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  "body": "\n## Statement\n\nLet $A$ be a complex Banach algebra with unit $1 \\ne 0$, let $a \\in A$, and let $p \\in \\C[z]$ be a [[polynomial|0SDC934R52]].\nLet $\\sigma(b)$ denote the [[spectrum|EH0CZ19RKP]] of $b \\in A$.\nThen\n$$\\sigma(p(a)) = p(\\sigma(a)) = \\set{p(\\lambda) : \\lambda \\in \\sigma(a)}.$$\n\n## Remarks\n\nProof idea: if $p$ is constant equal to $c$, both sides are $\\set{c}$, using the [[nonemptiness of the spectrum|QWQC5SV0SD]].\nOtherwise fix $\\mu \\in \\C$ and factor $p(z) - \\mu = c (z - \\lambda_1) \\cdots (z - \\lambda_k)$ with $c \\ne 0$.\nThen $p(a) - \\mu 1 = c (a - \\lambda_1 1) \\cdots (a - \\lambda_k 1)$, a product of commuting factors.\nA product of commuting elements is invertible if and only if each factor is invertible.\nSo $\\mu \\in \\sigma(p(a))$ if and only if some root $\\lambda_i$ of $p - \\mu$ lies in $\\sigma(a)$, that is, if and only if $\\mu = p(\\lambda)$ for some $\\lambda \\in \\sigma(a)$.\n\nExample: if $a$ is a $2 \\times 2$ complex matrix with eigenvalues $1$ and $-1$, then $a^2$ has the single eigenvalue $1$, matching $p(\\sigma(a)) = \\set{1}$ for $p(z) = z^2$.\n\n## Explanations\n\n### L1 human\n\nIf you know the spectrum of $a$, you know the spectrum of every polynomial in $a$: just apply the polynomial to each point.\nFor matrices this is the familiar fact that $p(A)$ has eigenvalues $p(\\lambda)$ for the eigenvalues $\\lambda$ of $A$.\n\n### L2 human\n\nThe theorem gives $\\sigma(a^n) = \\set{\\lambda^n : \\lambda \\in \\sigma(a)}$, which is the easy half of [[Gelfand's spectral radius formula|T4RFYT8HDN]].\nIt extends to holomorphic functions in the holomorphic functional calculus, and to continuous functions of normal elements in the [[continuous functional calculus|JB7S6WPZ77]] of C*-algebras.\n",
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