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  "body": "\n## Definition\n\nLet $\\mathbb{K}$ be a field, let $A$ be an associative algebra over $\\mathbb{K}$ with unit $1$, and let $a \\in A$.\nThe *resolvent set* of $a$ is\n$$\\rho(a) = \\set{\\lambda \\in \\mathbb{K} : a - \\lambda 1 \\text{ is invertible in } A} = \\mathbb{K} \\setminus \\sigma(a),$$\nwhere invertible means being a [[unit|0XVEZDE9J8]] of $A$, and $\\sigma(a)$ is the [[spectrum|EH0CZ19RKP]] of $a$.\nFor $\\lambda \\in \\rho(a)$, the element $R(\\lambda) = (a - \\lambda 1)^{-1}$ is the *resolvent* of $a$ at $\\lambda$.\n\nLet $T$ be a closed linear operator with domain $D(T)$ in a complex [[Banach space|06GZ956657]] $X$.\nThe resolvent set of $T$ is the set of $\\lambda \\in \\C$ such that $T - \\lambda$ maps $D(T)$ bijectively onto $X$ with bounded inverse.\n\n## Remarks\n\nSome authors use $(\\lambda 1 - a)^{-1}$ as the resolvent; this changes its sign only.\nThe resolvent satisfies the resolvent identity $R(\\lambda) - R(\\mu) = (\\lambda - \\mu) R(\\lambda) R(\\mu)$ for $\\lambda, \\mu \\in \\rho(a)$.\nIt follows from $R(\\lambda) - R(\\mu) = R(\\lambda) \\big( (a - \\mu 1) - (a - \\lambda 1) \\big) R(\\mu)$.\n\nExample: in $C([0,1])$ with complex values, let $f(t) = t$.\nThen $\\rho(f) = \\C \\setminus [0,1]$, and $R(\\lambda)(t) = 1/(t - \\lambda)$.\n\nNon-example: for the same $f$, the number $\\lambda = 1/2$ is not in $\\rho(f)$, because $f - 1/2$ vanishes at $t = 1/2$ and has no continuous inverse.\n",
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