most recent 30 from http://mathoverflow.net 2013-06-19T05:41:19Z http://mathoverflow.net/feeds/question/120014 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/120014/eigenvalues-of-a-diagonal-matrix-times-a-matrix Jason Spoon 2013-01-27T11:36:07Z 2013-01-27T22:50:17Z

<p>Suppose we are multiplying matrix $A$ with a diagonal matrix $D$ from left, i.e.,</p> <p>$X=D A$</p> <p>where $D$ is a diagonal matrix with elements </p> <p>$$d_{ii}=\frac{1}{2} \text{ for } i=1,n$$</p> <p>$$d_{ii}=\frac{1}{3} \text{ for } i =2, \dots, n-1.$$</p> <p>Is there any relation with the set of eigenvalues of $X$ and $A$ (like any lower bounds on the eigenvalues)?</p>

http://mathoverflow.net/questions/120014/eigenvalues-of-a-diagonal-matrix-times-a-matrix/120064#120064 Ralph 2013-01-27T22:50:17Z 2013-01-27T22:50:17Z

<p>I suppose $A=(a_{ij}) \in M_n(\mathbb C)$. By <a href="http://en.wikipedia.org/wiki/Gershgorin_circle_theorem" rel="nofollow">Gerschgorin's circle theorem</a> 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 $$|z-\frac{a_{ii}}{2}| \le \frac{1}{2}\sum_{j\neq i}|a_{ij}|\qquad(i=1,n)$$ $$\qquad\quad |z-\frac{a_{ii}}{3}| \le \frac{1}{3}\sum_{j\neq i}|a_{ij}|\qquad(i=2,\ldots,n-1)$$</p>