= Solution
{c}
In the <finite field> $\mathbb F_5=\mathbb Z/5\mathbb Z$, the <characteristic polynomial> becomes
$$
\chi_A(\lambda)=(\lambda-1)(\lambda-3)^2,
$$
because $(\lambda-3)^2=\lambda^2-\lambda-1$ modulo $5$. Splitting into linear factors does not by itself guarantee <diagonalizability>: the repeated <eigenvalue> needs enough <eigenvectors>.
Solving $(A-3I)(x,y,z)^T=0$ gives
$$
2x+y=0,\qquad x+3y=0,\qquad y+3z=0,
$$
so $y=3x$ and $z=4x$. Hence
$$
\ker(A-3I)=\operatorname{span}_{\mathbb F_5}\{(1,3,4)^T\}
$$
has <dimension> one, although $3$ has <algebraic multiplicity> two. The <geometric multiplicity> is too small, and \b[$A$ is not diagonalizable over $\mathbb F_5$]. This illustrates <diagonalizability depends on the field>: characteristic zero gave three distinct <eigenvalues>, while reduction modulo $5$ merges two of them.
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