When can two matrices be simultaneously skewsymmetrized?

Question

Let $A$ and $B$ be two real, diagonalizable matrices with purely imaginary eigenvalues. What extra conditions does the pair $A,B$ need to satisfy in order for a nonsingular real matrix $U$ to exist such that both $U^{-1}AU$ and $U^{-1}BU$ are skew symmetric?
For $2 \times 2$ matrices their simultaneous anti symmetrization is equivalent to $ A$ is a scalar multiple of $B.$ A necessary condition for the general case is that any linear combination $$aA+bB, \,\ a,b \in \mathbb{R}$$ is to have purely imaginary eigenvalues. Are there any sufficient conditions?

Answer 1

A sufficient condition for every $U$ that makes $A$ skew-symmetric to also make $B$ skew-symmetric is that $B = f(A)$ (in the holomorphic functional calculus) where $f$ is an odd function analytic in a neighbourhood of the spectrum of $A$. Thus if both $A$ and $B$ are such functions of a third matrix $C$ (not necessarily real), any $U$ that makes $C$ skew-symmetric will make both $A$ and $B$ skew-symmetric.

Answer 2

Here is one criterion that is necessary and sufficient: There exists a positive-definite, symmetric matrix $S$ satisfying the system of linear equations $$ AS + SA^T = BS + SB^T = 0,\tag1 $$ if and only if $A$ and $B$ can be simultaneously conjugated to skew-symmetric matrices. (Proof: Given a positive definite symmetric $S$ satisfying $(1)$, write $S = U^2$ where $U$ is also positive definite and symmetric (there always exists such a $U$ and it is unique). Then $(1)$ implies that $U^{-1}AU$ and $U^{-1}BU$ are skew-symmetric. Conversely, if there exists an invertible $U$ such that $U^{-1}AU$ and $U^{-1}BU$ are skew-symmetric, then $S = UU^T$ is a positive definite symmetric matrix that satisfies the linear system $(1)$.)

Whether or not a positive definite $S$ satisfying the system $(1)$ exists can usually be easily checked, since the above system of linear equations is overdetermined and, for most pairs $(A,B)$ that satisfy your conditions, the only solution for symmetric $S$ is $S=0$. It is only for special pairs $(A,B)$ that there are nonzero solutions $S$ and, for them, usually, the space of solutions has dimension $1$ and it will be easy to tell whether there is a positive definite $S$ in the solution space.

You may not regard the above criterion as an answer, but, generally, as the rank of the matrices goes up, there will be more and more degenerate cases, and it will be harder and harder to write down more explicit criteria.

However, in the low dimensions, it is not too bad. For example, when the matrices $A$ and $B$ are $3$-by-$3$ matrices, consider the matrix $C = [A,B] = AB-BA$. Then a necessary and sufficient condition that $A$ and $B$, assumed diagonalizable with purely imaginary eigenvalues, be simultaneously skew-symmetrizable is that the commutators $[C,A]$ and $[C,B]$ be linear combinations of $A$, $B$, and $C$ and that any linear combination of $A$, $B$, and $C$ be diagonalizable with purely imaginary eigenvalues. (This criterion does not work for larger matrices, though.)

Answer 3

From my comment. This answers the original question, which was, if $A,B$ are $\mathbf{C}$-diagonalizable and only have purely imaginary eigenvalues, does there always exist $U$ conjugating both $A$ and $B$ to real skew-symmetric matrices.

If the conclusion for $(A,B)$ holds, then $tA+B$ is conjugate to a skew-symmetric matrix for every real $t$, hence has imaginary eigenvalues. Counterexample: $$A=\begin{pmatrix}0 & 1\\ -1 & 0\end{pmatrix},\quad B=\begin{pmatrix}-1 & -2\\ 1 & 1\end{pmatrix},$$ (note that $B$ is conjugate to $A$; their eigenvalues are $\pm i$); $$A+B=\begin{pmatrix}-1 & -1\\ 0 & 1\end{pmatrix},$$ which has real nonzero eigenvalues.