{
  "aliases": null,
  "backlinks": [
    {
      "field": "uses",
      "target": "NT564Y315M"
    }
  ],
  "body": "\n## Statement\n\nLet $A$ be a complex Banach algebra with unit $1 \\ne 0$, and let $a \\in A$.\nLet $\\sigma(a)$ be the [[spectrum|EH0CZ19RKP]] of $a$, and let $r(a) = \\max \\set{\\abs{\\lambda} : \\lambda \\in \\sigma(a)}$ be its [[spectral radius|9V7YQA85PC]].\nThen the [[limit|05M5SWSZZD]] of $\\norm{a^n}^{1/n}$ exists, and\n$$r(a) = \\lim_{n \\to \\infty} \\norm{a^n}^{1/n} = \\inf_{n \\ge 1} \\norm{a^n}^{1/n}.$$\n\n## Remarks\n\nThe maximum exists because $\\sigma(a)$ is compact and, by the [[nonemptiness of the spectrum|QWQC5SV0SD]], nonempty.\nLower bound: by the [[spectral mapping theorem for polynomials|Q71WX8HFYT]], $\\lambda^n \\in \\sigma(a^n)$ for $\\lambda \\in \\sigma(a)$, and every $\\mu \\in \\sigma(b)$ satisfies $\\abs{\\mu} \\le \\norm{b}$; so $r(a)^n \\le \\norm{a^n}$ for every $n$.\nUpper bound: for $\\abs{\\lambda} \u003e r(a)$ the resolvent $(\\lambda 1 - a)^{-1}$ is holomorphic, and its Laurent series $\\sum_{n \\ge 0} a^n \\lambda^{-n-1}$ converges there; applying functionals and the uniform boundedness principle gives $\\limsup_n \\norm{a^n}^{1/n} \\le r(a)$.\n\nExample: on $C([0,1])$ with the sup norm, $\\norm{f^n} = \\norm{f}^n$, so the formula gives $r(f) = \\norm{f} = \\max \\abs{f}$, the largest absolute value of a point of the range.\nFor a nilpotent element, $a^n = 0$ for large $n$, so $r(a) = 0$ and $\\sigma(a) = \\set{0}$, even if $\\norm{a}$ is large.\nFor matrices this is [[Gelfand's formula|3KX0KQ7V05]].\n\n## Explanations\n\n### L1 human\n\nThe spectral radius measures the largest size of a point of the spectrum.\nThe norm of $a$ may be bigger than this, but the norms of the powers $a^n$ grow at the exponential rate $r(a)^n$.\nTaking $n$-th roots and letting $n$ grow recovers $r(a)$ from norms alone.\n\n### L2 human\n\nThe formula shows that $r(a)$ depends only on the norms of the powers of $a$.\nSo it does not change when the spectrum is computed in a closed subalgebra containing $a$ and $1$.\nIn a C*-algebra, the C* identity gives $\\norm{a^{2^k}} = \\norm{a}^{2^k}$ for self-adjoint $a$, so $r(a) = \\norm{a}$; this is a key step in the commutative Gelfand–Naimark theorem.\nIt also shows that the Gelfand transform satisfies $\\norm{\\hat a}_\\infty = r(a)$ in a commutative unital Banach algebra.\n",
  "commit": "5fc425f97f1f6b11da5892c2948ef8c5d8805665",
  "front": {
    "about": [
      "9V7YQA85PC"
    ],
    "kind": "statement",
    "msc": [
      "46H05",
      "47A10"
    ],
    "name": "Gelfand's spectral radius formula",
    "role": "theorem",
    "sources": [
      {
        "url": "https://en.wikipedia.org/wiki/Spectral_radius"
      }
    ],
    "tag": "T4RFYT8HDN",
    "uses": [
      "EH0CZ19RKP",
      "QWQC5SV0SD",
      "Q71WX8HFYT",
      "05M5SWSZZD"
    ]
  },
  "kind": "statement",
  "link": "https://math2.tamnd.com/node/T4RFYT8HDN/",
  "links": [
    {
      "field": "about",
      "target": "9V7YQA85PC"
    },
    {
      "field": "uses",
      "target": "EH0CZ19RKP"
    },
    {
      "field": "uses",
      "target": "QWQC5SV0SD"
    },
    {
      "field": "uses",
      "target": "Q71WX8HFYT"
    },
    {
      "field": "uses",
      "target": "05M5SWSZZD"
    }
  ],
  "markdown": "https://math2.tamnd.com/node/T4RFYT8HDN.md",
  "msc": [
    "46H05",
    "47A10"
  ],
  "name": "Gelfand's spectral radius formula",
  "prov": null,
  "role": "theorem",
  "source": "https://github.com/tamnd/math2/blob/main/nodes/T4R/T4RFYT8HDN.md",
  "status": null,
  "tag": "T4RFYT8HDN",
  "uri": "m2://T4RFYT8HDN",
  "url": "https://math2.tamnd.com/wiki/gelfands-spectral-radius-formula/",
  "writing": {
    "tag": "T4RFYT8HDN",
    "kind": "statement",
    "level": "W0",
    "next": [
      {
        "what": "an inventory row",
        "note": "the row of the inventory that the node accounts for",
        "have": 0,
        "want": 1
      },
      {
        "what": "a source with a locator",
        "have": 0,
        "want": 1
      }
    ]
  }
}