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I'm trying to find information on the eigenvalues of an $n \times n$ matrix A such that

$A = D + J$

Where $D$ is some complex valued diagonal matrix, and $J$ is an matrix consisting of all $1$'s.
When $D$ has identical values, the problem is equivalent to finding the eigenvalues of $J$.

So my question is this:
If $D$ has non-identical values (specifically, non-identical imaginary components), is there an elementary way to compute the eigenvalues of $A$ ?

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.

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1 Answer 1

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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 $\phi(D,t)-\phi'(D,t)$.

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  • $\begingroup$ Thanks very much, this was exactly what i was looking for. $\endgroup$
    – Peter Cudmore
    Commented Jun 16, 2011 at 13:11
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    $\begingroup$ Well, $J$ is a rank-one matrix. So this calculation is nothin but a special case of the Sherman--Morrison formula. $\endgroup$
    – Denis Serre
    Commented Jun 16, 2011 at 14:22
  • $\begingroup$ @Denis: Could you say another word or two about making the connection to the Sherman--Morrison formula? I see that $J$ here is a particular rank-one matrix, but I don't see what this has to do with calculating an inverse $(D+J)^{-1}$ where $D^{-1}$ is already known. I can generalize the above answer for arbitrary rank-one $J$, but then the sum has no obvious interpretation. @Chris, is there intuition for the RHS, $\phi(D,t)-\phi'(D,t)$? What's the usefulness of a derivative of a characteristic function? $\endgroup$
    – Jess Riedel
    Commented Jul 18, 2012 at 14:38
  • $\begingroup$ @Jess: I have no intuition for $\phi(D,t)-\phi'(D,t)$. The ratio $\phi'(M,t)/\phi(M,t)$ turns up in combinatorics in various places. I am not sure I really understand what "usefulness" means here. $\endgroup$
    – Chris Godsil
    Commented Jul 19, 2012 at 22:54
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    $\begingroup$ To avoid confusing future readers: note that this answer does not provide a way to compute the eigenvalues, as asked, but only an expression for the characteristic polynomial. $\endgroup$
    – Federico Poloni
    Commented Apr 29, 2018 at 17:28

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