Diagonalisable iff the minimal polynomial has distinct roots

Statement

Let $F$ be a field and let $A$ be an $n \times n$ matrix over $F$ with minimal polynomial $m_A$. Then $A$ is diagonalisable over $F$ if and only if $$m_A(t) = (t - \lambda_1)(t - \lambda_2) \cdots (t - \lambda_k)$$ for some pairwise distinct $\lambda_1, \dots, \lambda_k \in F$.

Remarks

In words: $m_A$ splits over $F$ and has no repeated roots. If $A^2 = A$, then $m_A$ divides $t^2 - t = t(t - 1)$, so each projection matrix is diagonalisable. The matrix $\begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}$ has $m_A = (t - 1)^2$, so it is not diagonalisable.

Explanations

L1 human

The minimal polynomial decides whether a matrix is diagonalisable. The matrix is diagonalisable exactly when this polynomial is a product of distinct linear factors. A repeated factor, like $(t - 1)^2$, signals a Jordan block of size at least $2$.