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Solution_5

# Solution_5 - THE CHINESE UNIVERSITY OF HONG KONG Department...

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THE CHINESE UNIVERSITY OF HONG KONG Department of Mathematics MAT2310E Linear Algebra and Applications (Fall 2007) Suggested Solutions For Homework 5 1. (a) By definition of L , we have L ( y 1 ) = y 1 + 2 y 2 , L ( y 2 ) = y 1 - 2 y 3 , L ( y 3 ) = y 1 + y 2 - y 3 . Hence the matrix A representing L with respect to the ordered basis [ y 1 y 2 y 3 ] is 1 1 1 2 0 1 0 - 2 - 1 (b)Solving the following linear system: 1 1 1 0 1 1 0 0 1 c 1 c 2 c 3 = 1 2 3 We obtain ( c 1 , c 2 , c 3 ) = ( - 1 , - 1 , 3). Hence the vector x is equal to - y 1 - y 2 + 3 y 3 . And the coordinate of the vector L ( x ) with respect to the ordered basis [ y 1 y 2 y 3 ] can be obtained by A ( - 1 , - 1 , 3) T , that is (1 , 1 , - 1) T . 2. According to the definition of the transition matrix, we obtain the matrix V should be as follows. 2 2 0 2 4 - 4 2 0 2 By theorem 0 . 9 (cf.Lecture note 14,page 105), the matrix B should be V - 1 AV . Using [ V | I ] [ I | D ], where D = V - 1 , the inverse of V can be written as follows. - 1 1 / 2 1 3 / 2 - 1 / 2 - 1 1 - 1 / 2 - 1 / 2 . Let C be the matrix satisfying [ L ( x )] E = C [ x ] E ( * ) , where E = { e 1 , e 2 , e 3 } , since we have [ x ] E = V [ x ] V , by ( * ) , we have V [ L ( x )] V = CV [ x ] V , that is, [ L ( x )] V = V - 1 CV [ x ] V , hence B = V - 1 CV, and C = [[ L ( e 1 )] E [ L ( e 2 )] E [ L ( e 3 )] E ] = 3 - 1 - 2 4 0 - 4 2 - 1 - 1 . Hence, B = - 1 1 / 2 1 3 / 2 - 1 / 2 - 1 1 - 1 / 2 - 1 / 2 3 - 1 - 2 4 0 - 4 2 - 1 - 1 2 2 0 2 4 - 4 2 0 2 = 0 2 - 2 0 - 1 2 0 - 2 3 . 3. (a)By definition, we have [ L ( v 1 )] T = 1 2 - 1 and [ L ( v 2 )] T = 0 1 - 2 (b)By (a), L ( v 1 ) = ( t 2 + 1) + 2 t - ( t - 1) = t 2 + t + 2 and L ( v 2 ) = t - 2( t - 1) = - t + 2. (c)Since 2 t + 1 = 3 2 v 1 + 1 2 v 2 and by property of L , L (2 t + 1) = 3 2 L ( v 1 ) + 1 2 L ( v 2 ) = 3 2 ( t 2 + t + 2) + 1 2 ( - t + 2) = 3 2 t 2 + t + 4. (d)Since at + b = ( a + b ) 2 v 1 + ( a - b ) 2 v 2 ,

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2 we have L ( at + b ) = ( a + b ) 2 L ( v 1 ) + ( a - b ) 2 L ( v 2 ) = ( a + b ) 2 t 2 + bt + 2 a. 4. Solution: (a) Since S = Span { x , y } , for any z = z 1 z 2 z 3 S x · z = x 1 z 1 + x 2 z 2 + x 3 z 3 = 0 , y · z = y 1 z 1 + y 2 z 2 + y 3 z 3 = 0 . We get x 1 x 2 x 3 y 1 y 2 y 3 z 1 z 2 z 3 = 0 0 . i.e. A z = 0 , which implies that the vector z N ( A ). On the other hand, for any z = z 1 z 2 z 3 N ( A ), which satisfies A z = 0 . We have x 1 z 1 + x 2 z 2 + x 3 z 3 = 0 , y 1 z 1 + y 2 z 2 + y 3 z 3 = 0 . i.e. x · z = 0 and y · z = 0 thus z S . Hence S = N ( A ). (b) Let A = 1 2 1 1 - 1 2 , by (a), we only need to find N ( A ). Transforming A to reduced row echelon form, we obtain B = 1 0 5 3 0 1 - 1 3 .
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Solution_5 - THE CHINESE UNIVERSITY OF HONG KONG Department...

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