Let $A$ be a matrice with real entries and $A^T=-A$ with eigenvalues $\pm i\lambda$ with $\lambda\in\mathbb R_{\gt0}$ with respective eigenspaces of dimension $1$ (i.e. $\dim \operatorname{Eig}_{i\lambda,A}=1$). Let $u,v$ be real vectors of norm $1$ s.t. $u\bot v$ and $A^2u=-\lambda^2u$ and $A^2v=-\lambda^2v$.
Q1: Is it true that now $Au=-\lambda v$ and $Av=\lambda u$ holds? (Up to permutation of $u$ and $v$.)
Q2: Is it possible to express a possible pair of $u$ and $v$ in terms of the (complex) eigenvectors of $A$?
I came to this question while trying to prove this: https://en.wikipedia.org/wiki/Skew-symmetric_matrix#Spectral_theory.
Edit: I just got the answer to the first part (Q1) after asking. $(Au)^Tu=-u^TAu$ so $u\bot Au$.
If $u$, $v$ are of norm one then it is true (up to a global sign).
Q1: A better approach is perhaps to take one of them, say $u=-\lambda^{-2} A^2u$ and define $v=\lambda^{-1} Au$. Then the couple $u,v$ verify the wanted properties (almost by definition).
Q2:We then also have $A(u\pm i v) = \pm i \lambda (u \pm v)$.