Solutions to Axler, Linear Algebra Done Right

# Solutions to Axler, Linear Algebra Done Right - Solutions...

This preview shows pages 1–3. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Solutions to Axler, Linear Algebra Done Right 2nd Ed. Edvard Fagerholm [email protected] { 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....
View Full Document

{[ snackBarMessage ]}

### Page1 / 35

Solutions to Axler, Linear Algebra Done Right - Solutions...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online