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  "body": "\n## Statement\n\nLet $F$ be a field and let $A$ be an $n \\times n$ matrix over $F$ with [[minimal polynomial|03NDVCSHSB]] $m_A$.\nThen $A$ is [[diagonalisable|07SQCNV233]] over $F$ if and only if\n$$m_A(t) = (t - \\lambda_1)(t - \\lambda_2) \\cdots (t - \\lambda_k)$$\nfor some pairwise distinct $\\lambda_1, \\dots, \\lambda_k \\in F$.\n\n## Remarks\n\nIn words: $m_A$ splits over $F$ and has no repeated roots.\nIf $A^2 = A$, then $m_A$ divides $t^2 - t = t(t - 1)$, so each [[projection|033DA1WE8Y]] matrix is diagonalisable.\nThe matrix $\\begin{pmatrix} 1 \u0026 1 \\\\ 0 \u0026 1 \\end{pmatrix}$ has $m_A = (t - 1)^2$, so it is not diagonalisable.\n\n## Explanations\n\n### L1 human\n\nThe minimal polynomial decides whether a matrix is diagonalisable.\nThe matrix is diagonalisable exactly when this polynomial is a product of distinct linear factors.\nA repeated factor, like $(t - 1)^2$, signals a Jordan block of size at least $2$.\n",
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