2
$\begingroup$

I know that eigenvalue estimates involving products of matrices are in general tricky, but probably this question has some hope:

Let $A$ and $B$ be two real symmetric positive semi-definite $n\times n$ matrices (with $A+B$ positive definite if needed). Let $\lambda(X)$ denote any eigenvalue of $X$.

I am fairly sure that $$ |1-\lambda\big( (I+A+B+AB)^{-1}(A+B)\big)| < 1 $$ but I would like ask:

  • Do you know a slick proof of this?
  • Do you know finer estimates (using,e.g., all eigenvalues/vectors of $A$, $B$ and $A+B$) or techniques that could be helpful?
$\endgroup$
4
  • $\begingroup$ what do you mean by 'some'? That there exists an eigenvalue between 0 and 2? Or that ll eigenvalues are between 0 and 2 (and all are real?)? $\endgroup$ Feb 15, 2017 at 17:54
  • $\begingroup$ I actually mean "any eigenvalue", so that all eigenvalues are in that range, and yes, I hope are all real. $\endgroup$
    – Dirk
    Feb 15, 2017 at 18:13
  • 1
    $\begingroup$ Sheesh, I messed up, the eigenvalues are actually not real - sorry. I adapt the question… $\endgroup$
    – Dirk
    Feb 15, 2017 at 18:15
  • 1
    $\begingroup$ $I+A+B+AB = (I+A)(I+B)$ is invertible. $\endgroup$ Feb 15, 2017 at 18:28

1 Answer 1

4
$\begingroup$

Yes, and more may be said.

Assume that $(I+A+B+AB)^{-1}(A+B)z=\lambda z$, then $(I+c(A+B)+AB)z=0$, where $c=1-\lambda^{-1}$.

Denote $Bz=u$, where $u$ may be arbitrary vector such that $(u,z)>0$. Next, $A(cz+u)=-z-cu$ and this is possible whenever $0<(cz+u,-z-cu)=-c(z,z)-(1+|c|^2)(u,z)-{\bar c}(u,u)$. This is possible if and only if the real part of $c$ is negative, i.e., $\Re \lambda^{-1}>1$, $|\lambda-1/2|<1/2$.

$\endgroup$
1
  • $\begingroup$ Thanks! I suspected something "more general" going on, but dealing with the eigenvalue equation directly is pretty straight forward here. $\endgroup$
    – Dirk
    Feb 17, 2017 at 16:07

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.