Diagonalizability and the minimal polynomial

Forward direction. If $A=PDP^{-1}$ with $D=\textrm{diag}(\lambda_1,\ldots,\lambda_n)$, then $A$ is killed by $\prod_i(X-\lambda_i)$ where the product is over distinct eigenvalues. The minimal polynomial is the smallest such polynomial, so it is a product of distinct linear factors.

Converse. Let $\mu_A(X)=\prod_{i=1}^r(X-\lambda_i)$ with distinct $\lambda_i$. For each $i$, let $E_i=\ker(A-\lambda_i I)$. We claim that the sum $E_1+\cdots+E_r$ is $\mathbb{R}^n$.

Indeed, the polynomials $P_i(X)=\prod_{j\neq i}\frac{X-\lambda_j}{\lambda_i-\lambda_j}$ form a Lagrange interpolation basis: $\sum_i P_i=1$ and $P_i$ is zero at all $\lambda_j$ for $j\neq i$, and $P_i(\lambda_i)=1$.

For any $v\in\mathbb{R}^n$: $v=\sum_i P_i(A)v$ (since $\sum_i P_i=1$). But $(A-\lambda_i I)P_i(A)v = \mu_A(A)\cdot(\text{something})=0$, so $P_i(A)v\in E_i$. Hence $v$ decomposes as a sum of eigenvectors, and $\mathbb{R}^n$ is spanned by the eigenspaces. Thus $A$ is diagonalizable.

Conclusion. $A$ is diagonalizable iff its minimal polynomial has only simple roots.