Can even-dimensional complex skew-symmetric matrices be made block diagonal by orthogonal matrices?

Question

Suppose an even dimensional matrix Q is complex and skew-symmetric. Can Q be written as $Q= O^T \Sigma O$, where O is orthogonal (i.e. $O^T O = O O^T=1$) and $\Sigma$ is block diagonal with $2 \times 2$ blocks?

Answer 1

With such a $\Sigma$, the size of every Jordan block of $Q=O^\top \Sigma O=O^{-1}\Sigma O$ must also be at most $2\times2$. However, according to Lemma 5.2.1 on p.35 of Olga Ruff's Master thesis, for any natural number $k$ and any $\lambda\in\mathbb{C}$, the matrix $J_k(\lambda)\oplus J_k(-\lambda)$ is similar to some complex skew-symmetric matrix. Hence there exists a complex skew-symmetric matrix $Q$ whose Jordan form contains a $3\times3$ Jordan block. In other words, the answer to your question is negative when the dimension of $Q$ is at least $6\times6$.