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Unformatted text preview: MATH 4377, ADVANCED LINEAR ALGEBRA, SUMMER 2011, Key to HW#4 1. Problem 9 on Page 75. Solution : (a): By definition, T (2 , 2) = (1 , 2), 2 T (1 , 1) = 2(1 , 1) = (2 , 2). So T (2(1 , 1)) 6 = 2 T (1 , 1). Hence T is not lienar. Remark : To show that T is NOT linear, you only need to provide a counterexample. (b)(e) are omitted. 2. Problem 11 on Page 75. Solution : Since { (1 , 1) , (2 , 3) } is linear independent. It serves a basis for R 2 and every ( x,y ) can be written as ( x,y ) = (3 x 2 y )(1 , 1) + ( y x )(2 , 3) . Hence, we can define T ( x,y ) = (3 x 2 y ) T (1 , 1) + ( y x ) T (2 , 3) = (3 x 2 y )(1 , , 2) + ( y x )(1 , 1 , 4) = (2 x y,x y, 2 x ) . It is easy to see that T is linear and T (8 , 11) = (5 , 3 , 16). 3. Problem 15 on Page 75. Solution : The proof of T is linear is omitted. We now determine N ( T ) and R ( T ). Take any f ( x ) = a + a 1 x + + a n x n N ( T ), i.e. T ( f ) = 0, or Z x ( a + a 1 t + + a n t n ) dt = a x + 1 2 a 1 x 2 + + a n 1 n + 1 x n +1 = 0 , hence a = 0 ,...,a n = 0, so f ( x ) = 0. This shows that N ( T ) = { } . So T is one to one. To determine R ( T ), by definition, denote by Z + the set of all nonnegative integers, R ( T ) = { T ( f )  f P ( R ) } = { Z x ( a + a 1 t + + a n t n ) dt  a ,...,a n R ,n Z + } = { a x +...
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This note was uploaded on 02/21/2012 for the course MATH 4377 taught by Professor Staff during the Summer '08 term at University of Houston.
 Summer '08
 Staff
 Linear Algebra, Algebra

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