TITLE: Eigenvalue problems for matrices over finite fields
QUESTION [9 upvotes]: Suppose I have a symmetric matrix A with entries in a finite field. In particular, I have the case in mind where $A \in \{0,1\}^{n \times n}$ and want to treat the entries as elements of $GF(2)$. How much is known about the eigenvalue problem in this case? Is there a spectral theorem? Are there fast algorithms for computing eigenvectors?

REPLY [9 votes]: I'm not quite sure what you mean by a spectral theorem in this context. So I will list some thoughts/facts that come to mind, and may be related to what you want to ask about.

More often than not the eigenvalues belong to an extension field of $GF(2)$. Therefore finding eigenvectors and eigenvalues forces us to first define those extensioin fields. That is straightforward.
Jordan canonical form of any matrix is still there. Nominally it exists over an algebraic closure $K$ of $GF(2)$, but we obviously get away with a finite subfield of $K$ generated over $GF(2)$ by the eigenvalues.
You specifically ask about symmetric matrices. Here there is a marked difference to the case of symmetric real matrices. Symmetry of a matrix does not mesh at all well with it being over a field of characteristic two. There is no reason to expect such a matrix to be diagonalizable (not even over $K$). As examples I proffer
$$
A=\left(\begin{array}{cc}0&1\\1&0\end{array}\right)\quad\text{and}\quad
B=\left(\begin{array}{cc}1&1\\1&1\end{array}\right).
$$
The characteristic polynomial of $A$ is $x^2+1=(x+1)^2$. But we see that the eigenspace of the sole eigenvalue $\lambda=1$ is just 1-dimensional, and hence $A$ is not diagonalizable. Similarly the characteristic polynomial of $B$ is $x^2$, and again the eigenspace of the sole eigenvalue $\lambda=0$ is 1-dimensional. In both cases the eigenspace is spanned by $(1,1)^T$.
Algorithmically finding a generalized eigenspace decomposition (=the Jordan blocks) should not be any harder than factoring the characteristic polynomial. There is an algorithm due to Berlekamp for doing that. I have not thought this through in detail, so please take this with a grain of salt.