# Eigenvalues of symmetric endomorphisms of abelian varieties

Let$$f\in End(X)$$ for an abelian variety X and let $$f'=\phi_H^{-1}\hat{f}\phi_H$$ be the usual Rosati involution. An endomorphism of abelian varieties is called symmetric if $$f=f'$$.

Now I want to show that symmetric endomorphisms have real eigenvalues. This is supposed to be similar to the proof that symmetric matrices only posses real eigenvalues.

Any advice appreciated.

• I assume you are working over $\mathbb{C}$, your $X$ is the quotient of a vector space $V$ by a lattice, you have a polarization on $X$ given by a positive hermitian form $H$ on $V$, and the eigenvalues you are considering are those of the endomorphism $\tilde{f}$ of $V$ induced by $f$. Then $f$ is symmetric if and only if $\tilde{f}$ is self-adjoint w.r.t. $H$, in which case it has real eigenvalues. – abx Aug 3 at 19:47
• Yes your assumptions are correct. If we denote the analytic representation of $\tilde f$ by A and by using the fact that symmetric endomorphisms are self-adjoint w.r.t to H we get hat $A=A'=(H^t)^{-1}\bar A^t H^t$. Then we get for an eigenvalue $\lambda$ of A to the eigenvector x:$\lambda x^t x = x^t(\lambda x) = x^t(Ax) = (A^t x)^t x = (H\bar A H^{-1}x)^t x$ If I can now show that $\bar \lambda$ is an eigenvalue of $H \bar A H^{-1}$ we get $\lambda x^tx= (\bar \lambda x)^t x = \bar \lambda x^t x$ and therefore $\lambda = \bar \lambda$. Is there any theorem about eigenvalues of conjugations? – Christopher Bunse Aug 4 at 8:33