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  "body": "\n## Proof\n\nWe look at the row of $A x = \\lambda x$ where the eigenvector $x$ has its largest coordinate.\n\n1. Let $\\lambda$ be an [[eigenvalue|0ZND7NQ970]] of $A$. By definition, there is $x = (x_1, \\dots, x_n) \\in \\C^n$ with $x \\ne 0$ and $A x = \\lambda x$.\n2. Choose an index $i$ with $\\abs{x_i} \\ge \\abs{x_j}$ for all $j$. Since $x \\ne 0$, some $x_j \\ne 0$, so $\\abs{x_i} \u003e 0$.\n3. By the definition of the [[matrix product|076JGP6R3T]], coordinate $i$ of $A x = \\lambda x$ reads $\\sum_{j=1}^{n} a_{ij} x_j = \\lambda x_i$.\n4. Subtract $a_{ii} x_i$ from both sides. This gives $(\\lambda - a_{ii}) x_i = \\sum_{j \\ne i} a_{ij} x_j$.\n5. Take [[absolute values|00S7C68D51]]. By the [[multiplicativity of the absolute value|0GMV7GP8W0]] and the [[triangle inequality|03HF5PMY5T]], $\\abs{\\lambda - a_{ii}} \\abs{x_i} \\le \\sum_{j \\ne i} \\abs{a_{ij}} \\abs{x_j}$.\n6. By the choice of $i$ in step 2, $\\abs{x_j} \\le \\abs{x_i}$ for all $j$. So the right side of step 5 is at most $\\sum_{j \\ne i} \\abs{a_{ij}} \\abs{x_i} = R_i \\abs{x_i}$.\n7. Divide by $\\abs{x_i} \u003e 0$. This gives $\\abs{\\lambda - a_{ii}} \\le R_i$.\n8. So $\\lambda \\in D_i$. Since $\\lambda$ was any eigenvalue, each eigenvalue of $A$ lies in at least one of the discs $D_1, \\dots, D_n$.\n\n## Idea\n\nTake an eigenvector and its coordinate of largest size. In that row, the diagonal term must balance the other terms of the row. These other terms are at most $R_i$ times the largest coordinate, so $\\lambda$ is within $R_i$ of $a_{ii}$.\n",
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