taking the transpose (recall that   rank (A) =   rank(At ) by a previous exercise), we have that dim(E λ ) = 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 −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 =  Q t At (Q−1 )t . Clear,y B t is diagonal so by definition, we have that  A t 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 ,...,v n }  such that  T (vi ) =  λv i and  U (vi ) =  αv i  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. 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). 5.4.13.

Proof. (⇒) Suppose w ∈  W   and assume dim(W ) =  n . Let β  =  { v,Tv,...,T n−1 v}  be an ordered basis for W . Then, by definition, w =  a 0 v +  a 1 T v  +  ...  +  a n−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) =  a 0 + 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  +  a 1 t  +  ...  +  a n tn and  w  =  a 0 v + a1 T v + ... + an T n v  for some scalars  a 0 , a1 ,...,a n . Since   v,Tv,...,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  T W  . 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  T W  . By part (a), we know that f   splits. Pick λ  to be the root of  f . Then, f (λ) =  det (T W  − λI ) = 0 . But this means that  T W  − 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 . 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). 5.4.20.

2

Proof. (⇒) Suppose  U T  =  T U . Then, since  V   =  span(v, T (v), T 2 (v ), ...),  U (v) =  a 0 + a1 T (v)+ ... + an T n (v) for some scalars  a 0 , a1 ,...,a n . So,  U (v ) =  g (T )(v ) where  g (v) =  a 0 + a1 x + ... + an xn . Suppose  x  ∈  V  . Then, x  =  b 0  + b1 T (v) + ... + bm T m (v ) for some scalars  b 0 , b1 ,...,b m . It follows that U (x) =  U (b0  + b1 T (v) + ... + bm T m (v)) =  b 0 U (T 0 (v)) + b1 U (T (v)) + ... + bm U (T m (v ))

=  b 0 T 0 (U (v)) + b1 T (U (v)) + ... + bm T m (U (v )) =  b 0 T 0 (g(T )(v)) + b1 T (g (T )(v )) + ... + bm T m (g (T )(v)) =  b 0 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) =  a 0 + a1 x + ... + an xn . Then, U T (x) =  a 0 T 0 (T (x)) + a1 T (T (x)) + ... + an T n (T (x))

=  a 0 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).

3

Partial Solutions for Linear Algebra by Friedberg et al. Chapter 6 John K. Nguyen December 7, 2011 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  , . If  T  :  V    W    is linear, prove that  x, y  = T (x), T (y )  defines an inner product on  V    if and only if  T   is one-to-one. 6.1.18.

 · ·

→

    



Proof. ( ) Suppose x, y  = T (x), T (y )  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 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 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, 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  so it 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. Let  {v , v ,...,v }  be an orthogonal set in  V  , and let  a , a ,...,a   be scalars. Prove that  

6.1.12.

1

2

k

1

 

2

   | | || ||          k

ai vi

k

2

k

=

i=1

2

ai

vi

2

.

i=1

Proof.  We apply the definition of inner product space and  Theorem 6.2  to get the following 2

k

k

=

ai v i

i=1

as required.

k

aj v j ,

j =1

k

ai v i

k

=

k

aj a ¯i < vj , vi >=

i=1

j =1 i=1

 | | || || ai

2

vi

2

.

i=1

Let  V    be an inner product space, S  and  S 0  be subsets of  V  , and  W  be a finite-dimensional subspace  of  V  . Prove the following results. (a) S 0  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) 6.2.13.

 ⊆  ⊆

⊆  ⊆

 ⊕

Proof of (a).   Suppose S 0 S . Let x S ⊥ . Then, by definition, for all y that y  S 0 . But S 0  S , so x  S 0⊥ . Thus, S ⊥  S 0⊥  as required.

 ∈

Proof of (b). S   ( S ⊥ )⊥ .

 ⊆

 ⊆ Let x  ∈ S .

 ⊆

 ∈

 ∈

⊆

Then,   < x , y >= 0 for all y

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

⊆

⊥

 ∈ S 

 ∈ S ,  < x, y >= 0 which implies

. But this also means that x

⊥ ⊥

 ∈ (S 

) . So,

⊥ ⊥

 ⊆  (W 

) .

( ) We will prove the contrapositive so assume that x

 ∈ W . Then, by Exercise 6.2.6 , there exists y ∈ W 

⊇

1

⊥

such that  < x, y >= 0. This implies that x  ( 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 ⊥ .

 ∩

 ∈

 ⊕

 { }  ∩

 ∈

 { }

 ∈

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

∩

 ∩

Proof of (a). ( ) Let x (W 1  +  W 2 )⊥ . By definition, it follows immediately that x (W 1  + W 2 )⊥  W 1⊥ W 2⊥ .

