HW5solutions_3

HW5solutions_3 - MATH 115 SOLUTION SET 5 ANSWERS TO...

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MATH 115 SOLUTION SET 5 ANSWERS TO SELECTED PROBLEMS § 2.2: 9. The fact that T is linear follows from standard facts about complex numbers: If z 1 ,z 2 C then z 1 + z 2 = z 1 + z 2 . Also, if c R then cz 1 = c z 1 . The matrix [ T ] β is ± 1 - 1 . 10. This is the matrix with 1s along the main diagonal, 1s along the diagonal above that, and 0s everywhere else. 15. a. For if T 1 ,T 2 S 0 , and c is a scalar, then cT 1 + T 2 also lies in S 0 because whenever x S we have ( cT 1 + T 2 )( x ) = cT 1 ( x ) + T 2 ( x ) = c 0 + 0 = 0. b. Suppose f S 0 2 , so that f kills everything in S 2 . But S 1 S 2 , so of course f kills everything in S 1 as well. Thus f S 0 1 . c. Suppose f ( V 1 + V 2 ) 0 . Then f kills everything in V 1 + V 2 . In particular it kills both V 1 , so f V 0 1 . Similarly f V 0 2 . Thus f V 0 1 V 0 2 . Conversely, suppose f V 0 1 V 0 2 . Then f V 0 1 and f V 0 2 . Thus f kills both V 1 and V 2 . So if v 1 V 1 and v 2 V 2 , then f ( v 1 + v 2 ) = 0. Thus f kills V 1 + V 2 and therefore f ( V 1 + V 2 ) 0 . / 16. Let { v 1 ,...,v k } be a basis for N ( T ). Extend this to a basis { v 1 ,...,v n } for V . We claim that { T ( v k +1 ) ,...,T ( v n ) } is linearly independent. Indeed, if c k +1 T ( v k +1 ) + ··· + c n T ( v n ) = 0 for scalars c k +1 ,...,c n , then T ( c k +1 v k +1 + ··· + c n v n ) = 0, so that c k +1 v k +1 + ··· + c n v n N ( T ). But then this element must be a linear combination of the vectors v 1 ,...,v k : c k +1 v k +1 + ··· + c n v n = c 1 v 1
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HW5solutions_3 - MATH 115 SOLUTION SET 5 ANSWERS TO...

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