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It's proved here that every real antisymmetric matrix is orthogonally similar to its transpose?

Now let $A,B$ a pair of symmetric and antisymmetric matrices $(A^T=- A,B^T=B)$.

Is it true that $A,B$ are simultaneously orthogonally similar to their transposes? i.e Does there exists a matrix $S \in O(n)$ such that

$S^TAS=-A,S^TBS=B$


By this paper, it seems the answer is positive. However, the paper uses some sophisticated arguments whichI would like to avoid, and the criterion it gives for simultaneous similarity is non-trivial to check. (based on some series of trace equalities, see Corollary 2.3 there).

Is there a more elementary approach to see this?

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It's false when $n\geq 3$.

Proof: let $B=diag(b_j)$ where the $(b_j)$ are distinct. Note that $S^{-1}BS=B$ iff $SB=BS$ and $S$ is diagonal, that is $S=diag(e^{i\theta_j})$. Let $A$ be skew symmetric with, for every $j\not= k, a_{j,k}\not= 0$; then $S^{-1}AS=-A$ iff $e^{-i\theta j}a_{j,k}e^{i\theta k}=-a_{j,k}$ iff $\theta_k-\theta_j=\pi\; mod(2\pi)$. In particular, $\theta_1-\theta_2=\pi\; mod(2\pi)$ and $\theta_2-\theta_3=\pi\; mod(2\pi)$ imply $\theta_1-\theta_3=0\; mod(2\pi)$, a contradiction.

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  • $\begingroup$ Thanks. You answer is very nice! (and it even works for the complex case). $\endgroup$ Jan 10 '16 at 14:08

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