04-23-03 - Lecture 29 Andrei Antonenko April 23, 2003 1...

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Unformatted text preview: 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 1 ,x 2 ,...,x n ) in the new basis { e 1 ,e 2 ,...,e n } and C is a change-of-basis matrix such that ( e 1 ,e 2 ,...,e n ) = ( e 1 ,e 2 ,...,e n ) C then x 1 x 2 . . . x n = C x 1 x 2 . . . x 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 1 ,e 2 ,...,e n } be the new basis, and the change-of-basis matrix from the old basis to the new one is C , such that ( e 1 ,e 2 ,...,e n ) = ( e 1 ,e 2 ,...,e n ) C. (1) Multiplying this equation by C- 1 we have ( e 1 ,e 2 ,...,e 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 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 (...
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This note was uploaded on 05/28/2011 for the course AMS 2010 taught by Professor Andant during the Spring '03 term at SUNY Stony Brook.

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04-23-03 - Lecture 29 Andrei Antonenko April 23, 2003 1...

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