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  "body": "\n## Statement\n\nLet $A = (a_{ij})$ be an $n \\times n$ complex matrix that is strictly [[diagonally dominant|89P5GTKW87]] by rows:\n$$\\abs{a_{ii}} \u003e \\sum_{j \\ne i} \\abs{a_{ij}} \\quad \\text{for } i = 1, \\dots, n.$$\nThen $A$ is [[invertible|0EB9N0S33C]], that is, $\\det A \\ne 0$.\n\n## Remarks\n\nSuppose $Ax = 0$ with $x \\ne 0$, and choose $i$ with $\\abs{x_i}$ maximal, so $\\abs{x_i} \u003e 0$.\nRow $i$ gives $a_{ii} x_i = -\\sum_{j \\ne i} a_{ij} x_j$, so $\\abs{a_{ii}} \\abs{x_i} \\le \\sum_{j \\ne i} \\abs{a_{ij}} \\abs{x_i}$.\nThis contradicts strict dominance in row $i$.\nEquivalently, $0$ lies in no disc of the [[Gershgorin circle theorem|0ETGP0XVCJ]].\n\nExample: $\\begin{pmatrix} 3 \u0026 1 \u0026 1 \\\\ 0 \u0026 2 \u0026 1 \\\\ 1 \u0026 -1 \u0026 4 \\end{pmatrix}$ is strictly diagonally dominant, so it is invertible without any computation.\n\nStrictness is necessary: $\\begin{pmatrix} 1 \u0026 1 \\\\ 1 \u0026 1 \\end{pmatrix}$ satisfies $\\abs{a_{ii}} \\ge \\sum_{j \\ne i} \\abs{a_{ij}}$ with equality, and it is singular.\nThe theorem also holds for strict dominance by columns, by applying it to $A^T$.\n\n## Explanations\n\n### L1 human\n\nIf each diagonal entry is larger in absolute value than all other entries of its row together, the matrix cannot send a nonzero vector to zero.\nThe diagonal wins against the rest of the row, so it can never be cancelled.\n\n### L2 human\n\nThe theorem gives a cheap test for invertibility, used for the matrices that arise when differential equations are discretised.\nFor strictly diagonally dominant matrices, the Jacobi and Gauss–Seidel iterations for $Ax = b$ converge.\nIt is the special case \"$0$ is not an eigenvalue\" of the Gershgorin circle theorem.\n",
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