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  "body": "\n## Statement\n\nLet $A$ be a real symmetric [[positive definite|0KX66CZMBB]] $n \\times n$ matrix, let $b \\in \\R^n$, and let $\\omega \u003e 0$.\nThe *modified Richardson iteration* for the [[linear system|0GT4WWGWH2]] $A x = b$ starts from any $x^{(0)} \\in \\R^n$ and sets\n$$x^{(k+1)} = x^{(k)} + \\omega \\left( b - A x^{(k)} \\right).$$\nLet $\\lambda_{\\min}$ and $\\lambda_{\\max}$ be the least and greatest [[eigenvalues|0BQAA2BK3Q]] of $A$.\n\n1. The iterates converge to $A^{-1} b$ for every $b$ and every $x^{(0)}$ if and only if $0 \u003c \\omega \u003c 2 / \\lambda_{\\max}$.\n2. The [[spectral radius|9V7YQA85PC]] of the iteration matrix $I - \\omega A$ is least for $\\omega = 2 / (\\lambda_{\\min} + \\lambda_{\\max})$. Its value there is $(\\kappa - 1) / (\\kappa + 1)$, where $\\kappa = \\lambda_{\\max} / \\lambda_{\\min}$.\n\n## Remarks\n\nThe error satisfies $e_{k+1} = (I - \\omega A) e_k$, and $I - \\omega A$ has the eigenvalues $1 - \\omega \\lambda$ for the eigenvalues $\\lambda$ of $A$.\nAll of them lie in $(-1, 1)$ exactly when $0 \u003c \\omega \\lambda_{\\max} \u003c 2$, which gives clause 1.\nThe largest of $\\abs{1 - \\omega \\lambda_{\\min}}$ and $\\abs{1 - \\omega \\lambda_{\\max}}$ is least when the two are equal, which gives clause 2.\nExample: if $A$ has eigenvalues $1$ and $3$, the best choice is $\\omega = 1/2$, and the error shrinks by the factor $1/2$ at each step.\nThe number $\\kappa$ is the [[condition number|AB3S6MHJ8X]] of $A$, so the method is slow when $A$ is ill-conditioned.\n\n## Explanations\n\n### L1 human\n\nThe residual $b - A x$ tells how far the current guess is from solving the system.\nRichardson's method moves the guess a fixed fraction $\\omega$ of the residual and repeats.\nThe step must not be too large, or the guess overshoots and the error grows.\n\n### L2 human\n\nRichardson iteration is the simplest stationary [[iterative method|2RPE77ACER]], and the Jacobi method is Richardson iteration applied to the system scaled by the diagonal.\nWith a [[preconditioner|KE98SW3SBD]] $P$ it becomes $x^{(k+1)} = x^{(k)} + \\omega P^{-1} (b - A x^{(k)})$, and the same analysis applies to $P^{-1} A$.\nFor a symmetric positive definite $A$ it is gradient descent with a fixed step on $\\frac12 x^T A x - b^T x$.\nLetting $\\omega$ vary with $k$ and choosing the steps as reciprocal Chebyshev nodes leads to Chebyshev iteration.\n",
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