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I am thinking about the question of the title..

Suppose we know $Av = \lambda v$ and all eigenvalues of $A$ are different (so diagonalizable). What do the eigenvectors and eigenvalues look like of $A^TP-PA$?

I know that $A^TP - PA$ is a linear map.

Obviously, we can let $A = V\Lambda V^{-1}$, where $\Lambda$ is diagonal. So we have $$(V\Lambda V^{-1})^TP-P(V\Lambda V^{-1}) = V^{-T}\Lambda V^TP - PV\Lambda V^{-1}$$

then how do I go further? I suppose to get a form like $M\tilde{\Lambda}M^{-1}$ and I can say something. How to get that point?

Denny
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    It might be worth looking at a very simple case, where $A$ is diagonal, and $P$ is the identity. The evals of your matrix are all zeroes in that case, and apparently unrelated to those of $A$. If $A$ is not diagonal, can you say anything useful about the eigenvalues of $A^t - A$? It's skew-symmetric, so all its eigenvalues are pure imaginary (or $0$). But how are they related to evalues $A$, even in the $2 \times 2$ case? Have you tried working that out? – John Hughes Jan 11 '18 at 01:53
  • Is $P$ a particular matrix, or are you looking at the linear map $P \mapsto A^T P - P A$? – Robert Israel Jan 11 '18 at 02:07
  • @RobertIsrael $P \in \mathbb{R}^{n\times n}$. Yes I am looking at the linear map. – Denny Jan 11 '18 at 02:08

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There was a very similar recent question about $X \mapsto A X A^T$. Coincidence?

The answer is also essentially the same. If $A^T v = \lambda v$ and $A^T w = \mu w$ are two eigenvalue-eigenvector pairs for $A^T$, then $P = v w^T$ is an eigenvector of your linear map, with eigenvalue $\lambda - \mu$. If $A$ is diagonalizable, this gives you a full set of $n^2$ linearly independent eigenvectors of your linear map.

Robert Israel
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