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 $k$ ?

  • $\begingroup$ Is the field finite? $\endgroup$ – Mick Mar 13 at 15:34
  • $\begingroup$ @Mick : Not necessarily, but any info on finite fields is welcome ! $\endgroup$ – Boccherini Mar 13 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.

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    $\begingroup$ 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? $\endgroup$ – lonza leggiera Mar 13 at 22:32
  • $\begingroup$ @lonzaleggiera Good point, I'll mention it. $\endgroup$ – Robert Israel Mar 13 at 23:07

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