Let \(F\) be a field and \(M\in M_n(F)\) with minimal polynomial \(m_M\in F[x]\). Then \(M\) is diagonalizable over \(F\) if and only if \(m_M\) splits over \(F\) into distinct linear factors.
If \(M=PDP^{-1}\) with \(D\) diagonal whose distinct diagonal entries are \(\lambda_1,\ldots,\lambda_r\), then \(\prod_i(x-\lambda_i)\) annihilates \(D\) and hence \(M\), so \(m_M\) divides a product of distinct linear factors.
Conversely, suppose \(m_M=\prod_{i=1}^r(x-\lambda_i)\) with the \(\lambda_i\in F\) distinct. The factors are pairwise coprime, so by the kernel decomposition \(F^n=\ker m_M(M)=\bigoplus_{i=1}^r\ker(M-\lambda_i I)\). Choosing a basis of each eigenspace \(\ker(M-\lambda_i I)\) gives a basis of \(F^n\) of eigenvectors of \(M\).
A matrix whose minimal polynomial splits over \(F\) need not be diagonalizable over \(F\). Over any field \(F\), \(M=\begin{bmatrix} 1 & 1 \\ 0 & 1\end{bmatrix}\) has \(m_M = (x-1)^2\), which splits over \(F\), but \(\ker(M-I)\) is one-dimensional, so \(M\) is not diagonalizable.