Any skew-symmetric matrix can be written as $A=UQU^\dagger$ where $U$ is unitary and
\begin{align*}
Q=\begin{bmatrix}
0 & \lambda_1 & \\
-\lambda_1 & 0 & \\
& & 0 & \lambda_2\\
& & -\lambda_2 &0\\
& & & & \ddots\\
& & & & & 0 & \lambda_r\\
& & & & & -\lambda_r & 0
\end{bmatrix}
\end{align*}
(https://en.wikipedia.org/wiki/Skew-symmetric_matrix#Spectral_theory)
Moreover $\{\lambda_i\}$ are the eigenvalues of $iA$.
If $V$ diagonalizes the hermitian matrix $iA=VDV^\dagger$, what is the relation between $U$ and $V$ ?
Best Answer
Note that there are multiple matrices satisfying the properties of the desired matrix $V$, so it doesn't make sense to talk about "the" matrix that diagonalizes $iA$. However, with the following hint, we can use $U$ to generate one such matrix $V$.
Hint: Let $J,K$ denote the matrices $$ J = \frac 1{\sqrt{2}}\pmatrix{1&i\\i&1}, \quad K = \pmatrix{J\\ & \ddots \\ & &J}, $$ where $K$ has the same shape as $Q$. We find that $K$ is unitary and that $K^\dagger QK$ is diagonal.