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  "body": "\n## Statement\n\nLet $X \\ne \\set{0}$ be a complex [[Banach space|06GZ956657]], and let $T \\colon X \\to X$ be a [[continuous linear map|03TM135G0N]] with [[spectrum|BMKATJ802A]] $\\sigma(T)$.\nThen $\\sigma(T)$ is nonempty and compact, and its *spectral radius* $r(T) = \\max \\set{\\abs{\\lambda} : \\lambda \\in \\sigma(T)}$ satisfies\n$$r(T) = \\lim_{n \\to \\infty} \\norm{T^n}^{1/n} = \\inf_{n \\ge 1} \\norm{T^n}^{1/n},$$\nwhere $\\norm{\\cdot}$ is the [[operator norm|06EKW1AKXF]].\n\n## Remarks\n\nProof idea: if $\\lambda \\in \\sigma(T)$, then $\\lambda^n \\in \\sigma(T^n)$, and the [[Neumann series|8QJ91709V2]] shows $\\abs{\\lambda}^n \\le \\norm{T^n}$; this gives $r(T) \\le \\inf_n \\norm{T^n}^{1/n}$.\nFor $\\abs{\\lambda} \u003e \\norm{T}$, the [[resolvent|J5YNNKSB9P]] has the expansion $R(\\lambda, T) = -\\sum_{n \\ge 0} T^n / \\lambda^{n+1}$.\nThe resolvent is analytic on $\\abs{\\lambda} \u003e r(T)$, so this Laurent series converges there, and the uniform boundedness principle with the Cauchy estimates gives $\\limsup_n \\norm{T^n}^{1/n} \\le r(T)$.\nNonemptiness of $\\sigma(T)$ follows from Liouville's theorem applied to $\\lambda \\mapsto f(R(\\lambda, T))$ for continuous linear functionals $f$.\n\nExample: the Volterra operator $(Vf)(x) = \\int_0^x f(t) \\, dt$ on $C[0, 1]$ with the supremum norm satisfies $\\norm{V^n} = 1/n!$, attained at $f = 1$.\nSo $r(V) = \\lim (1/n!)^{1/n} = 0$ and $\\sigma(V) = \\set{0}$, although $V \\ne 0$.\n\nThe field must be $\\C$.\nThe rotation $T(x, y) = (-y, x)$ of the real plane has $\\norm{T^n} = 1$ for all $n$, but $T - \\lambda I$ is invertible for every real $\\lambda$, so its real spectrum is empty.\n\nFor matrices this is [[Gelfand's formula|3KX0KQ7V05]] for the [[spectral radius|9V7YQA85PC]].\nThe same statement, with the same proof, holds for every element of a unital complex Banach algebra.\n\n## Explanations\n\n### L1 human\n\nThe spectral radius is the size of the largest point in the spectrum.\nThe formula says it can be read off from the norms of the powers $T^n$: these grow, or shrink, like $r(T)^n$.\nSo the long-run growth rate of $T^n$ is fixed by the spectrum, even though a single norm $\\norm{T}$ can be much larger than $r(T)$.\n\n### L2 human\n\nThe formula shows that $T^n \\to 0$ in norm if and only if $r(T) \u003c 1$, and that $\\sum T^n$ converges in that case, which sharpens the Neumann series.\nIt also shows that $r(T)$ depends only on the algebra, not on the choice of an equivalent norm.\nA quasinilpotent operator, with $r(T) = 0$, need not be nilpotent, as the Volterra operator shows.\nFor normal operators on a Hilbert space, $\\norm{T^{2^k}} = \\norm{T}^{2^k}$, so $r(T) = \\norm{T}$; this is a key step in the theory of C*-algebras.\n",
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