2.1_2.2 - 115A HW 4 SELECTED SOLUTIONS Sec 2.1. 2. Consider...

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115A HW 4 SELECTED SOLUTIONS Sec 2.1. 2. Consider T : R 3 R 2 given by T ( a 1 ,a 2 ,a 3 ) = ( a 1 - a 2 , 2 a 3 ). We see that the vector (1 , 1 , 0) is a basis for N ( T ). By the dimension theorem, rank( T ) = 3 - 1 = 2, which means that R ( T ) = R 2 (and thus T is onto). However, T cannot be one-to-one since nullity( T ) > 0. 5. Consider T : P 2 ( R ) P 3 ( R ) given by T ( f ( x )) = xf ( x ) + f 0 ( x ). Let f = a 2 x 2 + a 1 x + a 0 P 2 ( R ). Then T ( f ) = x ( a 2 x 2 + a 1 x + a 0 ) + (2 a 2 x + a 1 ) = a 2 x 3 + a 1 x 2 + (2 a 2 + a 0 ) x + a 1 , from which we see that T ( f ) = 0 if and only if f = 0. That is, N ( T ) = { 0 } , and so T is one-to-one. Thus, by the dimension theorem, rank( T ) = 3. A basis for R ( T ) is { T (1) ,T ( x ) ,T ( x 2 ) } . (It should be noted that by Exercise 14, this set is linearly independent, since T is one-to-one.) 14. Let V and W be vector spaces and T : V W be linear. (a) Prove that
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This note was uploaded on 11/30/2009 for the course MATH 115A taught by Professor Liu during the Winter '07 term at UCLA.

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2.1_2.2 - 115A HW 4 SELECTED SOLUTIONS Sec 2.1. 2. Consider...

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