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It's well known that the following set of real square matrices is dense: those matrices $M$ for which there exists an invertible matrix $P$ such that $P M P^{-1}$ is diagonal. My question is can this statement be strengthened so that $P = P^{-1}$?

The motivation for this question comes from the spectral theorem, whose statement is: Given a matrix $M$ such that $M = M^T$, there exists a matrix $P$ such that $P M P^{-1}$ is diagonal and $P P^T = I$. I'm wondering what happens if you drop the transpose operation. The resulting statement is clearly false, but the question remains "how true" it is.

Note that the set of matrices $P$ such that $P=P^{-1}$ is the same as the set of matrices that are diagonalisable and whose eigenvalues are $\pm 1$.

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No

Consider $2\times2$ matrices $M$ whose eigenvalues are not real (they form an open set). The eigenvalues are complex conjugate, as well as the eigenvectors. If $M=P^{-1}DP$, then the columns of $P$ are eigenvectors, thus $$P=\begin{pmatrix} aw & b\bar w \\ a z & b\bar z \end{pmatrix}.$$ Writing $P^2=I_2$ gives you that $wz$ is real, which is unlikely.

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    $\begingroup$ This answer boils down to the claim that the following set is non-empty and open: $\Omega = \{M \in M_2(\mathbb R) \mid \Delta(\chi_M) < 0 \text{ and } \forall z \in S^1. \forall k \in [0,1]. M(z,k\bar z)^T \times (z, k\bar z)^T \neq 0\}$, where $\chi_M$ denotes the characteristic polynomial of $M$, $\Delta$ denotes the discriminant, $S^1$ denotes the set of unit complex numbers, and $\times$ denotes the cross product (for detecting non-parallelism). By the argument above, no element of $\Omega$ can be diagonalised by a square root of the identity $\endgroup$
    – wlad
    Commented Nov 8, 2020 at 12:02
  • $\begingroup$ The set is $\Omega$ is clearly non-empty because $M = \begin{pmatrix} 0 & 1 \\ -1 & 0\end{pmatrix}$ is an element of it. But I'm not sure how you show it's open. (By the way, there is a language called Abstract Stone Duality in which one can prove that this set is open, but this is not widely known) $\endgroup$
    – wlad
    Commented Nov 8, 2020 at 12:03
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    $\begingroup$ $F(M) := \Delta(\chi_M)$ is a polynomial in the entries of $M$, right? You can write it down. So clearly $F$ is continuous, and thus the set $F^{-1}((-\infty, 0))$ is open. $\endgroup$
    – Nate Eldredge
    Commented Nov 8, 2020 at 16:13
  • $\begingroup$ @NateEldredge I'm actually concerned with the "$\forall z \in S^1. \forall k \in [0,1]. M(z,k\bar z)^T \text{ not parallel to } (z, k\bar z)^T$" bit. (Sorry, edited a bit) $\endgroup$
    – wlad
    Commented Nov 8, 2020 at 17:55
  • $\begingroup$ A follow-up question is what happens if we replace $\mathbb R$ with $\mathbb C$. The above argument doesn't work any more $\endgroup$
    – wlad
    Commented Nov 8, 2020 at 22:17

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