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  "body": "\n## Statement\n\nLet $T = (t_{i-j})_{1 \\le i, j \\le N}$ be an $N \\times N$ [[Toeplitz matrix|GKY2WSHE26]] over $\\mathbb{K}$, where $\\mathbb{K}$ is $\\R$ or $\\C$, and let $y \\in \\mathbb{K}^N$.\nFor $n \\le N$, let $T^n$ be the leading $n \\times n$ submatrix of $T$, and assume that every $T^n$ is [[invertible|0EB9N0S33C]].\nLet $f^n, b^n, x^n \\in \\mathbb{K}^n$ solve $T^n f^n = e_1$, $T^n b^n = e_n$ and $T^n x^n = (y_1, \\dots, y_n)^T$, where $e_1, e_n$ are the first and last unit vectors.\nPut\n$$\\varepsilon_f^n = \\sum_{j=1}^n t_{n+1-j} f^n_j, \\qquad \\varepsilon_b^n = \\sum_{j=1}^n t_{-j} b^n_j, \\qquad \\varepsilon_x^n = \\sum_{j=1}^n t_{n+1-j} x^n_j.$$\nThen $1 - \\varepsilon_f^n \\varepsilon_b^n \\ne 0$, and\n$$f^{n+1} = \\frac{1}{1 - \\varepsilon_f^n \\varepsilon_b^n} \\left( \\begin{pmatrix} f^n \\\\ 0 \\end{pmatrix} - \\varepsilon_f^n \\begin{pmatrix} 0 \\\\ b^n \\end{pmatrix} \\right), \\qquad b^{n+1} = \\frac{1}{1 - \\varepsilon_f^n \\varepsilon_b^n} \\left( \\begin{pmatrix} 0 \\\\ b^n \\end{pmatrix} - \\varepsilon_b^n \\begin{pmatrix} f^n \\\\ 0 \\end{pmatrix} \\right),$$\n$$x^{n+1} = \\begin{pmatrix} x^n \\\\ 0 \\end{pmatrix} + (y_{n+1} - \\varepsilon_x^n) \\, b^{n+1}.$$\nStarting from $f^1 = b^1 = 1/t_0$ and $x^1 = y_1 / t_0$, the *Levinson recursion* computes the solution $x^N$ of $T x = y$ in $O(N^2)$ arithmetic operations.\n\n## Remarks\n\nThe key fact is that both the leading and the trailing $n \\times n$ blocks of $T^{n+1}$ equal $T^n$.\nSo $T^{n+1} (f^n, 0)^T = (1, 0, \\dots, 0, \\varepsilon_f^n)^T$ and $T^{n+1} (0, b^n)^T = (\\varepsilon_b^n, 0, \\dots, 0, 1)^T$, and the stated combinations give $e_1$ and $e_{n+1}$.\nSimilarly $T^{n+1} (x^n, 0)^T = (y_1, \\dots, y_n, \\varepsilon_x^n)^T$, and adding $(y_{n+1} - \\varepsilon_x^n) b^{n+1}$ fixes the last entry.\nIf $1 - \\varepsilon_f^n \\varepsilon_b^n$ were $0$, the nonzero vector $(f^n, 0)^T - \\varepsilon_f^n (0, b^n)^T$ would lie in the kernel of $T^{n+1}$.\nThis vector is nonzero because its last entry is $-\\varepsilon_f^n b^n_n$, with $\\varepsilon_f^n \\ne 0$ and $b^n_n = \\det T^{n-1} / \\det T^n \\ne 0$ by Cramer's rule.\nEach step costs $O(n)$ operations, which gives the total $O(N^2)$, against $O(N^3)$ for Gaussian elimination.\nThe hypothesis on all leading [[minors|0BT9BYWVRY]] is needed: for $T = \\begin{pmatrix} 0 \u0026 1 \\\\ 1 \u0026 0 \\end{pmatrix}$ the matrix is invertible but $t_0 = 0$, and the recursion cannot start.\nFor symmetric positive definite $T$ every leading submatrix is invertible, and the special case of the [[Yule–Walker equations|GA46NC0CJZ]] is the Levinson–Durbin recursion.\n\n## Explanations\n\n### L1 human\n\nA Toeplitz matrix is constant along each diagonal, so it is described by only $2N - 1$ numbers.\nThe Levinson recursion uses this structure to solve a Toeplitz system by growing the solution one size at a time.\nEach growth step reuses the previous answer and costs only a few vector operations.\n\n### L2 human\n\nToeplitz systems arise from stationary time series, in linear prediction, and in signal processing, where $T$ is a matrix of autocovariances.\nThe recursion solves them in $O(N^2)$ time and $O(N)$ memory, which made it a standard tool in speech coding.\nIt can be unstable when some leading submatrix is nearly singular, even if $T$ itself is well conditioned; superfast $O(N \\log^2 N)$ algorithms also exist.\n",
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