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  "body": "\n## Statement\n\nLet $K$ be a commutative ring, let $n \\ge 1$, and let $A$ be an $n \\times n$ matrix over $K$ with [[characteristic polynomial|0VSSPAFM2C]] $p_A(t) = \\det(t I_n - A)$.\nThen\n$$p_A(A) = 0,$$\nwhere $p_A(A)$ is the [[value of $p_A$ at $A$|0E3MV6E6MG]].\n\n## Remarks\n\nFor $n = 2$, it says $A^2 - (\\tr A) A + (\\det A) I_2 = 0$.\nSubstituting $t = A$ into $\\det(t I_n - A)$ is not a valid proof, because $p_A(A)$ is a matrix and $\\det(A - A)$ is a scalar.\nOne proof uses the [[adjugate|07W9H8QF38]] identity $(t I_n - A) \\operatorname{adj}(t I_n - A) = p_A(t) I_n$.\nIf $A$ is invertible, then the theorem expresses $A^{-1}$ as a polynomial in $A$.\n\n## Explanations\n\n### L1 human\n\nEvery square matrix satisfies its own characteristic polynomial.\nFor $A = \\begin{pmatrix} 1 \u0026 2 \\\\ 3 \u0026 4 \\end{pmatrix}$, the polynomial is $t^2 - 5t - 2$, so $A^2 = 5A + 2I$.\nSo every power of $A$ is a combination of $I$ and $A$.\n",
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  "name": "Cayley–Hamilton theorem",
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