# Calculate the eigenvalues of a matrix over a finite field

eigenvalues-eigenvectorsfinite-fieldslinear algebramatrices

Is there any way to calculate the eigenvalues (or eigen vectors) of a matrix over a finite field rather than brute force? I created $$5\times5$$ matrices over $$\mathbb{F}_5$$ and calculated eigenvalues. I just tried all the $$5 \times 1$$ vectors over $$\mathbb{F}_5$$ (there are only $$5^5$$ $$5\times1$$ vectors).

Some have eigenvalues and the others do not (or I wasn't able to find eigenvalues).
The following matrix $$A$$ does not have an eigenvalue, for example. The rank is 5 and it has the inverse matrix too… Do some matrices over a finite field have no eigenvalue? If it's not a finite field, it has 5 eigenvalues in $$\mathbb{C}$$

$$A = \begin{bmatrix} 2&1&1&1&4\\ 2&4&2&4&1\\ 1&0&1&4&0\\ 4&4&4&3&2\\ 2&0&0&4&1\\ \end{bmatrix}$$

The characteristic polynomial of that matrix factors over $$GF(5)$$ as $$\chi_A(x)=\det(xI_5-A)=\left(x^2+2 x+4\right) \left(x^3+2 x^2+2 x+2\right).$$ Therefore the matrix $$A$$ has two eigenvalues in $$GF(5^2)$$ and three eigenvalues in $$GF(5^3)$$. From the first factor $$x^2+2x+4=(x+1)^2-2$$ we see that $$-1\pm\sqrt2$$ are the eigenvalues in $$GF(5^2)=GF(5)(\sqrt2)$$.
If you want a single field to contain them all you need to go to $$GF(5^6)$$.