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 $1$?
Best Answer
Well, in a way, yes. The real Jordan normal form theorem yields that every real matrix has an invariant subspace $W$ (i.e. $AW\subseteq W$) such that $1\le\dim W\le 2$. Since $\dim W^\perp +\dim W=n$ and, if $A$ is orthogonal, the orthogonal complement of an $A$-invariant subspace is $A$-invariant, there is an orthonormal basis $B=(b^1,\cdots, b^n)$ such that $B^{-1}AB=B^TAB=\begin{pmatrix}U &0\\ 0&U'\end{pmatrix}$, for some $U\in O(2)$ and $U'\in O(n-2)$ - or, repsectively, $O(1)$ and $O(n-1)$. Now, the eigenvalues of $A$ are either eigenvalues of $U$ or eigenvalues of $U'$. The form and eigenvalues of a $2\times 2$ orthogonal matrix can be calculated explicitly, and the rest can be done by induction.
That being said, as far as I know the real Jordan normal form theorem is proved by complexifying the endomorphism of $A$, using the machinery of Jordan normal form in $\Bbb C^n$, and then bringing it back to $\Bbb R^n$. So one could argue that this is just hiding the issue under the carpet.