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Unformatted text preview: Math 431  Assignment 3 Solutions Section 2.2 5(c). Since dim ( M 2 2 ( F )) = 4 and dim ( F ) = 1, we are looking for a 1 4 matrix. We have T 1 0 0 0 = 1, T 0 1 0 0 = 0, T 0 0 1 0 = 0, and T 0 0 0 1 = 1. Therefore, [ T ] = ( 1 0 0 1 ) 9. Let c be a real number, and let z 1 = a 1 + b 1 i and z 2 = a 2 + b 2 i be two complex numbers. Then cz 1 + z 2 = ( ca 1 + cb 1 i ) + ( a 2 + b 2 i ) = ( ca 1 + a 2 ) + ( cb 1 + b 2 ) i, so T ( cz 1 + z 2 ) = ( ca 1 + a 2 ) ( cb 1 + b 2 ) i = c ( a 1 b 1 i ) + ( a 2 b 2 i ) = cT ( z 1 ) + T ( z 2 ) . Therefore, T is linear. Since dim R ( V ) = 2, [ T ] is a 2 2 matrix. We have T (1) = 1 = 1 1 + i and T ( i ) = i = 1 0 + i ( 1) Therefore [ T ] = 1 1 13. Let O be the zero transformation in L ( V,W ). Let a,b be in F such that aT + bU = O . We want to show that a = b = 0. Without loss of generality assume to the contrary that a 6 = 0. Since T is nonzero, there exists an element v in V , such that w := T ( v ) 6 = 0. Since aT ( v ) + bU ( v ) = 0 , solving for T ( v ) we obtain w = T ( v ) = a 1 bU ( v ) . From the linearity of U , we see that w = U ( a 1 bv ) That is, w is in both R ( T ) and R ( U ). However, w is nonzero, while R ( T ) R ( U ) = { } , a contradiction. Therefore, a contradiction....
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 Winter '07
 Liu
 Math, Linear Algebra, Algebra

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