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Matrix $A$ and $B$ are symmetric positive semidefinite. I wish to lower bound the minimal eigenvalue of the matrix product:

$$B(A+B)^{-1}A$$

The first lower bound I came up with is:

$$\lambda_{\min} \left( B (A+B)^{-1} A \right) \geq \frac{\lambda_{\min}(A) \cdot \lambda_{\min}(B)}{\lambda_{\max}(A+B)}\geq\frac{\lambda_{\min}(A) \cdot \lambda_{\min}(B)}{\lambda_{\max}(A)+\lambda_{\max}(B)}$$

According to this lower bound, if we do rank-one update on $A$ and $B$ in such a way that $\lambda_{\min}(A)$ and $\lambda_{min}(B)$ remain the same, but $\lambda_{\max}(A)$ and $\lambda_{\max}(B)$ increase (by updating A and B using eigenvectors corresponding to their maximum eigenvalues), then the lower bound will decrease.

However, my simulation results show that the minimum eigenvalue of this matrix product actually always increases as I rank-one updating $A$ and $B$.

Then my question is, whether there is any tighter lower bound of the minimum eigenvalue of this product?

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  • $\begingroup$ You assume that $A+B>0$. Show that i) $B(A+B)^{-1}A$ is symmetric, ii) the eigenvalues of $B(A+B)^{-1}A$ are $\geq 0$ and iii) your first inequality. $\endgroup$
    – user91684
    Jan 6 '20 at 17:50
  • $\begingroup$ Moreover, if $A$ or $B$ is not $>0$, then the minimal eigenvalue is $0$. Have you really thought about this problem? $\endgroup$
    – user91684
    Jan 6 '20 at 17:55
  • $\begingroup$ @loupblanc Thanks for your comment! What do you mean by the three steps? $\endgroup$ Jan 6 '20 at 18:30
  • $\begingroup$ So far you haven't shown anything. So it would be good if you could prove the $3$ above points. $\endgroup$
    – user91684
    Jan 6 '20 at 21:55
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Assume both $A$ and $B$ are invertible.

To lower bound the minimum eigenvalue, we need the following result. $$B(A+B)^{-1}A=(A^{-1}+B^{-1})^{-1}$$ Then it directly leads to: $$\lambda_{\min} \left( B (A+B)^{-1} A \right) = \frac{1}{\lambda_{max}(A^{-1}+B^{-1})}\geq \frac{1}{\lambda_{max}(A^{-1})+\lambda_{max}(B^{-1})}=\frac{1}{\frac{1}{\lambda_{min}(A)}+\frac{1}{\lambda_{min}(B)}}$$

To show that $B(A+B)^{-1}A=(A^{-1}+B^{-1})^{-1}$, we start from: $$A^{-1}(A+B)B^{-1}=A^{-1}AB^{-1}+A^{-1}BB^{-1}=B^{-1}+A^{-1}$$ Then taking inverse of both sides gives the result.

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