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Let $A=\left(a_{i j}\right)$ be an $n \times n$ matrix with the property that its absolute row sums are at most 1 , that is, $\left|a_{i 1}\right|+\cdots+\left|a_{\text {in }}\right| \leq 1$ for all $i=1, \ldots, n$. Show that all of its (possibly complex) eigenvalues are in the unit disk: $|\lambda| \leq 1$.
[SUGGESTION: Let $v=\left(v_{1}, \ldots, v_{n}\right) \neq 0$ be an eigenvector and say $v_{k}$ is the largest component, that is, $\left|v_{k}\right|=\max _{j=1, \ldots, n}\left|v_{j}\right|$. Then use the $\mathrm{k}^{\text {th }}$ row of $\lambda v=A v$, that is, $\left.\lambda v_{k}=a_{k 1} v_{1}+\cdots+a_{k n} v_{n}\right] .$

REMARK: This is a special case of: "for any matrix norm, if $\|A\|<1$ then $I-A$ is invertible." However, the proof of this special case can be adapted to give a deeper estimate for the eigenvalues of a matrix.
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