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  "body": "\n## Statement\n\nLet $F$ be a field and let $A$ be an $n \\times n$ matrix over $F$.\nThen $A$ is [[diagonalisable|07SQCNV233]] if and only if $F^n$ has a [[basis|0KCQ9Y3QKA]] of [[eigenvectors|0N6960W0BV]] of $A$.\nIf $v_1, \\dots, v_n$ is such a basis with $A v_i = \\lambda_i v_i$, and $P$ has the columns $v_1, \\dots, v_n$, then $P^{-1} A P$ is the diagonal matrix with entries $\\lambda_1, \\dots, \\lambda_n$.\n\n## Remarks\n\nThe equation $AP = PD$ says column by column that $A v_i = \\lambda_i v_i$.\nEquivalently, $F^n$ is the direct sum of the eigenspaces of $A$.\nEquivalently, the sum of the [[geometric multiplicities|099TWVJ909]] of the eigenvalues is $n$.\n\n## Explanations\n\n### L1 human\n\nTo diagonalise a matrix, we need $n$ independent eigenvectors.\nWe put them as columns into a matrix $P$.\nThen $P^{-1} A P$ is diagonal, with the eigenvalues on the diagonal.\n",
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