In my research on linear algebra and its applications, I have come across the following problem which has stumped me:

Let $ A $ be a positive definite matrix and let $ D $ be a positive diagonal matrix with entries on the main diagonal: $ d_1,d_2,...,d_n $, both $ A $ and $ D $ have the same dimension $ n \times n $. I was interested in understanding how the eigenvalues of the sum $ A + ADA $ qualitatively behave with respect to the eigenvalues of $ A $ and the entries $ d_1,...,d_n $.

I thought since this sum has a special form, one could hopefully say a bit more analytically than by exclusively using the fact that the two summands are Hermitian. I thought about using some other techniques, perhaps expressing the product of $A$ and $ D $ as a polynomial of $ A $, but unfortunately, I am stuck. I certainly appreciate all help on this.

  • 1
    $\begingroup$ why do you say that "the two summands commute" ? $A$ and $ADA$ do not commute, do they? $\endgroup$ – Carlo Beenakker Apr 30 '18 at 20:07
  • 1
    $\begingroup$ Because $A>0$ on a finite dimensional space we have no problems to build the (bounded) operator / matrix $A^{-1}>0$. Then $$A+ADA=A(A^{-1}+D)A\ ,$$ so the problem can be restated: Given two diagonalizable matrices $B,D>0$, (and without restriction, let $D$ be already diagonal,) what can be said about the spectrum of the sum in terms of the spectra of $B,D$? Answer: Well, not so much. Does this answer the question? $\endgroup$ – dan_fulea Apr 30 '18 at 20:20
  • 1
    $\begingroup$ @dan_fulea I don't follow your argument --- how is $B$ related to $A$ and $D$? $\endgroup$ – Federico Poloni Apr 30 '18 at 22:10
  • 1
    $\begingroup$ @DimaPasechnik But then the eigenvalues of $B+D$ aren't related in a simple way to those of $A(B+D)A$... $\endgroup$ – Federico Poloni May 1 '18 at 20:02
  • 1
    $\begingroup$ Yes, $B$ was supposed to be the inverse of $A$. Let then $B+D$ have an eigenvector $w$ for some eigenvalue $\lambda$, so $(B+D)w=\lambda w$. We write $w=Av$ for the one $v$. Then $(B+D)Av = \lambda Av$. And there is no "algebraic" way to relate $\lambda$ to an eigenvalue of $A(B+D)A$, sorry. Thanks for the comment, Federico! $\endgroup$ – dan_fulea May 2 '18 at 10:55

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.