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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.
in Mathematics by Platinum (164,236 points) | 243 views

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MathsGee Android Q&A

MathsGee Android Q&A