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?
Best Answer
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.