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  "body": "\n## Definition\n\nLet $H$ be a complex [[Hilbert space|HM2HHE3J51]].\nLet $T \\colon D(T) \\to H$ be a [[densely defined|BCJHNHKQFG]] linear operator, where the domain $D(T)$ is a dense linear subspace of $H$.\nThe operator $T$ is *symmetric* when\n$$\\ip{Tx}{y} = \\ip{x}{Ty} \\quad \\text{for all } x, y \\in D(T).$$\n\n## Remarks\n\nExample: on $H = L^2(0, 1)$ let $D(T)$ be the smooth functions with compact support in $(0, 1)$ and $Tf = -i f'$.\nFor $f, g \\in D(T)$, integration by parts with vanishing boundary terms gives $\\int_0^1 -i f' \\bar g = \\int_0^1 f \\, \\overline{-i g'}$, so $T$ is symmetric.\n\nNon-example: on the same domain, $Sf = f'$ satisfies $\\ip{Sf}{g} = -\\ip{f}{Sg}$, so $\\ip{Sf}{g} \\ne \\ip{f}{Sg}$ whenever $\\ip{f}{Sg} \\ne 0$; it is skew-symmetric, not symmetric.\n\nAn equivalent form: $T$ is symmetric if and only if its [[adjoint|86B7HF79PQ]] $T^*$ extends $T$, written $T \\subseteq T^*$.\n$T$ is self-adjoint when moreover $T^* = T$, including equality of domains.\nBy the [[Hellinger–Toeplitz theorem|ZXJ7FJ3WCE]], a symmetric operator with $D(T) = H$ is bounded, and then it is self-adjoint.\nSome authors, and Mathlib, call a linear map $T \\colon E \\to E$ on an inner product space symmetric when the identity holds on all of $E$, without a domain.\n",
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