Show: If the adjoint of T is -T, all eigenvalues are purely imaginary Homework question.
Let $V$ be a finite dimensional inner-product space over $\mathbb{C}$.  Let $T \in L(V,V)$ satisfy $T^*=-T$.  Show that all eigenvalues of $T$ are purely imaginary, i.e., if $\lambda$ is an eigenvalue of $T$, then $\lambda = ia$ with $a \in \mathbb{R}$.

I can recall a proof we did in class that if $T$ is self-adjoint then all eigenvalues are real, and the basic gist of it was let $\lambda$ be an eigenvalue, then lambda is equal to its complex conjugate.  I understand this; it implies that if $\lambda = a+bi = a-bi$ then $b = 0$.
It follows (in my mind, at least) that I need to show that given that $T^* = -T$ that $\lambda = -\lambda$.
I'm kind of stuck here, though.  I've tried adapting the proof I have for self-adjoint operators, but I'm unable to do it without having the conjugate in the result, which is not what I want. (that seems to imply that all eigenvalues are 0, I don't think that's true).

Showing that $\lambda=-\lambda$ would only let you say that $\lambda=0$, not that $\lambda$ was imaginary, so that is not what you need to show.
Instead, look at what exactly was originally done for $S$ symmetric and $\lambda,v$ an eigenpair:
$$\lambda \langle v,v\rangle =\langle Sv,v\rangle=\langle v,Sv\rangle = \overline{\langle Sv,v\rangle} =\overline{\lambda \langle v,v\rangle}=\overline{\lambda}\langle v,v\rangle.$$
The first equality is by linearity, the second by $S=S^*$, the third by conjugate symmetry of the inner product, the fourth by linearity again, and the last by $\overline{ab}=\bar{a}\,\bar{b}$ along with the fact $\langle v,v\rangle$ is purely real and so is invariant under complex conjugation. From this we deduce $\lambda = \lambda^*$.
So what happens if instead $T=-T^*$?
