If an endomorphism satisfies $\alpha^* = -\alpha$, then its eigenvalues are purely imaginary This is another exercise from Golan's book.
Problem: Let $V$ be an inner product space over $\mathbb{C}$ and let $\alpha$ be an endomorphism of $V$ satisfying $\alpha^*=-\alpha$, where $\alpha^*$ denotes the adjoint. Show that every eigenvalue of $\alpha$ is purely imaginary.
My proposed solution is below.
 A: Let me show another argument which applies to a more general setting: if $\alpha$ is a linear operator on a Hilbert space satisfying $\alpha^*=-\alpha$, then the spectrum of $\alpha$ is purely imaginary (i.e. real part equal zero). 
Indeed, one simply needs to notice that $\alpha-\lambda\,\text{id}$ is invertible if and only if $(\alpha-\lambda\,\text{id})^*$ is invertible. As $$(\alpha-\lambda\,\text{id})^*=\alpha^*-\overline\lambda\,\text{id}=-\alpha-\overline\lambda\,\text{id}=-(\alpha+\overline\lambda\,\text{id}),$$ we conclude that any $\lambda$ in the spectrum of $\alpha$ satisfies $\overline\lambda=-\lambda$.
A: Suppose $c$ is an eigenvalue of $\alpha$ and $v$ is a corresponding eigenvector. We have
$$c \langle v,v\rangle= \langle \alpha(v),v\rangle =\langle v, \alpha^*(v)\rangle =\langle v,-\alpha(v)\rangle=\langle v, -cv\rangle=\overline{\langle -cv, v\rangle} = \overline{-c\langle v,v\rangle}$$
Because $\langle v,v\rangle$ is real, as it is the norm of a vector, we have $c=\overline{-c}$, which immediately shows $c$ is imaginary (write $c$ out in terms of imaginary and real parts to see this).
