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What are the conditions for the product of two skew-symmetric real matrices A and B to be sign-definite (either positive or negative definite)? Here, it is assumed that the cross-commutation [A,B]=AB-BA=0 and both of the matrices are nonsingular (and necessarily of even dimension).

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Let $A,B\in M_n(\mathbb{R})$ be skew-symmetric matrices. According to Stenzel (dated 1922), generically, the eigenvalues of $AB$ have (each) multiplicity $2$ and may be real or complex.

Now, if $AB=BA$ and $n$ is even, then $AB$ is symmetric and the eigenvalues are non-zero real and (generically) have multiplicity $2$. Unfortunately, there are instances s.t. $AB$ is not $>0$ and not $<0$ as in this example:

Let $U=\begin{pmatrix}0&-1\\1&0\end{pmatrix}$, $A=diag(2U,1/2.U)$ and $B=A+A^3$. Then $spectrum(AB)=\{12,12,-0.1875,-0.1875\}$.

EDIT. I think that (generically) $AB>0$ or $AB<0$ when $B$ is an alternating odd polynomial in $A$, that is $B=u_1A-u_3A^3+u_5A^5-u_7A^7+\cdots$ where the $(u_i)$ have the same signum.

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  • $\begingroup$ Thank you very much for your answer. Your comment on the definiteness of the product for the case when B is an odd polynomial in A is interesting. $\endgroup$ – Arash Jan 15 '16 at 0:18

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