= Solution
In the <finite field> $\mathbb F_2$, the <matrix> becomes
$$
A=\begin{pmatrix}0&1&0\\0&0&0\\0&0&1\end{pmatrix}.
$$
On the first two coordinate vectors it is a nonzero <nilpotent> <Jordan block> of size two, and on the last it is the <identity map>. Consequently
$$
\boxed{m_A(x)=x^2(x-1)\quad\text{over }\mathbb F_2.}
$$
The repeated zero root prevents <diagonalizability>. Directly, the only possible <eigenvalues> are zero and one, so <diagonalizability> would imply $A^2=A$, whereas the displayed off-diagonal entry disappears on squaring.
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