# Matrix norm inequality proof: inverse of two p.s.d matrices sum

I wonder if the following matrix norm inequality holds: Let $A$ and $B$ are both strictly symmetric positive definite matrix $\|(A+B)^{-1}\|_2\leq \|A^{-1}\|_2$ ?

• Which norm are you referring to as $\| . \|_2$? Unfortunately, there are several different norms that this notation is used for. – Robert Israel Dec 8 '15 at 0:45
• $\|\cdot\|$ denotes the euclidean norm – YoooHan Dec 8 '15 at 10:30
If $\|\cdot\|_2$ denotes the maximum singular value, then it can be proved as follows.
The inequality in the original problem is equivalent to showing $$\sigma_{\max}((A+B)^{-1})\le \sigma_{\max}(B^{-1}) \Leftrightarrow \frac{1}{\sigma_{\min}(A+B)}\le \frac{1}{\sigma_{\min}(B)} \Leftrightarrow \sigma_{\min}(B)\le\sigma_{\min}(A+B)$$ Since $A$ is PD, the last inequality above is easy to prove (consider the eigenvector for the minimum eigenvalue of $A+B$, then $\sigma_{\min}(A+B)=x^T(A+B)x\ge x^TBx\ge \sigma_{\min}(B))$.