change of matrix

change of matrix - m × n matrix A to be L v 1 . . . v m =...

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Let L be the linear map from R 2 to R 2 , e 1 ,e 2 be a basis of R 2 . L ( e 1 ) = 2 e 1 - e 2 , L ( e 2 ) = 3 e 1 + 2 e 2 . Denote ¯ e 1 = e 1 + e 2 , ¯ e 2 = - e 1 + e 2 . The basis change matrix B from ( e 1 ,e 2 ) to (¯ e 1 , ¯ e 2 ) is defined to be B ± e 1 e 2 ² = ± ¯ e 1 ¯ e 2 ² Hence B is ± 1 1 - 1 1 ² The associated matrix A of L under the basis ( e 1 ,e 2 ) and ( e 1 ,e 2 ) is defined to be L ± e 1 e 2 ² = A ± e 1 e 2 ² That is ± 2 e 1 - e 2 3 e 1 + 2 e 2 ² = A ± e 1 e 2 ² Then A = ± 2 - 1 3 2 ² The associated matrix A new of L under the basis (¯ e 1 , ¯ e 2 ) and ( e 1 ,e 2 ) is defined to be L ± ¯ e 1 ¯ e 2 ² = A new ± e 1 e 2 ² That is L ± B ± e 1 e 2 ²² = A new ± e 1 e 2 ² Since L is linear, we have B ± L ± e 1 e 2 ²² = A new ± e 1 e 2 ² B ± A ± e 1 e 2 ²² = A new ± e 1 e 2 ² Then A new = BA = ± 1 1 - 1 1 ²± 2 - 1 3 2 ² = ± 5 1 1 3 ² 1
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Let us verify A new by a direct calculation: L e 1 ) = L ( e 1 ) + L ( e 2 ) = 2 e 1 - e 2 + 3 e 1 + 2 e 2 = 5 e 1 + e 2 L e 2 ) = - L ( e 1 ) + L ( e 2 ) = - 2 e 1 + e 2 + 3 e 1 + 2 e 2 = e 1 + 3 e 2 It shows that L ± ¯ e 1 ¯ e 2 ² = ± 5 1 1 3 ²± e 1 e 2 ² = A new ± e 1 e 2 ² We have the following theorem: Theorem 0.0.1. Suppose L is a linear map from a vector space V to another vector space W . Let ( v 1 ,...,v m ) to be a basis of V and ( w 1 ,...,w n ) to be another basis of W . Define the associated
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Unformatted text preview: m × n matrix A to be L v 1 . . . v m = L ( v 1 ) . . . L ( v m ) = A w 1 . . . w n Let (¯ v 1 ,..., ¯ v m ) to be a new basis of V . Define the associated m × n matrix A new to be L ¯ v 1 . . . ¯ v m = L (¯ v 1 ) . . . L (¯ v m ) = A new w 1 . . . w n Then A new = BA, where B is the m × m basis change matrix from ( v 1 ,...,v m ) to (¯ v 1 ,..., ¯ v m ) , i.e. B v 1 . . . v m = ¯ v 1 . . . ¯ v m 2...
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This note was uploaded on 10/03/2010 for the course MATH 223 taught by Professor Loveys during the Spring '07 term at McGill.

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change of matrix - m × n matrix A to be L v 1 . . . v m =...

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