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  "body": "\n## Statement\n\nLet $n \\ge 1$ and let $K$ be an $n \\times n$ matrix of functions in $L^1(\\R)$.\nWrite $\\hat K(\\xi) = \\int_{\\R} K(x)\\, e^{i\\xi x}\\, dx$ entrywise, the [[Fourier transform|DPVM34ADZV]], and let $A(\\xi) = I - \\hat K(\\xi)$.\nAssume that $\\det A(\\xi) \\ne 0$ for all $\\xi \\in \\R$, so that $A(\\xi)$ is an [[invertible matrix|0EB9N0S33C]] for every real $\\xi$.\nLet $r(\\xi) = (\\xi - i)/(\\xi + i)$.\nThen:\n\n1. There are $n \\times n$ matrices $K_+$ and $K_-$ of $L^1(\\R)$ functions, with $K_+ = 0$ on $(-\\infty, 0)$ and $K_- = 0$ on $(0, \\infty)$. There are integers $\\kappa_1 \\ge \\kappa_2 \\ge \\cdots \\ge \\kappa_n$ such that\n$$A(\\xi) = \\bigl(I - \\hat K_-(\\xi)\\bigr)\\, \\operatorname{diag}\\bigl(r(\\xi)^{\\kappa_1}, \\ldots, r(\\xi)^{\\kappa_n}\\bigr)\\, \\bigl(I - \\hat K_+(\\xi)\\bigr), \\quad \\xi \\in \\R.$$\nHere $\\det(I - \\hat K_+)$ has no zeros in the closed upper half-plane and $\\det(I - \\hat K_-)$ has no zeros in the closed lower half-plane.\n2. The integers $\\kappa_1, \\ldots, \\kappa_n$, the *partial indices*, are uniquely determined by $A$, and their sum is the [[winding number|0X7SBS8E0F]] of $\\xi \\mapsto \\det A(\\xi)$ about $0$.\n3. For $1 \\le p \u003c \\infty$, the operator $\\varphi \\mapsto \\varphi(x) - \\int_0^\\infty K(x - t)\\, \\varphi(t)\\, dt$ on $L^p(0, \\infty)^n$ is invertible if and only if $\\kappa_1 = \\cdots = \\kappa_n = 0$.\n\n## Remarks\n\nThe factors themselves are not unique, but the partial indices are.\nFor $n = 1$ this is the scalar [[Wiener–Hopf factorization|BYKP1V2AQN]], with the single index $\\kappa_1$ equal to the winding number.\nThe proof cannot take logarithms, because matrices do not commute; one instead reduces to a matrix Riemann–Hilbert problem or works in the algebra of matrix functions with $L^1$ Fourier data.\n\nThe total index alone does not decide invertibility.\nExample: $A = \\operatorname{diag}\\bigl(r, r^{-1}\\bigr)$ has $\\det A = 1$, so the winding number of the determinant is $0$.\nIts partial indices are $1$ and $-1$, so the corresponding system of [[Wiener–Hopf equations|QYY4NMDY5H]] is not invertible.\nThe first equation has a range of codimension $1$ and the second a kernel of dimension $1$.\nEach entry is admissible, since $r(\\xi) = 1 - \\frac{2i}{\\xi + i}$ and $\\frac{2i}{\\xi + i}$ is the Fourier transform of the function equal to $2e^{-x}$ for $x \u003e 0$ and to $0$ for $x \\le 0$.\nThe inverse $r^{-1} = 1 + \\frac{2i}{\\xi - i}$ is admissible in the same way, with $\\frac{2i}{\\xi - i}$ the transform of a function supported on $(-\\infty, 0]$.\n\n## Explanations\n\n### L1 human\n\nA system of Wiener–Hopf equations has a matrix symbol instead of a scalar one.\nThe symbol can still be split into a forward-looking part, a backward-looking part and a simple diagonal middle part.\nThe middle part is described by a list of whole numbers, the partial indices, and these decide whether the system has exactly one solution.\nUnlike the scalar case, the winding number of the determinant is not enough: the indices can cancel in the sum.\n\n### L2 human\n\nThe partial indices are the matrix version of the index of a Fredholm operator, but they carry more information than the index.\nThey are unstable: a small perturbation of $A$ can change them, while their sum stays the same.\nThe same factorization governs the matrix [[Riemann–Hilbert problem|RY8YQ9JZF9]] on a closed contour, with the circle in place of the real line.\nIn applications, such as diffraction by a wedge or elasticity with mixed conditions, finding the factors explicitly is the main difficulty, and only special classes of matrices admit closed forms.\n",
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