= Solution
Let $Av=\lambda v$ with $v\ne0$, using the standard complex <Hermitian inner product>. Then
$$
v^\dagger Av=\lambda v^\dagger v.
$$
Taking the <complex conjugate> and using $A^\dagger=-A$ yields $\overline\lambda v^\dagger v=v^\dagger A^\dagger v=-\lambda v^\dagger v$. Since $v^\dagger v>0$,
$$
\boxed{\overline\lambda=-\lambda,\qquad \operatorname{Re}\lambda=0.}
$$
Thus every <eigenvalue> of an <anti-Hermitian matrix> lies on the imaginary axis, including the possible zero <eigenvalue>; each nonzero <eigenvalue> is purely imaginary.
Now take $Au=\lambda u$ and $Av=\mu v$. On one hand $u^\dagger Av=\mu u^\dagger v$. On the other hand
$$
u^\dagger Av=-(Au)^\dagger v=-\overline\lambda u^\dagger v=\lambda u^\dagger v.
$$
Consequently $(\mu-\lambda)u^\dagger v=0$. Distinct <eigenvalues> therefore imply
$$
\boxed{u^\dagger v=0,}
$$
which is the required orthogonality of the <eigenvectors>.
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