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Let $A\in \mathcal{M}_{m\times m}(\mathbb R)$ , $det(A)=1$ , $A$ is positively definite. Which matrices $P$ satisfy the equation $$P^TAP=A$$ In fact I am interested in sequences of traces $tr P^n$ of the iterations of such solutions.

In dimension $2$ one can show that $$P^n=\left( \begin{array}{cc} \cos n\phi& -\beta \sin n\phi \\ \alpha \sin n\phi & \cos n\phi\\ \end{array}\right)$$ for some $\phi, \alpha, \beta$ satisfying$\alpha\beta=1$ hence $tr P^n= 2\cos n\phi$.

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  • $\begingroup$ Two comments: All invertible $P$ with this property is a group. The second comment:the expresion $P^TAP$ instead of $P^{-1}AP$ remind me of the following: after change of coordinate with linear part $P$ the first fundamental form of the surface would be replaced by $P^T\mathcal{F}P$ instead of $P^{-1}\mathcal{F}P$. $\endgroup$
    – Ali Taghavi
    Commented Jun 23, 2021 at 21:30

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As $A=L^\top L$, for some $L\in M_{m\times m}(\mathbb{R})$, $\det L=1$, you can rewrite $P^\top L^\top LP=L^\top L$ and then multiply both sides by $L^{-1}$, etc., obtaining $(L^{\top})^{-1}P^\top L^\top LPL^{-1}=(LPL^{-1})^\top LPL^{-1}=I$, i.e. each $LPL^{-1}$ must be orthogonal.

As traces are preserved under conjugation, you can assume $L=I$, and so your question is reduced to studying traces of orthogonal matrices. The trace is the sum of eigenvalues, which are all modulus 1 complex numbers for orthogonal matrices.

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  • $\begingroup$ It would be interesting to compare the two lie groups: centralizer of A and the other one mentioned in the question. $\endgroup$
    – Ali Taghavi
    Commented Jun 23, 2021 at 22:13
  • $\begingroup$ Centraliser of $A$, i.e. $\{X \in GL_m\mid AX=XA\}$ is a different beast - it depends on eigenvalues of $A$. It's quite small if they are all different, and the whole $GL_m$ is they are all the same. $\endgroup$
    – Dima Pasechnik
    Commented Jun 24, 2021 at 12:08
  • $\begingroup$ Yes I see. On the other hand the Lie algebra of the other group is $V^T A+AV=0$ $\endgroup$
    – Ali Taghavi
    Commented Jun 24, 2021 at 14:32
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Using the vec operator, any matrix $X$ such that $vec A$ is an eigenvector of $(X\otimes X)^T$ corresponding to the eigenvalue $1$ would work.

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