Is it possible to consider complex eigenvalues without a Hermitian (i.e. sesquilinear) inner product over a complex vector space?
For instance: let $A$ be a real orthogonal matrix (so $A^TA = I$). Without referencing a Hermitian inner product, is it possible to show that the complex eigenvalues of $A$ have magnitude $1$?
The usual proof of this fact is as follows: if $x,\lambda$ is an eigenpair of $A$, then we have $$ \|x\|^2 = x^*x = x^*(A^*A)x = (Ax)^*(Ax) = \lambda\overline{\lambda} (x^*x) = |\lambda|^2 \|x\|^2 $$ from which it follows that $|\lambda| = 1$. Note: this proof required the use of the sesquilinear inner product $\langle x,y \rangle = y^*x$.
A rephrasing of the original question: consider $\Bbb C^n$ with the bilinear from $$ \langle x,y \rangle = y^Tx $$ note that this bilinear form is not an inner product. The complex-orthogonal matrices are those matrices $A$ that satisfy $A^TA = I$, where $T$ is the entrywise transpose. Notably, the complex-orthogonal matrices preserve the above bilinear form. How can we show that if $A$ is complex-orthogonal with real entries, then the eigenvalues of $A$ have magnitude $1$?