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04-23-03

# 04-23-03 - Lecture 29 Andrei Antonenko 1 Change of the...

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Lecture 29 Andrei Antonenko April 23, 2003 1 Change of the matrix of an operator Last time we studied what happens with the coordinates of the vector if we change the basis from the “old” to the “new” one. We figured out that there exists a change-of-basis matrix, and if vector x had coordinates ( x 1 , x 2 , . . . , x n ) in the “old” basis { e 1 , e 2 , . . . , e n } and coordinates ( x 0 1 , x 0 2 , . . . , x 0 n ) in the “new” basis { e 0 1 , e 0 2 , . . . , e 0 n } and C is a change-of-basis matrix such that ( e 0 1 , e 0 2 , . . . , e 0 n ) = ( e 1 , e 2 , . . . , e n ) C then x 1 x 2 . . . x n = C x 0 1 x 0 2 . . . x 0 n On this lecture we will study what happens with the matrix of a linear operator if we take another basis in the space. Let V be a vector space, and let { e 1 , e 2 , . . . , e n } be the “old” basis and { e 0 1 , e 0 2 , . . . , e 0 n } be the “new” basis, and the change-of-basis matrix from the “old” basis to the “new” one is C , such that ( e 0 1 , e 0 2 , . . . , e 0 n ) = ( e 1 , e 2 , . . . , e n ) C. (1) Multiplying this equation by C - 1 we have ( e 0 1 , e 0 2 , . . . , e 0 n ) C - 1 = ( e 1 , e 2 , . . . , e n ) . (2) Let A be a linear operator in the vector space V , and its matrix with respect to the “old” basis is A . Then by definition of the matrix of the operator, ( A ( e 1 ) , A ( e 2 ) , . . . , A ( e n )) = ( e 1 , e 2 , . . . , e n ) A. (3) 1

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Now we can get the expression for the matrix of A is “new” basis. Applying A to both sides of the equality (1), we get ( A ( e 0 1 ) , A ( e 0 2 ) , . . . , A ( e 0 n )) = ( A ( e 1 ) , A ( e 2 ) , . . . , A ( e n )) C by (1) = ( e 1 , e 2 , . . . , e n ) AC
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04-23-03 - Lecture 29 Andrei Antonenko 1 Change of the...

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