Prove that the eigenvalues of skew-Hermitian matrices are purely imaginary I would like some help on proving that the eigenvalues of skew-Hermitian matrices are all pure imaginary. I have gotten started on it, but am getting stuck.

Attempt at proof: $Av=\lambda v \implies A \bar{v}=\bar{\lambda}\bar{v}.$

Also, $v^TA^T=\lambda v^T \implies v^TA^T\bar{v}=\lambda v^T \bar{v} \implies ?$

Should I conjugate both sides next?
Let $x$ an eigenvector of $A$ associated to the eigenvalue $\lambda$ then 

$$\langle Ax,x\rangle=\overline \lambda||x||^2=\langle x,A^*x\rangle=-\langle x,Ax\rangle=-\lambda\langle x,x\rangle$$
so 
$$\overline\lambda=-\lambda$$
hence $\lambda$ is pure imaginary complex.
$AX=λX$
so $X^H AX=X^H λX$

$-A^H X=λX$, so $X^H A^H X=-X^H λX$

$X^H AX=-X^H λ^H X$

Hence, $λ=-λ^H$