Where does $V$ from the spectral decomposition $A = VDV^*$ lie, if $A$ has only imaginary entries?

Question

The spectral theorem says that for every Hermitian matrix $A \in \mathbb{C}^{n \times n}$ there is a unitary matrix $V \in U(n)$ and a diagonal matrix $D \in \mathbb{R}^{n \times n}$ such that $A = VDV^*$.

If $A$ is assumed to have only real entries, then $V$ must come from the set of orthogonal matrices $O(n) \subset U(n)$.

Question: If $A \in i\mathbb{R}^{n \times n}$ is assumed to have only imaginary entries, can we say anything about the subset of $U(n)$ in which $V$ can lie?

Any help is much appreciated!

Answer 1

The $n\times n$ imaginary matrix $A$ satisfies $A^\top=-A$, so it is skew-symmetric. The Youla decomposition is $$A=iO\Sigma O^\top,$$ where $O$ is a real orthogonal matrix and $\Sigma$ is a real block-diagonal matrix of the form $$\Sigma = \begin{pmatrix} \begin{matrix}0 & \lambda_1 \\ -\lambda_1 & 0\end{matrix} & 0 & \cdots & 0 \\ 0 & \begin{matrix}0 & \lambda_2 \\ -\lambda_2 & 0\end{matrix} & & 0 \\ \vdots & & \ddots & \vdots \\ 0 & 0 & \cdots & \begin{matrix}0 & \lambda_{n/2}\\ -\lambda_{n/2} & 0\end{matrix} \end{pmatrix} $$ for $n$ even. If $n$ is odd a row and column of zeroes is appended.