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Unformatted text preview: 2.4. (2) R = T 1 Since T is invertible, T 1 : V V T 1 = T 1 1 V = T 1 ( TR ) = ( T 1 T ) R = IR = R R = T 1 11. T : P 3 ( R ) M 2 2 ( R ) is linear by T ( f ) = f (1) f (2) f (3) f (4) We are going to show that T ( f ) = 0 f = 0(the zero polynomial) In this case, f (1) = 0( i.e.f ( c ) = b ) ,f (2) = 0 ,f (3) = 0 ,f (4) = 0 f ( c i ) = 0 ,i = 0 , 1 , 2 , 3 f ( x ) = 3 i =0 b i f i ( x ) = 3 i =0 f ( c i ) f i ( x ) = 0 f is the zero polynomial T is onetoone 12. ( v i ) = [ v i ] = e i = (0 , , 1 , , 0) t [ ] = I n , when is the standard basis for F n is an isomorphism or : V F n is onto is an isomorphism because dimV = dimF n 13. is an equivalence relation on the class of vector space over F (i) is reflexive 107 PNUMATH 2.4. V C ,V V ( ) I V : V V s.t I V = v, v V is an isomorphism (ii) is symmetric If V W , then W V ( ) If T : V W is isomorphic then T 1 : W V is isomorphic W V (iii) is transitive If V W and W Z , then V Z ( ) Let T : V W and U : W Z are isomorphic, then UT is isomorphic 14. Let V = { a a + b c  a,b,c F } T : V F 3 s.t T a a + b c = ( a,b,c ) For the basis for V, { v 1 = 1 1 0 0 ,v 2 = 0 1 0 0 ,v 3 = 0 0 0 1 } ! T ( v i ) = w i is linear, w i F 3 i = 1 , 2 , 3 and since dimV = dimF 3 , V is isomorphic to F 3 T is an isomorphism from...
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 Spring '09
 RUDYAK

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