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Suppose we are multiplying matrix $A$ with a diagonal matrix $D$ from left, i.e., $X=D A$ where $D$ is a diagonal matrix with elements $$d_{ii}=\frac{1}{2} \text{ for } i=1,n$$ $$d_{ii}=\frac{1}{3} \text{ for } i =2, \dots, n-1.$$ Is there any relation with the set of eigenvalues of $X$ and $A$ (like any lower bounds on the eigenvalues)? |
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I suppose $A=(a_{ij}) \in M_n(\mathbb C)$. By Gerschgorin's circle theorem the eigenvalues of $A$ lie in the union of the discs $$|z-a_{ii}| \le \sum_{j\neq i} |a_{ij}|\qquad (i=1,...,n)$$ Hence the eigenvalues of $DA$ lie in the union of the discs $$|z-\frac{a_{ii}}{2}| \le \frac{1}{2}\sum_{j\neq i}|a_{ij}|\qquad(i=1,n)$$ $$\qquad\quad |z-\frac{a_{ii}}{3}| \le \frac{1}{3}\sum_{j\neq i}|a_{ij}|\qquad(i=2,\ldots,n-1)$$ |
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