# Lower bound $\langle (I + A)^{-1}x, x \rangle$ given that $\sigma_\text{max}(A) < 1$

Let $$A$$ a square real matrix such that the largest singular value $$\sigma_\text{max}(A) = \sigma < 1$$. I want to find a lower bound on $$\langle (I + A)^{-1}x, x\rangle$$ where $$x$$ is a vector of euclidean norm $$1$$: $$\langle x, x\rangle=1$$.

I empirically find that a seemingly tight lower bound is $$\langle (I + A)^{-1}x, x\rangle \geq \frac{1}{1+\sigma}$$ which is reached for $$A= \sigma I$$. I cannot prove the above result.

Note that it is pretty straightforward to prove that $$\sigma_\text{min}((I + A)^{-1}) \geq \frac1{1+\sigma}$$ but that does not suffice to conclude, since I do not assume that $$A$$ is symmetric.

PS: $$\sigma_\text{max}$$ and $$\sigma_\text{min}$$ are the largest and smallest singular values: $$\sigma_\text{max} = \sqrt{\lambda_\text{max}(AA^T)}$$ is the operator norm of $$A$$.

• Suggestions for better question: Explain notation $\sigma_{\text{max}}, \sigma_{\text{min}}$. What norm used for vectors? Aug 29 at 13:31

The map $$f(z)=(1+z)^{-1} - (1+\sigma)^{-1}$$ maps the disk of radius $$\sigma$$ into the right half plane as a function of one complex variable.

Therefore, essentially by von Neumann's inequality, we get that $$\frac{f(A)+f(A)^*}{2}=\mathrm{Re }f(A)\geq 0$$ since $$\|A\|\leq \sigma.$$ Assuming $$A$$ has real entries, this implies the claim as $$\langle (1+A)^{-1}x,x\rangle = \langle \mathrm{Re} (1+A)^{-1}x,x\rangle.$$

To see the calculation with von Neumann's inequality more explicitly, let $$\psi(z) = \frac{z-1}{z+1}.$$ Note $$\psi$$ takes the right half plane to the disk. So, $$\psi \circ f$$ takes the disk of radius $$\sigma$$ into the unit disk. Therefore, von Neumann's inequality states that $$\|\psi \circ f(A)\|\leq \sup_{z\in \sigma\mathbb{D}} |\psi\circ f(z)| \leq 1.$$ Note that $$1-(\psi \circ f(A))^*(\psi \circ f(A)) \geq 0.$$ (Here by $$T\geq 0$$ we mean that $$T$$ is positive semi-definite.) Writing out what that means $$1-(f(A)^*+1)^{-1}(f(A)-1)^*(f(A)-1)(f(A)+1)^{-1} \geq 0.$$ So, $$(f(A)^*+1)(f(A)+1)-(f(A)^*-1)(f(A)-1)=2(f(A)+f(A)^*)\geq 0.$$

Results of the above form (positivity of noncommutative rational functions) always have to have "algebraic proofs," many of which can be done algorithmically. See, e. g., Helton, Klep, and McCullough - The convex Positivstellensatz in a free algebra and Pascoe - Positivstellensätze for noncommutative rational expressions.

• Thanks! I am unfamiliar with von Neumann's inequality: are you talinkg about en.wikipedia.org/wiki/Von_Neumann%27s_inequality ? If so, could you explain how it implies $f(A) + f(A)^T \geq 0$ ? (Sorry I forgot to mention that everything is real here).
– PAb
Aug 29 at 14:01
• @PAb I added an explanation at the bottom. Hopefully that helps. Aug 29 at 14:19
• @PAb and yes that was the theorem I was talking about. Aug 29 at 14:28
• Small typo: $\psi$ rather maps the right half plane to the unit disk. Thanks for the neat trick !
– PAb
Aug 30 at 13:22