I want to determine the Jordan normal form of the following matrix:
$$\begin{bmatrix}0 & 1\\1 & 1\ \end{bmatrix}$$ in $M_2(\mathbb F_2) $ and $M_2(\mathbb F_4) $.
I know how to generally determine the Jordan normal form of a matrix, but I'm somehow stuck on how to do it in $M_2(\mathbb F_2) $ or $M_2(\mathbb F_4) $. I first wanted to show that the matrix given is diagonalisable, but then I end up with eigenvalues $\frac{-1 \pm \sqrt{-3}}{2}$. Can I simply ignore the fact that those are not $\in \mathbb F_2$ nor $\mathbb F_4$?
If someone could show me how to solve it in $\mathbb F_2$, then I think I could do the other one. I'd be happy about every help.
The technique of completing the square for solving a quadratic involves multiplying by $4$ or dividing by $2$, which cannot be done in a field of characteristic $2$.
Indeed the characteristic polynomial of the matrix is $$ \det\begin{bmatrix}-x & 1 \\ 1 & 1-x\end{bmatrix} =-x(1-x)-1=x^2-x-1=x^2+x+1 $$ which is irreducible over $\mathbb{F}_2$, having no roots. Since it has degree $2$, it has its roots in $\mathbb{F}_4$. If $\alpha$ is one of them, then the other one is $\alpha+1=\alpha^{-1}$.
So the matrix is diagonalizable over $\mathbb{F}_4$, having distinct eigenvalues. We can also find the eigenvectors: $$ \begin{bmatrix}1\\\alpha\end{bmatrix} \qquad \begin{bmatrix}\alpha\\1\end{bmatrix} $$ so $$ \begin{bmatrix}0 & 1 \\ 1 & 1\end{bmatrix}= \begin{bmatrix}1 & \alpha\\\alpha & 1\end{bmatrix} \begin{bmatrix}\alpha & 0 \\ 0 & \alpha^{-1}\end{bmatrix} \begin{bmatrix}\alpha^{-1} & 1 \\ 1 & \alpha^{-1}\end{bmatrix} $$ Note that $\alpha^2=\alpha^{-1}$ and $\alpha^{3}=1$, so the powers of the diagonal matrix are $$ D^{3k}=I,\quad D^{3k+1}=D=\begin{bmatrix}\alpha & 0 \\ 0 & \alpha^{-1}\end{bmatrix}\quad D^{3k+2}=D^2=\begin{bmatrix}\alpha^{-1} & 0 \\ 0 & \alpha\end{bmatrix} $$ which you can compute the powers of the original matrix from.