# absolute value of eigenvalues of product of positive semi-definite matrix and diagonally dominant matrix

I have a asymmetric diagonally dominant matrix $A=M+k*I$, where $M$ is bounded, non-negative and asymmetric, $k$ could be arbitrarily large so that $A$ is always diagonally dominant, and a bounded strictly positive semi definite matrix $B$.

I am interested in the range of absolute value of eigenvalues of $C=A^{-1}B$. My conjecture is that $$|\lambda_i|\in(0,\epsilon)$$ for any $\epsilon>0$, provided we let $k$ to infinity. However, due to $M$ matrix is asymmetric, we cannot conclude that A is strictly positive semi-definite.

Any idea on proving my conjecture, or showing that my conjecture is wrong?

• This is not true. If $B$ is singular, so is $C$ and hence some $\lambda_i$ is zero. – user1551 May 18 '18 at 6:21
• B is strictly PSD, so all eigenvalue are strictly positive – M.Z. May 18 '18 at 20:01

Use the Gershgorin theorem to say that all eigenvalues of $A$ lie in the disk $\{z\in\mathbb{C}:|z-k|\leq \max_{1\leq i\leq n}\sum_{j=1}^{n}|M_{i,j}|\}$, so that the smallest possible absolute value of any eigenvalue of $A$ is $k-\max_{1\leq i\leq n}\sum_{j=1}^{n}|M_{i,j}|$. Then the eigenvalues of $A^{-1}$ lie in the disk $\{z\in\mathbb{C}:|z|\leq (k-\max_{1\leq i\leq n}\sum_{j=1}^{n}|M_{i,j}|)^{-1}\}$. Clearly as $k\rightarrow\infty$, this radius shrinks to zero, and because $B$ is bounded, this means $A^{-1}B\rightarrow0.$ By continuity of the eigenvalues, all eigenvalues must also tend to 0 as $k\rightarrow\infty$.