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  "body": "\n## Statement\n\nLet $F$ be a field and let $A$ be an $n \\times n$ matrix over $F$.\nIf $A$ has $n$ pairwise distinct eigenvalues in $F$, then $A$ is [[diagonalisable|07SQCNV233]] over $F$.\n\n## Remarks\n\nEigenvectors for the $n$ distinct eigenvalues are [[linearly independent|0HNMFG8F8K]], so they form a basis of $F^n$.\nThe converse fails: $I_n$ is diagonal, but it has only one eigenvalue.\nFor example, $\\begin{pmatrix} 1 \u0026 2 \\\\ 0 \u0026 3 \\end{pmatrix}$ has eigenvalues $1$ and $3$, so it is diagonalisable.\n\n## Explanations\n\n### L1 human\n\nIf an $n \\times n$ matrix has $n$ different eigenvalues, then it has $n$ independent eigenvectors.\nSo it is diagonalisable.\nThis sufficient condition is easy to check with the characteristic polynomial.\n",
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