# Eigenvectors of Hermitian matrices over arbitrary fields

Fix a field $$k$$, and suppose $$\gamma$$ is an involutory automorphism of $$\gamma$$ (that is, $$\gamma \ne 1$$, but $$\gamma^2 = 1$$).

Call a matrix $$A$$ $$\gamma$$-Hermitian if $${(A^\gamma)}^T = A$$ (where the "$$T$$" denotes the transpose).

In case $$k = \mathbb{C}$$, we know that there is an orthogonal base of eigenvectors spanning $$\mathbb{C}^n$$, with $$A$$ an $$(n \times n)$$-matrix.

Question: is there any general information available about the (possible) eigenvalues of $$A$$ (for a general field $$k$$) ?

What about when (modest) assumptions are made about $$k$$ ?

• Is the field finite?
– Mick
Mar 13, 2020 at 15:34
• @Mick : Not necessarily, but any info on finite fields is welcome ! Mar 13, 2020 at 15:37

A property of the complex numbers that is used to establish that eigenvalues of Hermitian matrices are real is that $$v^* v = 0$$ only if $$v = 0$$, where $$v \in \mathbb C^n$$. If this is not the case in field $$k$$ for the involution $$\gamma$$ (i.e. there are nonzero vectors such that $$(v^\gamma)^T v = 0$$), you might have $$\gamma$$-Hermitian matrices with eigenvalues not invariant under $$\gamma$$.
For example, take $$k = \mathbb Q(\sqrt{2})$$ with the involution $$\gamma(a+b\sqrt{2}) = a - b \sqrt{2}$$. The vector $$v = \pmatrix{1\cr 1 + \sqrt{2}}$$ satisfies $$(v^\gamma)^T v = 1^2 + (1 - \sqrt{2})(1+\sqrt{2}) = 0$$ Correspondingly, the $$\gamma$$-Hermitian matrix $$A = \pmatrix{-1 & 1\cr 1 & 1\cr}$$ has $$v$$ as eigenvector for eigenvalue $$\sqrt{2}$$.
EDIT: However, any eigenvalue $$\lambda$$ which has an eigenvector $$v$$ such that $$(v^\gamma)^T v \ne 0$$ will be invariant under $$\gamma$$. See the comment below by lonza leggiera.
• Nevertheless, isn't it true that \begin{align} Av=\lambda v&\implies v^{\gamma T}Av=\lambda v^{\gamma T}v\\ &\implies v^{\gamma T}Av=\left(v^{\gamma T}Av\right)^{\gamma T}= \left(\lambda v^{\gamma T}v \right) ^{\gamma T}=\lambda^\gamma v^{\gamma T}v\ ? \end{align} So if $\ v^{\gamma T}v\ne0\$, then $\ \lambda=\frac{v^{\gamma T}Av}{v^{\gamma T}v}=\lambda^\gamma\$. Since this goes a little further towards answering the OP's question, would the fact (even without the calculation) be worth noting in your answer? Mar 13, 2020 at 22:32