Sol6 - SOLUTION TO ASSIGNMENT 6 - MATH 251, WINTER 2007 1....

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Unformatted text preview: SOLUTION TO ASSIGNMENT 6 - MATH 251, WINTER 2007 1. Let T : V W be a linear map and define T * : W * V * by ( T * ( g ))( v ) := g ( Tv ). Prove the following lemma: L EMMA 1. (1) T * is a well-defined linear map. (2) Let B,C be bases to V,W , respectively. Let A = C [ T ] B be the m n matrix representing T , where n = dim( V ) ,m = dim( W ) . Then the matrix representing T * with respect to the dual bases B * ,C * is the transpose of A : B * [ T * ] C * = C [ T ] B t . (3) If T is injective then T * is surjective. (Do NOT use Proposition 7.2.8 in the notes). (4) If T is surjective then T * is injective. Proof. (1) First, we need to show that the map v 7 g ( T ( v )) is a linear map. This is easy: T and g are linear maps and so is their composition. Next, we need to check that T * is linear. Namely, that for every v , a i F ,g i W * , we have ( T * ( a 1 g 1 + a 2 g 2 ))( v ) = ( a 1 T * g 1 + a 2 T * g 2 )( v ) . This follows from definitions: ( T * ( a 1 g 1 + a 2 g 2 ))( v ) = ( a 1 g 1 + a 2 g 2 )( Tv ) = a 1 g 1 ( Tv )+ a 2 g 2 ( Tv ) = a 1 T * g 1 ( v ) + a 2 T * g 2 ( v ) = ( a 1 T * g 1 + a 2 T * g 2 )( v ). (2) Let us write B = { b 1 ,...,b n } ,C = { c 1 ,...,c m } ,B * = { b * 1 ,...,b * n } ,C * = { c * 1 ,...,c * m } , where we have b * i ( b j ) = ij , c * i ( c j ) = ij . If C [ T ] B = A = ( a ij ) that means that [ Tb i ] C = a 1 i c 1 + + a mi c m . Let us write T * c * i = 1 i b * 1 + + ni b * n , so that B * [ T * ] C * = ( ij ) , and we wish to prove that ji = a ij . As we have seen before, the coefficient ji is equal to ( T * c * i )( b j ) (cf. the solution to question (2) below; we make use of B being the dual basis to B * . See also the notes, Example 7.1.4). But, by....
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This note was uploaded on 08/26/2008 for the course MATH 251 taught by Professor Goren during the Spring '08 term at McGill.

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Sol6 - SOLUTION TO ASSIGNMENT 6 - MATH 251, WINTER 2007 1....

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