It's well known that the following set of square real 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 $\pm 1$.

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 $\pm 1$.

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 $\pm 1$.

It's well known that the following set of square real 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 $\pm 1$.