# maximal singular value inequality

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 $$B$$.

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 $$C=(A+B+\lambda I)^{-1}A$$.
• 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? – Peng Zhao May 7 at 11:05
• @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. – user1551 May 7 at 11:08
• 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 $1$. – Peng Zhao May 7 at 15:55