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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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What is the reason for your interest in this problem? Why this particular form for D? I don't see how the choice of indices i=1 and n for the 1/2 elements can be relevant. Why not i=1 and 2? Why exactly two such indices? Why the values 1/2 and 1/3? What have you done to try to attack the problem? –  Ben Crowell Jan 27 '13 at 18:37

1 Answer 1

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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