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).
Best Answer
An alternative proof of this, probably not as enlightening but I like it: if $T^* = -T$, then $iT$ is self-adjoint (why?) and so the eigenvalues of $iT$ are real. It then follows that the eigenvalues of $T$ are purely imaginary. (again, why?)