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  "body": "\n## Statement\n\nLet $A$ be a Banach algebra over the [[complex numbers|09FTAET5Z4]] with unit $1 \\ne 0$.\nFor $a \\in A$, let $\\sigma(a) = \\set{\\lambda \\in \\C : a - \\lambda 1 \\text{ is not invertible in } A}$ be the [[spectrum|EH0CZ19RKP]] of $a$.\nThen $\\sigma(a)$ is nonempty for every $a \\in A$.\n\n## Remarks\n\nProof idea: suppose $\\sigma(a)$ is empty.\nThen the resolvent $R(\\lambda) = (a - \\lambda 1)^{-1}$ is defined on all of $\\C$.\nFor each continuous linear functional $\\varphi$ on $A$, the function $\\lambda \\mapsto \\varphi(R(\\lambda))$ is entire and tends to $0$ as $\\abs{\\lambda} \\to \\infty$.\nBy [[Liouville's theorem|0EHKVRF4J6]] it is zero, and by the [[Hahn–Banach theorem|QB5AE3VYZZ]] this forces $R(0) = a^{-1} = 0$, which is impossible since $1 \\ne 0$.\nTogether with the [[compactness of the spectrum|FA5V5KRWDF]], the spectrum of every element is a nonempty compact subset of $\\C$.\n\nExample: in the algebra $C([0,1])$ of continuous complex functions with the sup norm, the spectrum of $f$ is its range $f([0,1])$, which is nonempty.\n\nThe complex scalars are needed.\nIn the real Banach algebra of real $2 \\times 2$ matrices, the rotation $\\begin{pmatrix} 0 \u0026 -1 \\\\ 1 \u0026 0 \\end{pmatrix}$ minus $\\lambda 1$ has determinant $\\lambda^2 + 1 \u003e 0$ for every real $\\lambda$, so its real spectrum is empty.\nThe condition $1 \\ne 0$ is also needed: in the zero algebra every element is invertible.\n\n## Explanations\n\n### L1 human\n\nFor a square complex matrix, the spectrum is the set of eigenvalues, and every such matrix has an eigenvalue.\nThe theorem says the same thing for every element of a complex Banach algebra: there is always at least one number $\\lambda$ for which $a - \\lambda 1$ cannot be inverted.\nThe proof uses complex analysis instead of polynomials.\n\n### L2 human\n\nThis result is the first step of spectral theory in Banach algebras.\nIt gives the [[Gelfand–Mazur theorem|3D6XH41A3B]]: a complex Banach division algebra is $\\C$.\nIt also makes the spectral radius $\\max \\set{\\abs{\\lambda} : \\lambda \\in \\sigma(a)}$ well defined, which is the subject of [[Gelfand's spectral radius formula|T4RFYT8HDN]].\n",
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