Diagonalizability depends on the field ($A\text{ diagonalizable over }F⟺F^n\text{ has an eigenbasis of }A$)

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,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.