In a field $\mathbb{F}_4 = \{0, 1, a, a+1\}$ where $1 + 1 = 0$ and $a^2 = a + 1$,
I am to find the characteristic polynomial of $A = \begin{pmatrix} a+1 & 0 & 0 & a \\ 0 & 1 & 0 & 0 \\ a+1 & a & a & 1 \\ a & a & a+1 & a+1 \end{pmatrix}$.
I first swap row 1 with row 2, and take the determinant of $(A-\lambda I)$ which yields
$(A-\lambda I) = -(1-\lambda)\begin{pmatrix} a+1-\lambda & 0 & a \\ a+1 & a-\lambda & 1 \\ a & a+1 & a+1-\lambda \end{pmatrix}$,
but I just get stuck in very long expressions where I can't easily determine its eigenvalues (because I will later on write the Jordan normal form of $A$).
I arrive at, after using the fact that $1 + 1 = 0$ and $a^2 = a + 1$
$-(1-\lambda) \left( ((a+1-\lambda)(a-\lambda)(a+1-\lambda)-(a+1)) + a((a+1)^2-a(a-\lambda)) \right) = 0$
Which is the determinant of the above $3\times 3$ matrix,
and the first term I can see means one eigenvalue is $\lambda=1$, but then I get an expression I cannot make sense of inside the parentheses.
What am I doing wrong?
You can use the Addition and Multiplication in $F_4$ (I'll highlight in color where I'm using them).
Also, I think you misplaced a parenthesis in the following line:
$(a+1-\lambda)\Big((a-\lambda)(a+1-\lambda)-(a+1)\Big)=$
$\require{cancel} (a+1-\lambda)\Big(\color{blue}{a^2}+\cancel{a}-\color{red}{\bcancel{\color{black}{2\lambda a}}}-\lambda+\lambda^2-\cancel{a}+1\Big)=({a+1}-\lambda)\Big(\color{blue}{a+1}-\lambda+\lambda^2+1\Big)=\\\color{violet}{(a+1)^2}-\color{red}{\bcancel{\color{black}{2\lambda(a+1)}}}+\lambda^2(a+\cancel 1)+a+1-\cancel{\lambda^2}-\lambda^3-\lambda=\color{lightblue}{\bcancel{\color{violet}{a}}}+\lambda^2a+\color{lightblue}{\bcancel{\color{black}{a}}}+1-\lambda^3-\lambda=\\\color{lightgreen}-\lambda^3+\lambda^2a\color{lightgreen}-\lambda+1=\lambda^3+\lambda^2a\color{lightgreen}+\lambda+1$
On the other hand,
$a\Big(\color{violet}{(a+1)^2}-a(a-\lambda)\Big) = a\Big(\color{violet}a-\color{blue}{a^2}+\lambda a\Big)=a\Big(\cancel a-\color{blue}{(\color{black}{\cancel {\color{blue}{a}}}+1)}+\lambda a\Big)=-a+\lambda \color{darkblue}a=\\\color{lightgreen}-a+\lambda\color{darkblue}{(a+1)}=\lambda(a+1)\color{lightgreen}+a$
So we get
$\det(A-\lambda I)/(1-\lambda)=\big(\lambda^3+\lambda^2a+\color{red}{\bcancel{\color{black}{\lambda}}}+1\big)+\big(\lambda(a+\color{red}{\bcancel{\color{black}1}})+a\big)=\lambda^3+\lambda^2a+\lambda a+a+1$
Since there are only four elements in $\mathbb F_4$, it's easy to check that the resulting polynomial has $a$ as a single root and $a+1$ as a double root.