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}$$

## Best Answer

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 $GF(5^6)$.