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  "body": "\n## Definition\n\nLet $X$ be a complex [[Banach space|06GZ956657]], let $I$ be its identity map, and let $T \\colon X \\to X$ be a [[continuous linear map|03TM135G0N]].\nThe *spectrum* of $T$ is the set\n$$\\sigma(T) = \\set{\\lambda \\in \\C : T - \\lambda I \\text{ is not invertible}},$$\nwhere invertible means [[invertible as a bounded operator|CPAD5W1CGW]].\n\n## Remarks\n\nBy the bounded inverse theorem, $\\lambda \\in \\sigma(T)$ if and only if $T - \\lambda I$ is not bijective.\nIf $X \\ne \\set{0}$, then $\\sigma(T)$ is a nonempty compact subset of the closed disc $\\set{\\lambda : \\abs{\\lambda} \\le \\norm{T}}$.\nFor $X = \\C^n$, $\\sigma(T)$ is the [[spectrum of the matrix|0XPX5NC67E]] of $T$, its set of eigenvalues.\n\nExample: let $D(x_n) = (x_n / n)$ on $\\ell^2$.\nFor $\\lambda \\notin \\set{1/n : n \\ge 1} \\cup \\set{0}$, the numbers $1/(1/n - \\lambda)$ are bounded, so the diagonal operator with these entries is a bounded inverse of $D - \\lambda I$.\nEach $1/n$ is an eigenvalue, and $D$ is not surjective, so $\\sigma(D) = \\set{1/n : n \\ge 1} \\cup \\set{0}$.\n\nNon-example: the number $2$ is not in $\\sigma(D)$, since $\\abs{1/(1/n - 2)} \\le 1$ for all $n$, so $D - 2I$ has a bounded inverse.\n\nFor a [[closed operator|VM7B38PA87]] $T$ with domain $D(T)$, $\\lambda$ is outside $\\sigma(T)$ when $T - \\lambda I \\colon D(T) \\to X$ is bijective with bounded inverse.\nThe spectrum is then closed, but it can be empty or unbounded.\n",
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