Let \(F\) be a field and let \(M\in\Mat_n(F)\). The following are equivalent:
\(M\) is diagonalizable over \(F\).
The minimal polynomial \(\min_M(x)\) is a product of distinct monic linear factors in \(F[x]\).
There is a basis of \(F^n\) consisting of eigenvectors of \(M\).
Every elementary divisor of \(M\) has degree \(1\).
In particular, if \(M\) has \(n\) distinct eigenvalues in \(F\), equivalently if \(\min_M(x)\) has \(n\) distinct roots in \(F\), then \(M\) is diagonalizable over \(F\).
A matrix whose minimal polynomial splits over \(F\) need not be diagonalizable over \(F\). For \(M=\begin{bmatrix}1&1\\0&1\end{bmatrix}\) over any field \(F\), \(\min_M(x)=(x-1)^2\) splits over \(F\), and \(M\) is not diagonalizable, because its only eigenvalue is \(1\) and \(M\ne I\).