= Diagonalizability depends on the field
{title2=$A\text{ diagonalizable over }F\iff F^n\text{ has an eigenbasis of }A$}
For a <matrix> over a <field> $F$, <diagonalizability> requires an <eigenbasis> with entries in $F$ and <eigenvalues> in $F$. Passing to a larger <field> can supply missing <roots> of the <characteristic polynomial>. For example,
$$
A=\begin{pmatrix}0&1&0\\1&1&0\\0&1&1\end{pmatrix},\qquad
\chi_A(t)=(t-1)(t^2-t-1)
$$
is not a <diagonalizable matrix> over the <rational numbers>, since the quadratic factor has no rational <root>. It is a <diagonalizable matrix> over the <real numbers> and <complex numbers>, since its three <eigenvalues> are distinct. More generally, <diagonalizability> is equivalent to the <minimal polynomial> splitting into distinct linear factors over the specified <field>.
For an integer <matrix>, reduction modulo a <prime number> is a different operation from extension of its coefficient <field>, and can merge <eigenvalues>. In the same example modulo $5$, $\chi_A(t)=(t-1)(t-3)^2$, while $\ker(A-3I)$ is spanned by $(1,3,4)^T$. Thus the <geometric multiplicity> of $3$ is one and its <algebraic multiplicity> is two, preventing <diagonalizability> even though all <eigenvalues> belong to the <finite field>.
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