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I am interested in solving this problem.

Let $A\in \operatorname{GL}_n(\mathbb{C})$ and define $T:M_n(\mathbb{C})\rightarrow M_n(\mathbb{C})$ by $M\mapsto A^{-1}MA$. If the eigenvalues of $A$ are $\lambda_1,\dots,\lambda_n$, then what are the eigenvalues of the operator $T$?

In the case that the eigenvalues $\lambda_i$ are all distinct, then the result is clear: the $\lambda_j\lambda_i^{-1}$ are eigenvalues. The method of proof I used was to first show that to compute the eigenvalues, it suffices to replace $A$ by its Jordan Canonical Form. In the case of distinct eigenvalues, the Jordan form is diagonal and the eigenvectors can be chosen to be the elementary matrices $E_{i,j}$ which has entry 1 only at $(i,j)$ and zero elsewhere.

The same method can applied to the case where we know $A$ is diagonalizable but with eigenvalues repeating.

For the general case where the eigenvalues need not be distinct and $A$ not neccesarily diagonalizable, I am stuck and it appears to be a long computation. For instance, if we assume that $A$ is a single Jordan block with nonzero eigenvalue. Any hints?

After posting, the stackexchange suggested the following similar question. The solutions I see do not seem like the kinds of solution I was expecting. In particular, I was hoping for a straightforward solution utilizing JCF directly.

Shrugs
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    If the answer is $\lambda_j \lambda_i^{-1}$ in the diagonalizable case then that has to be the answer in all cases by a continuity argument. – Qiaochu Yuan Aug 03 '22 at 15:49
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    This can be also proven from the fact that that $A^{-1}MA=\lambda M$ if and only if $(A^T\otimes A^{-1})\mathrm{vec}(M)=\lambda \mathrm{vec}(M)$, which is a standard eigenvalue problem and we know that the eigenvalues of $A^T\otimes A^{-1}$ are $\lambda_i\lambda_j^{-1}$ for all $i,j=1,\ldots,n$. – KBS Aug 03 '22 at 16:19
  • @QiaochuYuan Could you elaborate on the continuity argument? – Shrugs Aug 03 '22 at 16:50
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    @AHappyMathematician: $T$ depends continuously on $A$, so its eigenvalues also depend continuously on $A$ (if this makes you nervous you can consider the characteristic polynomial instead). The diagonalizable matrices are dense, so if you know what the eigenvalues are in that case then you know what they are in the general case by density. – Qiaochu Yuan Aug 04 '22 at 10:36
  • @KBS What does $\operatorname{vec}(M)$ mean here? – Shrugs Aug 05 '22 at 12:04
  • @AHappyMathematician This is the vectorization operator https://en.wikipedia.org/wiki/Vectorization_(mathematics) – KBS Aug 05 '22 at 12:26

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