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Similar Diagonal Matrices |
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| Nov6-12, 09:02 PM | #1 |
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Similar Diagonal Matrices
As part of a larger problem involving classifying intertwining operators of two group representations, I came across the following question: If [itex]X[/itex] is an [itex]n \times n[/itex] diagonal matrix with [itex]n[/itex] distinct non-zero eigenvalues, then exactly which [itex]n \times n[/itex] matrices [itex]A[/itex] satisfy the following equality [itex]AXA^{-1} = X[/itex]? Does anyone know the answer to this question?
Edit: Nevermind. I found a better way of doing the problem that avoids this sort of argument. |
| Nov7-12, 07:04 AM | #2 |
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Those whose eigenvalues are the numbers on the diagonal of the original matrix.
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| Nov8-12, 01:54 PM | #3 |
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Is that true? I believe it is the set of operators with the same invariant subspaces. The eigenvalues don't have to be the same, they just have to be simultaneously diagonalizable.
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