# How to show for a PSD matrix $A$ that $\left \| \left ( A+I \right )^{-1} \right \| \leq \frac{1}{1+\sigma _{\min}\left ( A \right )}$?

If $A \in \mathbb{C}^{n \times n}$ is positive semidefinite, show that $\left \| \left ( A+I \right )^{-1} \right \| \leq \frac{1}{1+\sigma _{\min}\left ( A \right )}$, where $\sigma _{\min}\left ( A \right )$ is the smallest singular value of $A$, and $\left \| \cdot \right \|$ is any unitarily invariant induced matrix norm.

For a general unitarily invariant matrix norm $\|\cdot\|$ (that is not necessarily induced), we have $\|X\|\ge\|X\|_2$ (cf. Horn and Johnson, Matrix Analysis, p.308, corollary 5.6.35). Unitarily diagonalize $A$ and put $X=(A+I)^{-1}$, we see that you should have the inequality sign flipped.

However, on $\mathbb{C}^{n\times n}$, the only matrix norm that is both induced and unitarily invariant is the spectral norm $\|A\|_2=\sigma_\max(A)$ (again, see the above-mentioned corollary and its proof). And your inequality is actually an equality when the spectral norm is used. So, the direction of the inequality sign does not matter in this case.

Although the condition that $\|\cdot\|$ is induced is redundant, it is imposed here perhaps to make the proof easier. And the direction inequality sign is perhaps deliberate, so that one doesn't need to know or prove the fact that the spectral norm is the only unitarily invariant induced matrix norm for complex matrices.

• Thanks for your explanation, it was really helpful. So could you please tell me how to prove the equality with $\left \| \cdot \right \|_2$? – user79230 Mar 23 '13 at 21:02
• @Sam As was said in the answer, unitarily diagonalize $A$. Since the norm is unitarily invariant, you may assume that $A$ is a nonnegative diagonal matrix can calculate $\|(A+I)^{-1}\|_2$ directly. – user1551 Mar 23 '13 at 22:29

Are you completely sure about the $\leq$ sign, shouldn't it be $\geq$?
I tried to prove your statement but got a different answer. We can use some facts about singular values of positive definite matrices: $$\|(A+I)^{-1}\|_2\|A+I\|_2=\frac{\sigma_{\max}(A+I)}{\sigma_{\min}(A+I)}$$

$$\|A+I\|_2=\sqrt{\sigma_{max}((A+I)^{T}(A+I))}=\sqrt{\sigma_{max}(A+I) \circ\sigma_{max}(A+I)}$$ where $\circ$ - is Hadamard product. $$\|(A+I)^{-1}\|_2=\frac{\sigma_{\max}(A+I)}{\sigma_{\min}(A+I))\sqrt{\sigma_{max}(A+I) \circ\sigma_{max}(A+I)}}$$

$$\sigma_{\max}(A+I)=1+\sigma_{\max}(A)$$

$$\sigma_{\min}(A+I)=1+\sigma_{\min}(A)$$

Then we plug it into the first equation: $$\|(A+I)^{-1}\|_2=\frac{1+\sigma_{\max}(A)}{1+\sigma_{\min}(A))\sqrt{(1+\sigma_{\max}(A)) \circ(1+\sigma_{\min}(A))}}$$ or $$\|(A+I)^{-1}\|_2=\frac{(1+\sigma_{\max}(A))^{\frac{1}{2}}}{(1+\sigma_{\min}(A))^{\frac{3}{2}}}$$ and assuming that $\sigma_{\max}>\sigma_{\min}$ And we get that: $$\|(A+I)^{-1}\|_2\geq\frac{1}{1+\sigma_{\max}(A))}$$ $$\|(A+I)^{-1}\|_2\geq\frac{1}{1+\sigma_{\min}(A))}$$

• Would not $\sigma_\max(A+I)$ be a number? So, what is this Hadamard product? – Tapu Mar 23 '13 at 16:33
• @Tapu just miltiplication. In the "fact" I used standard notation as in textbooks. – Caran-d'Ache Mar 23 '13 at 16:34
• @Caran-d'Ache, why $\sigma_{max}((A+I)^{T}(A+I)) = \sigma_{max}(A+I) \circ\sigma_{max}(A+I)$? The inequality sign is as asked in the question. – user79230 Mar 23 '13 at 17:21