diagonalization of a matrix over finite fields I'm having a problem with determine whether a matrix is diagonalizable over $\mathbb F_{2}$, over $\mathbb F_{3}$, etc.
for example, for the following matrix:
$$        \begin{bmatrix}
        1 & 1  & 0 \\ 
        1 & 1  & 0 \\ 
        0 & 0  & 1 
        \end{bmatrix}$$
the characteristic polynomial is $P(x) = -x^{3}+3x^{2}+2x$ 
I understand that I need to "convert" it to the correct field in some way- but that part
I didn't get yet.
I'd be happy to get some help with this...
 A: If you compute the determinant by cofactor expansion along the third row, you see that the characteristic polynomial is
$$ p_A(\lambda) = |A - \lambda \cdot I| = (1 - \lambda) \cdot \left| \begin{matrix}
        1 - \lambda & 1 \\ 
        1 & 1 - \lambda  \\ \end{matrix} \right| =  (1 - \lambda) \cdot ((1 - \lambda)^2 - 1) = (1 - \lambda) \cdot (- \lambda) \cdot (2 - \lambda) = -\lambda (1-\lambda )(2-\lambda ).$$
Then, over any field in which $2 \neq 0$, the matrix is diagonalizable because it has three distinct eigenvalues. Over $\mathbb{F}_2$, $p_A$ takes the form of $p_A(\lambda) = \lambda^2 (1 - \lambda)$. The minimal polynomial of $A$ is $p_A$ and not $\lambda(1 - \lambda)$ and so $A$ is not diagonalizable.
Alternatively, since $A$ is already in block form, the matrix $A$ is diagonalizable if and only if the matrix
$$ B = \left( \begin{matrix}
        1 & 1 \\ 
        1 & 1 \\ \end{matrix}  \right) $$
is diagonalizable and this matrix is easier to analyze because it is $2 \times 2$. Over $\mathbb{F}_2$ (or any other field with characteristic $2$) this is a nontrivial nilpotent matrix ($A^2 = 0$) and so not diagonalizable, while over other fields it is diagonalizable.
A: In $\mathbb F_2$, $3=1$ and $2=0$. In $\mathbb F_3$, $3=0$. Does that help?
