2
$\begingroup$

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 $\pm 1$.

$\endgroup$
6
$\begingroup$

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.

$\endgroup$
5
  • 1
    $\begingroup$ 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 $\endgroup$
    – ogogmad
    Nov 8 '20 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) $\endgroup$
    – ogogmad
    Nov 8 '20 at 12:03
  • 2
    $\begingroup$ $F(M) := \Delta(\chi_M)$ 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. $\endgroup$ Nov 8 '20 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) $\endgroup$
    – ogogmad
    Nov 8 '20 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 $\endgroup$
    – ogogmad
    Nov 8 '20 at 22:17

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.