= Solution
A real <skew-symmetric matrix> satisfies $A^\dagger=A^T=-A$, so it is a <skew-Hermitian matrix>. If $Ax=\lambda x$, then
$$
\lambda\,x^\dagger x=x^\dagger Ax.
$$
Taking the <complex conjugate> and using $A^\dagger=-A$ shows that this scalar is the negative of its conjugate. Hence
$$
\boxed{\lambda^*=-\lambda},
$$
so every eigenvalue is purely imaginary or zero.
If $Ax=\lambda x$ and $Ay=\mu y$, then
$$
\lambda^*x^\dagger y=(Ax)^\dagger y=x^\dagger A^\dagger y=-\mu x^\dagger y.
$$
Since $\lambda^*=-\lambda$, this becomes $(\mu-\lambda)x^\dagger y=0$. Distinct eigenvalues therefore have orthogonal eigenvectors:
$$
\boxed{\lambda\ne\mu\implies x^\dagger y=0}.
$$
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