most recent 30 from http://mathoverflow.net 2013-05-23T01:24:26Z http://mathoverflow.net/feeds/question/67924 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/67924/eigenvalues-of-the-sum-of-a-diagonal-and-a-unit-matrix Peter Cudmore 2011-06-16T10:03:40Z 2011-06-16T12:39:12Z

<p>I'm trying to find information on the eigenvalues of an $n \times n$ matrix A such that </p> <p><code>$A = D + J$</code></p> <p>Where $D$ is some complex valued diagonal matrix, and $J$ is an matrix consisting of all $1$'s.<br> When $D$ has identical values, the problem is equivalent to finding the eigenvalues of $J$.</p> <p>So my question is this:<br> <em>If $D$ has non-identical values (specifically, non-identical imaginary components), is there an elementary way to compute the eigenvalues of $A$ ?</em> </p> <p>The problem comes from linearising about the origin of a system of $n$ near identical coupled resonators. $D$ relates to the behaviour of each resonator, $J$ relates to the coupling process. </p>

http://mathoverflow.net/questions/67924/eigenvalues-of-the-sum-of-a-diagonal-and-a-unit-matrix/67935#67935 Chris Godsil 2011-06-16T12:17:18Z 2011-06-16T12:39:12Z

<p>There is a formula. Recall that $\det(I-AB)=\det(I-BA)$ for any matrices $A$ and $B$ such that both products $AB$ and $BA$ are defined. Now $$ \det(tI-D-J) = \det(tI-D) \det(I-(tI-D)^{-1}J). $$ If $u$ is the column vector with each entry equal to 1 then $J=uu^T$ and $$ \det(I-(tI-D)^{-1}J) = \det(I-(tI-D)^{-1}uu^T) = \det(1- u^T(tI-D)^{-1}u) = 1-u^T(tI-D)^{-1}u. $$ If we write $\phi(M,t)$ for the characteristic polynomial off $M$, this yields $$ \phi(D+J,t) = \phi(D,t) \left(1-\sum_i \frac1{t-D_{i,i}}\right). $$ The sum is equal to $\phi'(D,t)/\phi(D,t)$ and therefore the right side equals $\phi(D,t)-\phi'(D,t)$.</p>