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ma1bAnMistermSol

# ma1bAnMistermSol - CALIFORNIA INSTITUTE OF TECHNOLOGY...

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CALIFORNIA INSTITUTE OF TECHNOLOGY Department of Mathematics Solutions to Math 1b Midterm, February 2008 1. Let B = 3 5 1 2 . By Problem 4 on HW1, B is invertible and B - 1 = 2 - 5 - 1 3 Now for each 2 by 2 matrix C , C · 0 = 0. Thus if A is a 2 by 2 matrix such that BA = 0, then 0 = B - 1 · 0 = B - 1 ( BA ) = ( B - 1 B ) A = I · A = A, so A = 0. 2. (1) Let f ( x ) = n i =0 a i x i V be of degree n ; in particular a n 6 = 0 if n > 0. Then T ( f ) = n X i =0 a i x i - n X i =0 ia i x i + n X i =0 3 i ( i - 1) a i x i - 2 - n X i =0 i ( i - 1) a i x i = n X i =0 a i [1 - i - i ( i - 1)] x i + n - 2 X i =0 3( i + 2)( i + 1) a i +2 x i = (1 - n 2 ) a n x n - n ( n - 2) a n - 1 x n - 1 + n - 2 X i =0 [(1 - i 2 ) a i + 3( i + 1)( i + 2) a i +2 ] x i . Thus if n = 0 then T ( f ) = f , so f N ( T ) iff f = 0. If n = 1 then T ( f ) = a 0 , so f N ( T ) iff f = a 1 x . Finally if n > 1 then a n 6 = 0 and the coefficient of x n in T ( f ) is (1 - n 2 ) a n 6 = 0, so f / N ( T ). Thus we have shown that N ( T ) = Fx is 1-dimensional. (2) First dim( V ) = 6 as { 1 , x, . . . , x 5 } is a basis for V . Thus by Theorem 2.3, dim( T ( V )) = dim( V ) - dim( N ( T )) = 6 - 1 = 5 , so dim( T ( V )) 6 = dim( V ) and hence V 6 = T ( V ).

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