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  "body": "\n## Definition\n\nLet $a \u003c b$ and let $K \\in L^2([a, b] \\times [a, b])$ be a [[square-integrable kernel|WWMWX7AD4V]].\nA number $\\lambda \\in \\C$ is a *characteristic value* of $K$ if the [[homogeneous equation|W6CYS54QXS]]\n$$\\varphi(x) = \\lambda \\int_a^b K(x, y) \\, \\varphi(y) \\, dy$$\nhas a solution $\\varphi \\in L^2[a, b]$ that is not $0$ almost everywhere.\n\n## Remarks\n\nA characteristic value is never $0$, since for $\\lambda = 0$ the equation forces $\\varphi = 0$.\nSo $\\lambda$ is a characteristic value exactly when $1/\\lambda$ is an eigenvalue of the integral operator $T_K$, and the nonzero solutions $\\varphi$ are the [[eigenfunctions|4NSACSWRJZ]] for $1/\\lambda$.\n\nExample: for $K(x, y) = xy$ on $[0, 1]$, the number $\\lambda = 3$ is a characteristic value with solution $\\varphi(x) = x$, since $3 \\int_0^1 x y \\cdot y \\, dy = 3 \\cdot x/3 = x$.\n\nNon-example: for the same kernel, $\\lambda = 1$ is not a characteristic value.\nEvery solution has the form $\\varphi(x) = c x$ with $c = \\int_0^1 y \\varphi(y) \\, dy = c/3$, so $c = 0$ and $\\varphi = 0$.\n\nMany classical texts call $\\lambda$ an eigenvalue of the kernel; the name characteristic value avoids confusion with the eigenvalue $1/\\lambda$ of the operator.\nA kernel may have no characteristic value at all: the Volterra kernel, equal to $1$ for $y \\le x$ and $0$ otherwise, is an example.\n",
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