Are there any special properties of real matrices (not necessarily symmetric) with "real" distinct eigenvalues, other than the well-known properties like being diagonalizable, which has nothing to do with the realness of the eigenvalues?

In short, I'm searching for the properties that comes from the realness of the eigenvalues. Can you at least suggest me a reference for the topic?

Note: This is a little different from the question What do real eigenvalues imply for a matrix as MorganRodgers pointed out in the comments. My question requires that the eigenvalues be distinct and the answer relies on that.

  • 1
    $\begingroup$ @MorganRodgers The duplicate you are suggesting did not require that the eigenvalues be distinct. Since the current answer relies in part on this requirement, I'm inclined to keep this question open. $\endgroup$ – Strants Aug 3 '15 at 21:10
  • $\begingroup$ @MorganRodgers It is very similar indeed, but I completely agree with Strants $\endgroup$ – obareey Aug 7 '15 at 8:31

Let $A\in M_n(\mathbb{R})$.

$A$ has $n$ distinct real eigenvalues iff there are $\lambda\in\mathbb{R}$ and $P,Q$ real symmetric $>0$ s.t. $(A+\lambda I)^2=PQ$ AND the dimension of the commutant of $A$ is $n$.

EDIT 1. Proof: $(\Rightarrow)$ Clearly $A$ is diagonalizable and the dimension of its commutant is $n$. Let $\lambda\in\mathbb{R}$ s.t. $A+\lambda I$ is invertible. Then $(A+\lambda I)^2$ is diagonalizable and has positive eigenvalues. Then it can be put in the form $PQ$.

$(\Leftarrow)$ $(A+\lambda I)^2=PQ$ implies that $(A+\lambda I)^2$ is invertible, diagonalizable and has positive eigenvalues. Then $A+\lambda I$ is invertible, diagonalizable and has real eigenvalues. Consequently $A$ is diagonalizable and has real eigenvalues; since its commutant has dimension $n$, its eigenvalues are simple.

EDIT 2. We can do better. $A$ has $n$ distinct real eigenvalues iff there are real symmetric $P,Q$ where $P>0$ s.t. $A=PQ$ AND the dimension of the commutant of $A$ is $n$.

Proof. $(\Rightarrow)$ $A=RDR^{-1}$, where $R,D$ are real and $D$ is diagonal, implies $A=PQ$ with $P=RR^T,Q=R^{-T}DR^{-1}$. (This proof is due to O. Taussky).

$(\Leftarrow)$ is easy.

  • $\begingroup$ This is useful, thanks. So what I understand by the dimension of the commutant is there exists $n$ linearly independent matrices $B_i$ such that $A B_i = B_i A, i=1,\dots,n$. Is that true? Also, is $\lambda$ related to the eigenvalues of $A$? Can we say there exists $P,Q>0$ for any given $\lambda \in \mathbb{R}$ or something? $\endgroup$ – obareey Aug 2 '15 at 9:32
  • $\begingroup$ 1. $\lambda$ is chosen s.t. $A+\lambda I$ is invertible. 2. The commutant is generated by $n$ matrices (its dimension is $\leq n$). $\endgroup$ – loup blanc Aug 2 '15 at 10:08
  • $\begingroup$ Can you provide a reference with the proof of this claim? $\endgroup$ – obareey Aug 2 '15 at 10:56
  • $\begingroup$ That is my idea, but I do not claim to originality. Try to solve this exercise. $\endgroup$ – loup blanc Aug 2 '15 at 11:20
  • $\begingroup$ Thank you for the proof. I've seen that for $P,Q>0, PQ$ has always positive real eigenvalues, but I didn't see the converse anywhere. Can you point a reference to this? $\endgroup$ – obareey Aug 7 '15 at 8:34

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