⊥ 1

⊥ 2

⊆  ∈  ∈ W  ∩ W  . So, ⊆ ∩ (⊇) Let x  ∈ W  ∩  W  . Then, x  ∈ W  and x  ∈ W    . By linearity, it follows that x  ∈ (W   +  W  ) . So, (W   + W  ) ⊇  W  ∩ W  . Thus, since (W   + W  ) ⊆  W  ∩ W    and (W   + W  ) ⊇  W  ∩ W  , (W   + W  ) =  W  ∩ W  . Proof of (b). (⊆) Let x ∈ ( W   ∩ W  ) . Then, since W  ⊆  W  + W  and W  ⊆  W  + W  by a previous exercise, it follows that  x  ∈ W  + W  so (W  ∩ W  ) ⊆  W  + W    (since  x  is perpendicular to vectors that ⊥ 1 ⊥

1

2

⊥ 2 ⊥ 1

⊥

1

2

⊥ 1

⊥ 2

⊥

1

2

⊥ 2

⊥ 1

⊥ 2

⊥

1

⊥

1

⊥ 1

2

⊥ 1

⊥ 2

⊥ 1

2

⊥

1

⊥ 1

⊥ 1

2

⊥ 2

⊥ 2

⊥

1

⊥ 1

2

⊥ 2

⊥ 1

⊥ 2

⊥ 2

⊥ 2

is in both W 1 and W 2 ).

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

( ) Let x

⊇

⊥ 1

⊆

⊥ 1

∈

 ∩

∈

⊆  ∩

 ∈

⊇ In conclusion, since (W   ∩ W  ) ⊆  W  ⊥

1

⊥ 1

2

⊥ 1

+ W 2⊥  and (W 1

⊥ 2

W  + W  .

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

⊥

⊥ 1

 ∩ W  ) ⊇ W 

 { √ }

(a) Find an orthonormal basis for  W .

2

2

1

  0

∩

∈ ⊆

+ W 2⊥ , we have that ( W 1

⊥

 ∩ W  ) 2

=

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

(b) Let  h(t) =  t 2 . Use the orthonormal basis obtained in (a) to obtain the ”best” (closest) approximation of  h in  W . Let  T  be a linear operator on an inner product space  V  . Let  U 1  =  T  + T ∗ and  U 2  =  T T ∗ . Prove that  U 1  =  U 1∗ and  U 2  =  U 2∗ . 6.3.6.

Proof. Let  T  be a linear operator on an inner product space  V    and suppose that  U 1  =  T  + T ∗ and  U 2  =  T T ∗ . We first prove that U 1  =  U 1∗ . By our assumption and Theorem 6.11, U 1∗ = (T  + T ∗ )∗ =  T ∗ + (T ∗ )∗ =  T ∗ + T  =  U 1 .

In a similar argument, we will show that U 2  =  U 2∗ . U 2∗ = ( T T ∗ )∗ =  T ∗ (T ∗ )∗ =  T ∗ T  =  U 2 .

Thus, we have that U 1  =  U 1∗ and U 2  =  U 2∗  as required. 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 )∗ . 6.3.8.

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,

 ∈

∗

−1 ∗

T  (T 

) (x), y  = (T −1 )∗ (x), T (y )

   = x, T  = x, y 

−1

T (y )





which implies that T ∗ (T −1 )∗ =  I . Therefore, (T ∗ )−1 = (T −1 )∗ . 6.3.9.  Prove

⊥

that if  V   =  W 

 ⊕ W 

⊥

N (T ) =  W  .

and  T  is the projection on  W   along  W ⊥ , then  T  =  T ∗ . Hint: Recall that 

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  v 1 , v2  V  . Then, there exists  w 1 , w2  W   and w˜1 , w˜2  W ⊥ such that  v 1  =  w 1 + w˜1 and v2  =  w 2  + w˜2 . Now, in consideration that T  is the projection on W   along W ⊥ ,

 ⊕  ∈

 ∈

 ⊕

 ∈

v , T (v ) = w  + w˜ , T (w  + w˜ ) = w  + w˜ , w  = w , w . 1

2

1

1

1

1

1

2

2

2

2

Similarly,

T (v ), v  = T (w  + w˜ ), w  + w˜  = w , w  + w˜  = w , w . T (v ), v    which means v , T (v ) = v , T  (v ). 1

So, it follows that v1 , T (v2 ) = T  =  T ∗





2

1

1

1

2

1

2

2

2

2

∗

1

2

1

2

1

2

This implies that

 For a linear operator  T   on an inner product space  V  , prove that  T ∗ T  =  T 0  implies  T  =  T 0 . Is the  same result true if we assume that  T T ∗ =  T 0 . 6.3.11.

Proof. Let T   be a linear operator on an inner product space V  . Suppose T ∗ T  = T 0 . Pick x V  . Then, T ∗ T (x), x  = T 0 (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  =  T 0 .



 

         || ||

 

We claim that the same result is true if we assume that T T ∗ =  T 0 . 3

 ∈   

