# Similar Diagonal Matrices

1. Nov 6, 2012

### jgens

As part of a larger problem involving classifying intertwining operators of two group representations, I came across the following question: If $X$ is an $n \times n$ diagonal matrix with $n$ distinct non-zero eigenvalues, then exactly which $n \times n$ matrices $A$ satisfy the following equality $AXA^{-1} = X$? 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.

Last edited: Nov 7, 2012
2. Nov 7, 2012

### HallsofIvy

Staff Emeritus
Those whose eigenvalues are the numbers on the diagonal of the original matrix.

3. Nov 8, 2012

### Vargo

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.