= Solution
For a real <antisymmetric matrix>, $A^T=-A$, so $(A^2)^T=A^2$. The previous result makes every <eigenvalue> $\eta$ of $A^2$ real, with a real nonzero <eigenvector> $v$. Moreover,
$$
\eta|v|^2=v^TA^2v=-(Av)^TAv=-|Av|^2\le0.
$$
Thus \b[every <eigenvalue> of $A^2$ is real and nonpositive].
Take real unit <eigenvectors> $u,w$ of $A^2$ for the respective distinct nonzero <eigenvalues> $-\lambda^2,-\mu^2$, where the real parameters $\lambda,\mu$ are nonzero. Define
$$
u'=\frac{Au}{\lambda},\qquad w'=\frac{Aw}{\mu}.
$$
The displayed quadratic identity gives $|u'|=|w'|=1$. Also $u\cdot Au=0$ and $w\cdot Aw=0$, because a real <antisymmetric matrix> has zero <quadratic form>. Thus $u\perp u'$ and $w\perp w'$. Since $A$ commutes with $A^2$, $u'$ lies in the same <eigenspace> of $A^2$ as $u$, and $w'$ in the same one as $w$. Distinct <eigenspaces> of the real <symmetric matrix> $A^2$ are <orthogonal>, so all four <vectors> are <orthonormal>. Finally,
$$
\boxed{Au=\lambda u',\quad Au'=-\lambda u,\qquad
Aw=\mu w',\quad Aw'=-\mu w.}
$$
Thus on each of the two perpendicular planes, $A$ acts as a scaled quarter-turn. This constructs the required <vectors>, rather than merely counting dimensions.
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