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Given any real-valued, symmetric, positive semi-definite matrix $A, B \in \mathbb{R}^{n \times n}$, $\lambda > 0$. Does the following inequality hold for some constant $c$? $$\sigma_{\max}\Big((A+B + \lambda I_n)^{-1}A\Big) \leq c,$$ where $\sigma_{\max}(\cdot)$ denotes the maximal singular value.

The constant $c$ might be dependent on $\lambda$ and some norm of $A$ or $B$.

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No. Let $A_0=\pmatrix{1&0\\ 0&0}, B_0=\pmatrix{1&1\\ 1&1}$ and $$ C_0=(A_0+B_0)^{-1}A_0=\pmatrix{1&-1\\ -1&2}\pmatrix{1&0\\ 0&0} =\pmatrix{1&0\\ -1&0}. $$ Then $\sigma_1(C_0)=\|C_0\|_2>1$ because the Euclidean norm of the first column of $C_0$ is greater than $1$.

It follows that if $A=A_0+tI$ and $B=B_0+\lambda I$ for some small $t,\lambda>0$, then $A$ is positive definite, $B\succeq\lambda I$ and $\|C\|_2>1$ for $C=(A+B)^{-1}A$.

Similarly, if $A=X_1X_1^T+\cdots+X_pX_p^T$ and $X_{p+1}X_{p+1}^T+\cdots+X_tX_t^T$ are close to $A_0$ and $B_0$ respectively and $\lambda>0$ is small, then $\|C\|_2>1$ when $C=(A+B+\lambda I)^{-1}A$.

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  • $\begingroup$ Many thanks! In my problem, the matrix $B$ actually has a better property that $B \geq \lambda I_n$ for some $\lambda > 0$. So can we avoid the counterexample and will the statement hold now? $\endgroup$ – Peng Zhao May 7 at 11:05
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    $\begingroup$ @PengZhao Still no. $\sigma_1\left((A+B)^{-1}A\right)$ is continuous in the entries of $B$. If you add a small $\lambda I$ to the $B$ above, you still get a counterexample. $\endgroup$ – user1551 May 7 at 11:08
  • $\begingroup$ Could you please offer some more elaborations or some references? This would be quite helpful. $\endgroup$ – Peng Zhao May 7 at 11:11
  • $\begingroup$ @PengZhao Sorry, I don't know any reference. $\endgroup$ – user1551 May 7 at 11:13
  • $\begingroup$ Thanks for the update! I believe the counterexamples can be avoided by requiring $\lambda > 0$ and $\| X_s \| \leq S$, and the threshold will be dependent on $\lambda$ and $S$, instead of the constant $1$. $\endgroup$ – Peng Zhao May 7 at 15:55

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