= Solution
Let $A\mathbf v=\lambda\mathbf v$ with $\mathbf v\ne0$ and $A^\dagger=-A$. Taking the conjugate of the scalar $\mathbf v^\dagger A\mathbf v$ gives
$$
\lambda^*\mathbf v^\dagger\mathbf v
=\mathbf v^\dagger A^\dagger\mathbf v
=-\lambda\mathbf v^\dagger\mathbf v.
$$
Since $\mathbf v^\dagger\mathbf v>0$, it follows that
$$
\boxed{\lambda^*=-\lambda,\qquad \operatorname{Re}\lambda=0.}
$$
Thus every nonzero <eigenvalue> of an <anti-Hermitian matrix> is purely imaginary; zero is allowed as well.
For $A\mathbf u=\lambda\mathbf u$ and $A\mathbf v=\mu\mathbf v$, compute the same mixed scalar in two ways:
$$
\lambda^*\mathbf u^\dagger\mathbf v
=(A\mathbf u)^\dagger\mathbf v
=\mathbf u^\dagger A^\dagger\mathbf v
=-\mu\mathbf u^\dagger\mathbf v.
$$
Using $\lambda^*=-\lambda$ gives $(\mu-\lambda)\mathbf u^\dagger\mathbf v=0$. Therefore
$$
\boxed{\lambda\ne\mu\ \Longrightarrow\ \mathbf u^\dagger\mathbf v=0.}
$$
The relevant orthogonality uses the complex <inner product>, not the unconjugated bilinear dot product. Repeated <eigenvalues> do not force every possible pair of eigenvectors within that one eigenspace to be orthogonal.
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