A matrix $ M $ is symmetrizable if $ M = D S $ with $ D $ a square diagonal matrix with positive entries, and $ S $ a symmetric matrix. What can be said about the spectrum of $ M $?
It seems like I can just define the inner product induced by $ D^{-1} $, ie $ \langle v, w \rangle = v^T D^{-1} w $, then recognize that $ M $ is hermitian under this inner product, and then apply the spectral theorem. But it has been a long day and I'm afraid I'm overlooking something, so a confirmation would be nice.
Yes, that's right, although I assume you mean that $S$ is symmetric and real; otherwise, any complex number $z$ can appear in the spectrum of $M$ by setting, for example, $D=I, S = zI.$
If $v$ is a unit eigenvector of $M$ with eigenvalue $\lambda$, then $$\lambda v^*D^{-1}v = v^*D^{-1}DSv = v^*Sv = v^*S^*v = v^*S^*D^*D^{-1}v = \lambda^*v^*D^{-1}v.$$
Since $D$ is Hermitian positive-definite, so is $D^{-1}$ and $v^*D^{-1}v \neq 0$, so $$\lambda = \lambda^*$$ and $\lambda$ is real. Notice that the signs of the entries on the diagonal of $D$ aren't really relevant, as long as they are nonzero and real.