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Unformatted text preview: Solutions to Axler, Linear Algebra Done Right 2nd Ed. Edvard Fagerholm edvard.fagerholm@ { helsinki.fi  gmail.com } Beware of errors. I read the book and solved the exercises during spring break (one week), so the problems were solved in a hurry. However, if you do find better or interesting solutions to the problems, Id still like to hear about them. Also please dont put this on the Internet to encourage copying homework solutions... 1 Vector Spaces 1. Assuming that C is a field, write z = a + bi . Then we have 1 /z = z/ zz = z/  z  2 . Plugging in the numbers we get 1 / ( a + bi ) = a/ ( a 2 + b 2 ) bi/ ( a 2 + b 2 ) = c + di . A straightforward calculation of ( c + di )( a + bi ) = 1 shows that this is indeed an inverse. 2. Just calculate ((1 + 3) / 2) 3 . 3. We have v + ( v ) = 0, so by the uniqueness of the additive inverse (prop. 1.3) ( v ) = v . 4. Choose a 6 = 0 and v 6 = 0. Then assuming av = 0 we get v = a 1 av = a 1 0 = 0. Contradiction. 5. Denote the set in question by A in each part. (a) Let v,w A , v = ( x 1 ,x 2 ,x 3 ), w = ( y 1 ,y 2 ,y 3 ). Then x 1 + 2 x 2 + 3 x 3 = 0 and y 1 + 2 y 2 + 3 y 3 = 0, so that 0 = x 1 + 2 x 2 + 3 x 3 + y 1 + 2 y 2 + 3 y 3 = ( x 1 + y 1 ) + 2( x 2 + y 2 ) + 3( x 3 + y 3 ), so v + w A . Similarly 0 = a 0 = ax 1 + 2 ax 2 + 3 ay 3 , so av A . Thus A is a subspace. (b) This is not a subspace as 0 6 A . (c) We have that (1 , 1 , 0) A and (0 , , 1) A , but (1 , 1 , 0)+(0 , , 1) = (1 , 1 , 1) 6 A , so A is not a subspace. (d) Let ( x 1 ,x 2 ,x 3 ) , ( y 1 ,y 2 ,y 3 ) A . If x 1 = 5 x 3 and y 1 = 5 y 3 , then ax 1 = 5 ax 3 , so a ( x 1 ,x 2 ,x 3 ) A . Similarly x 1 + y 1 = 5( x 3 + y 3 ), so that ( x 1 ,x 2 ,x 3 ) + ( y 1 ,y 2 ,y 3 ) A . Thus A is a subspace. 1 6. Set U = Z 2 . 7. The set { ( x,x ) R 2  x R }{ ( x, x ) R 2  x R } is closed under multiplication but is trivially not a subspace (( x,x ) + ( x, x ) = (2 x, 0) doesnt belong to it unless x = 0). 8. Let { V i } be a collection of subspaces of V . Set U = i V i . Then if u,v U . We have that u,v V i for all i because V i is a subspace. Thus au V i for all i , so that av U . Similarly u + v V i for all i , so u + v U . 9. Let U,W V be subspaces. Clearly if U W or W U , then U W is clearly a subspace. Assume then that U 6 W and W 6 U . Then we can choose u U \ W and w W \ U . Assuming that U W is a subspace we have u + w U W . Assuming that u + w U we get w = u + w u U . Contradiction. Similarly for u + w W . Thus U W is not a subspace. 10. Clearly U = U + U as U is closed under addition. 11. Yes and yes. Follows directly from commutativity and associativity of vector addition....
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 Fall '08
 Cooper,S
 Linear Algebra, Algebra

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