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  "body": "\n## Definition\n\nLet $H$ be a complex [[Hilbert space|HM2HHE3J51]], and let $T$ be a bounded operator on $H$ with [[adjoint|6HSJBJ8TTQ]] $T^*$.\nLet $\\abs{T} = (T^* T)^{1/2}$ be the [[square root|3JY5HNFRMF]] of the [[positive operator|77020CY5ZW]] $T^* T$.\nThe operator $T$ is *trace class* if for some [[orthonormal basis|0S9GGTR0EX]] $(e_i)_{i \\in I}$ of $H$,\n$$\\norm{T}_1 = \\sum_{i \\in I} \\ip{\\abs{T} e_i}{e_i} \u003c \\infty.$$\nThe number $\\norm{T}_1$ does not depend on the orthonormal basis; it is the *trace norm* of $T$.\n\n## Remarks\n\nIf $T$ is trace class, then for every orthonormal basis $(e_i)$ the sum $\\operatorname{tr} T = \\sum_i \\ip{T e_i}{e_i}$ converges absolutely, and its value does not depend on the basis.\nOn $\\C^n$, this is the usual [[trace|0XC1RDBY7A]] of a matrix.\nEquivalently, $T$ is trace class if and only if $T$ is compact and its [[singular values|QSJKPKPM7H]] satisfy $\\sum_n s_n(T) \u003c \\infty$, so the trace class is the [[Schatten class|153N5DW55Z]] $S_1$.\nExample: on $\\ell^2$, let $T e_n = e_n / n^2$; then $\\abs{T} = T$ and $\\norm{T}_1 = \\sum 1/n^2 \u003c \\infty$, so $T$ is trace class with $\\operatorname{tr} T = \\pi^2 / 6$.\nNon-example: $D e_n = e_n / n$ gives $\\sum_n \\ip{\\abs{D} e_n}{e_n} = \\sum 1/n = \\infty$, so $D$ is compact but not trace class.\nThe trace class operators form a two-sided ideal of the bounded operators, and $\\operatorname{tr}(AT) = \\operatorname{tr}(TA)$ for $T$ trace class and $A$ bounded.\n",
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