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  "body": "\n## Summary\n\nThe trace of an integral operator with a continuous kernel $K(s, t)$ on $[a, b]$ is the integral of the kernel along the diagonal, $\\int_a^b K(s, s) \\, ds$.\nIt is the continuous analogue of the sum of the diagonal entries of a matrix.\nIn Fredholm theory, minus the trace is the coefficient of $\\lambda$ in the Fredholm determinant $D(\\lambda) = 1 - \\lambda \\int_a^b K(s, s) \\, ds + \\cdots$.\nFor a [[trace class operator|Z2CP530WDE]] with continuous kernel, it equals the operator trace and the sum of the [[eigenvalues|0BQAA2BK3Q]].\n\n## Explanations\n\n### L1 human\n\nThink of the kernel $K(s, t)$ as a matrix with a continuum of rows $s$ and columns $t$.\nThe diagonal of this matrix is the set of points with $s = t$.\nSumming the diagonal entries becomes integrating $K(s, s)$ over $s$.\nFor example, the kernel $K(s, t) = st$ on $[0, 1]$ has trace $\\int_0^1 s^2 \\, ds = 1/3$.\nThis operator sends $x$ to $s \\mapsto s \\int_0^1 t x(t) \\, dt$, and its only nonzero eigenvalue is also $1/3$, with eigenfunction $s$.\n",
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