= Solution
In the <finite field> $\mathbb F_3$, reduce the <characteristic polynomial> to $(x-1)^3$. For $B=A-I$, reduction of the entries gives
$$
B=\begin{pmatrix}2&1&0\\0&2&1\\1&0&2\end{pmatrix},\qquad
B^2=\begin{pmatrix}1&1&1\\1&1&1\\1&1&1\end{pmatrix}\ne0,\qquad B^3=0.
$$
Hence
$$
\boxed{m_A(x)=(x-1)^3\quad\text{over }\mathbb F_3.}
$$
Its repeated <linear polynomial> prevents <diagonalizability>. Equivalently, a <diagonalizable> map with only the <eigenvalue> one would have to be $I$, and this <matrix> is not $I$.
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