I've read that any real skew-symmetric matrix $A$, where $A^T = -A$, can be brought into block diagonal form
$ A = Q \, \Sigma \, Q^T = \left( \begin{array}{ccccl} \vec{q}_1 & \vec{q}_2 & \vec{q}_3 & \vec{q}_4 & \dots \end{array} \right) \, \left( \begin{array}{ccccl} 0 & \lambda_1 & 0 & 0 & \dots \\ -\lambda_1 & 0 & 0 & 0 & \dots \\ 0 & 0 & 0 & \lambda_2 & \dots \\ 0 & 0 & -\lambda_2 & 0 & \dots \\ \vdots & \vdots & \vdots & \vdots & \ddots \end{array} \right) \, \left( \begin{array}{c} \vec{q}^T_1 \\ \vec{q}^T_2 \\ \vec{q}^T_3 \\ \vec{q}^T_4 \\ \vdots \end{array} \right)$
where $\vec{q}_i$ are real, orthogonal column vectors.
But also, the non-zero eigenvalues of $A$ are purely imaginary and occur in pairs $\pm i \lambda_i$, which are the same $\lambda_i$ as occur in the blocks of $\Sigma$. We could then also write $A$ in complex diagonal form
$ A = U \, D \, U^\dagger = \left( \begin{array}{ccccl} \vec{u}_1 & \vec{u}_2 & \vec{u}_3 & \vec{u}_4 & \dots \end{array} \right) \, \left( \begin{array}{ccccl} i\lambda_1 & 0 & 0 & 0 & \dots \\ 0 & -i\lambda_1 & 0 & 0 & \dots \\ 0 & 0 & i\lambda_2 & 0 & \dots \\ 0 & 0 & 0 & -i\lambda_2 & \dots \\ \vdots & \vdots & \vdots & \vdots & \ddots \end{array} \right) \, \left( \begin{array}{c} \vec{u}^\dagger_1 \\ \vec{u}^\dagger_2 \\ \vec{u}^\dagger_3 \\ \vec{u}^\dagger_4 \\ \vdots \end{array} \right)$
where $\vec{u}_i$ are complex, orthogonal column vectors.
My question is what is the relationship between the complex eigenvectors $\vec{u}_i$ and the real vectors $\vec{q}_i$ that bring $A$ into block diagonal form? And how can I show the link?