 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,Thus 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. DefineThe displayed quadratic identity gives $|u^{\prime }|=|w^{\prime }|=1$. Also $u\cdot Au=0$ and $w\cdot Aw=0$, because a real antisymmetric matrix has zero quadratic form. Thus $u\perp u^{\prime }$ and $w\perp w^{\prime }$. Since $A$ commutes with $A^2$, $u^{\prime }$ lies in the same eigenspace of $A^2$ as $u$, and $w^{\prime }$ in the same one as $w$. Distinct eigenspaces of the real symmetric matrix $A^2$ are orthogonal, so all four vectors are orthonormal. Finally,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.