Prove non-zero eigenvalues of skew-Hermitian operator are pure imaginary Just like the title:
Assume $T$ is a skew-Hermitian but not a Hermitian operator an a finite dimensional complex inner product space V. Prove that the non-zero eigenvalues of $T$ are pure imaginary.

Hint $$\langle \mathbf{v},\ A\mathbf{v}\rangle = \mathbf{v}^*A\mathbf{v}= \lambda\|\mathbf{v}\|^2$$
$$\langle A\mathbf{v},\ \mathbf{v}\rangle=\mathbf{v}^*A^*\mathbf{v}=-\mathbf{v}^*A\mathbf{v}=-\lambda\|\mathbf{v}\|^2$$
