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  "body": "\n## Definition\n\nLet $X$ and $Y$ be [[normed spaces|03J5RAKB30]] over $\\R$ or $\\C$, and let $T \\colon X \\to Y$ be a [[continuous linear map|03TM135G0N]].\nThe operator $T$ is *invertible* if there is a continuous linear map $S \\colon Y \\to X$ such that\n$$S T = \\id_X \\quad \\text{and} \\quad T S = \\id_Y.$$\nThe map $S$ is then unique; it is the *inverse* of $T$, written $T^{-1}$.\n\n## Remarks\n\nSo $T$ is an [[inverse element|0ZNZH0EXFX]] for composition, with the inverse required to be bounded.\n\nExample: on $\\ell^2$, the diagonal operator $D(x_n) = ((1 + 1/n) x_n)$ is invertible.\nIts inverse is $D^{-1}(x_n) = (\\tfrac{n}{n+1} x_n)$, which is bounded with norm at most $1$, and both compositions are the identity.\n\nNon-example: the diagonal operator $E(x_n) = (x_n / n)$ on $\\ell^2$ is injective, but it is not invertible.\nIt is not surjective, since $(1/n)$ lies in $\\ell^2$ but $(1)$ does not, so $(1/n)$ is not in the image of $E$.\n\nIf $X$ and $Y$ are [[Banach spaces|06GZ956657]], every bijective bounded operator is invertible, by the [[bounded inverse theorem|MR5FJXR348]].\nFor an operator $T$ on one space $X$, invertibility means invertibility in the algebra of bounded operators on $X$.\n",
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