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  "body": "\n## Statement\n\nLet $A = (a_{ij})$ be an $n \\times n$ matrix over $\\C$.\nFor $1 \\le i \\le n$, let $R_i = \\sum_{j \\ne i} \\abs{a_{ij}}$, and let $D_i = \\{z \\in \\C : \\abs{z - a_{ii}} \\le R_i\\}$ be the closed disc with centre $a_{ii}$ and radius $R_i$.\nThen each [[eigenvalue|0BQAA2BK3Q]] of $A$ lies in at least one of the discs $D_1, \\dots, D_n$.\n\n## Remarks\n\nThe discs are called *Gershgorin discs*.\nThe proof takes an eigenvector $x$ and an index $i$ with $\\abs{x_i}$ maximal, and estimates row $i$ of $Ax = \\lambda x$.\nSince $A$ and $A^T$ have the same eigenvalues, the same holds with column sums.\nA strictly diagonally dominant matrix, with $\\abs{a_{ii}} \u003e R_i$ for all $i$, has no eigenvalue $0$, so it is invertible.\n\n## Explanations\n\n### L1 human\n\nThe eigenvalues of a matrix lie near its diagonal entries, when the other entries are small.\nEach row gives a disc around its diagonal entry, with the radius equal to the sum of the other absolute values of the row.\nEvery eigenvalue lies in one of these discs.\n",
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