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assig4_solutions

# assig4_solutions - MATH 235 Assignment 4 Solutions 1(5 mks...

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MATH 235: Assignment 4 Solutions 1. (5 mks total) (a) (1 mk) We have 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) (2 mks) Note first that T ( p ( t )) = (1 + t 2 )( p ( t )). Now T ( p ( t ) + q ( t )) = (1 + t 2 )( p ( t ) + q ( t )) = (1 + t 2 )( p ( t )) + (1 + t 2 )( q ( t )) = T ( p ( t )) + T ( q ( t )) T ( kp ( t )) = (1 + t 2 )( kp ( t )) = k (1 + t 2 )( p ( t )) = kT ( p ( t )) for any p ( t ) , q ( t ) P 2 , k R . Thus T is a linear transformation. (c) (2 mks) Let B = { 1 , t, t 2 } , B 0 = { 1 , t, t 2 , t 3 , t 4 } . Now [ T ] B 0 ←B = [ T (1)] B 0 [ T ( t )] B 0 [ T ( t 2 )] B 0 = [1 + t 2 ] B 0 [ t + t 3 ] B 0 [ t 2 + t 4 ] B 0 = 1 0 0 0 1 0 1 0 1 0 1 0 0 0 1 2. (4 mks) A has eigenvalues 8 and - 2 with corresponding eigenvectors 1 - 1 and [ 3 7 ]. These two eigenvectors are linearly independent (as they must be, corresponding to different eigenvalues). Thus the set B = 1 - 1 , 3 7 is a basis for R 2 , and we have [ T ] B = 8 0 0 - 2 which is diagonal, as required. 3. (9 mks total) Let v = u + i w where u , w R 2 ; that is, u = Re v , w = Im v . 1

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(a) (1 mk) A v = A ( u + i w ) = A u + iA w Taking real and imaginary parts gives Re A v = A u = A (Re v ) , Im A v = A w = A (Im v ) . (b) (2 mks) Since v is an eigenvector, A v = λ v = ( a - bi )( u + i w ) = ( a u + b w ) + i ( - b u + a w ) Combining this with part (a) gives A (Re v ) = Re A v = a u + b w = a Re v + b Im v , A (Im v ) = Im A v = - b u
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assig4_solutions - MATH 235 Assignment 4 Solutions 1(5 mks...

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