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

Thanks in advance.

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    $\begingroup$ I wonder what you have attempted so far! $\endgroup$ – Xoque55 Dec 8 '15 at 0:21
  • $\begingroup$ It appears to me that instinctively it is true, but I wonder if there is a rigorous proof. $\endgroup$ – YoooHan Dec 8 '15 at 0:25
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    $\begingroup$ Which norm are you referring to as $\| . \|_2$? Unfortunately, there are several different norms that this notation is used for. $\endgroup$ – Robert Israel Dec 8 '15 at 0:45
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    $\begingroup$ What I'm getting at is that you will receive much more help (and avoid the closing of your question) if you demonstrate that you've tried something and explain where you got stuck. $\endgroup$ – Xoque55 Dec 8 '15 at 0:46
  • $\begingroup$ $\|\cdot\|$ denotes the euclidean norm $\endgroup$ – YoooHan Dec 8 '15 at 10:30
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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 $\sigma_{\min}(A+B)=x^T(A+B)x\ge x^TBx\ge \sigma_{\min}(B))$.

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