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Consider the matrix $$N=\left[\matrix{\mathbb{I}_n-\epsilon L & X\\ \epsilon Y & Z}\right]$$ where $\epsilon>0$ is a small positive parameter and $Z$ is a square $m\times m$ matrix with all eigenvalues within the unit circle. All eigenvalues of $n\times n$ matrix $L$ have positive real part. I am interested on obtaining upper bounds (if they exist) on $\epsilon$ so that all eigenvalues of $N$ lie within the unit circle or else find some counterexample such that the matrix has an eigenvalue with magnitude larger than 1 for all values of $\epsilon>0$.

I was thinking to use Bauer-Fike theorem but the unperturbed matrix $$ \left[\matrix{\mathbb{I}_n & X\\ 0 & Z}\right]$$ has exactly $n$ eigenvalues on the unit circle.

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    $\begingroup$ $n=m=1$, $L=X=1$, $Z=0$, $Y=2$ is a trivial counterexample ($\det(N-I)=-\varepsilon<0$, so one eigenvalue is real and above $1$). Try to see whether you can squeeze some extra conditions from your setting and, please, check $2\times 2$ yourself before posting. $\endgroup$
    – fedja
    Commented Mar 27, 2018 at 23:03
  • $\begingroup$ @fedja Thank you for your counterexample. You are right, I should have checked it. I have a very large matrix in this particular form that seems to have this property (verified through detailed simulations) but it appears that indeed extra conditions of the related matrices do come into play. $\endgroup$
    – CTNT
    Commented Mar 27, 2018 at 23:10
  • $\begingroup$ Have you tried using the discrete-time Lyapunov equation? $\endgroup$
    – Rodrigo de Azevedo
    Commented Apr 6, 2018 at 16:33
  • $\begingroup$ @RodrigodeAzevedo Thank you for your comment. I don't think it will help as these are very large matrices which do not have some explicit numerical form but do have some extra properties. I have managed to prove my result using the sensitivities of the eigenvalues lying in 1 for $\epsilon=0$. $\endgroup$
    – CTNT
    Commented Apr 6, 2018 at 16:41

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