
#1
Nov612, 09:02 PM

P: 1,623

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 nonzero 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. 



#2
Nov712, 07:04 AM

Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 38,886

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




#3
Nov812, 01:54 PM

P: 350

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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