WINTER 2007 assign8-sol

WINTER 2007 assign8-sol - Math 235 Assignment 8 Due 9:15am,...

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Math 235 Assignment 8 Due 9:15am, Wednesday March 21, 2007. 1. From the Text § 5.4 #6. Let T : P 2 P 4 be the transformation that maps a polynomial p ( t ) into the polynomial p ( t ) + t 2 p ( t ) . a. Find the image of p ( t ) = 2 - t + t 2 . b. Show that T is a linear transformation. c. Find the matrix for T relative to the bases { 1 , t, t 2 } and { 1 , t, t 2 , t 3 , t 4 } . d. Determine the rank of T. Is T injective? Is T surjective? Solution. a. T (2 - t + t 2 ) = 2 - t + t 2 + t 2 (2 - t + t 2 ) = 2 - t + 3 t 2 - t 3 + t 4 . b. We need to show that T ( a p 1 ( t ) + b p 2 ( t )) = aT ( p 1 ( t )) + bT ( p 2 ( t )) , where p 1 ( t ) , p 2 ( t ) P 2 and a and b are scalars. T ( a p 1 ( t ) + b p 2 ( t )) = a p 1 ( t ) + b p 2 ( t ) + t 2 ( a p 1 ( t ) + b p 2 ( t )) = ( a p 1 ( t ) + t 2 ( a p 1 ( t ))) + ( b p 2 ( t ) + t 2 ( b p 2 ( t ))) = a ( p 1 ( t ) + t 2 p 1 ( t )) + b ( p 2 ( t ) + t 2 p 2 ( t )) = aT ( p 1 ( t )) + bT ( p 2 ( t )) . c. A general element of P 2 is in the form p ( t ) = a + bt + ct 2 , and we are given that T ( a + bt + ct 2 ) = a + bt + ( a + c ) t 2 + bt 3 + ct 4 . Now coordinate vector for p ( t ) is a b c and coordinate vector for T ( p ( t )) is a b a + c b c . Thus we want to find a matrix A such that A a b c = a b a + c b c . 1
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It is easy to see that A = 1 0 0 0 1 0 1 0 1 0 1 0 0 0 1 . d. The rank of the matrix is 3. The transformation is injective since a 1 b 1 c 1 ± = a 2 b 2 c 2 implies a 1 b 1 a 1 + c 1 b 1 c 1 ± = a 2 b 2 a 2 + c 2 b 2 c 2 . The map is not surjective.
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This note was uploaded on 03/03/2010 for the course MATH 235 taught by Professor Celmin during the Winter '08 term at Waterloo.

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WINTER 2007 assign8-sol - Math 235 Assignment 8 Due 9:15am,...

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