# Relation between eigenvectors and singular vectors of complex skew-symmetric matrices

As shown in this answer, if $$A$$ is a real skew-symmetric matrix, and $$v,w$$ are a pair of orthogonal singular vectors with $$Av=sw \qquad\text{ and }\qquad Aw=-sv,$$ for some $$s>0$$, then the corresponding eigenvectors of $$A$$ are $$v\pm iw$$ (with eigenvalues $$\pm is$$).

However, as discussed for example in these notes (Link to pdf), one can state more generally that for any (possibly complex) skew-symmetric $$A$$ there is a unitary $$U$$ such that $$U^TAU$$ is block diagonal:

$$U^T A U = \operatorname{diag}\left\{\begin{pmatrix}0 & m_1 \\ -m_1 & 0\end{pmatrix}, \begin{pmatrix}0 & m_2 \\ -m_2 & 0\end{pmatrix},..., \begin{pmatrix}0 & m_n \\ -m_n & 0\end{pmatrix},\boldsymbol0\right\}.$$ In analogy with the above statement about the real case, this can be seen as equivalent to stating that the singular vectors of $$A$$ can be organised as pairs $$v_k,w_k$$ such that $$Av_k=s_kw_k \qquad\text{ and }\qquad Aw_k^*=-s_kv_k^*.$$ However, now $$v_k\pm iw_k$$ aren't always eigenvectors.

Is there a general relation between eigenvectors and singular vectors in the complex case?

$$\begin{pmatrix}0 & 1 & 0 \\ -1 & 0 & -i \\ 0 & i & 0 \end{pmatrix}$$ is not diagonalisable. Its singular values still pair as per the mentioned result: they are $$0$$ and $$\pm\sqrt2$$; however, the $$\pm\sqrt2$$ do not correspond to any eigenvector.