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Let $A$ and $B$ be positive definite real matrices. It seems obvious to me that it should be true that

$$\|A(A+B)^{-1}\|_2<1.$$

My heuristic argument is that

$$\|A(A+B)^{-1}\|_2< \|A(A+\lambda_{\min}(B) I)^{-1}\|_2 = \max\left\{\frac{\lambda_i(A)}{\lambda_i(A)+\lambda_{\min}(B)}\right\}<1.$$

However, I'm not able to justify the first inequality.

Lepidopterist
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1 Answers1

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This is actually false. If $\|x\|=1$, then $$ \|A(A+B)^{-1}x\|^2 = \langle A(A+B)^{-1} x, A(A+B)^{-1} x \rangle =\langle x, (A+B)^{-1}A^2(A+B)^{-1} x \rangle \\ = 1 - \langle y, ((A+B)^2 - A^2 ) y \rangle , $$ with $y=(A+B)^{-1}x$. So your claim is equivalent to $(A+B)^2-A^2$ being a positive definite matrix, and this will not follow from your assumptions.

The key fact is that the function $t\mapsto t^2$ is not operator monotone (there used to be a Wikipedia article on this, which I don't seem to be able to locate).

  • So if I add the assumption that $A$ and $B$ are symmetric, then $AB$ is positive semidefinite and my statement is true. Since $(A+B)^2-A^2 = B^2 +2AB$, where $B^2$ is positive definite and $AB$ has no negative eigenvalues by http://math.stackexchange.com/a/113859/10117. Do you agree? – Lepidopterist Jul 05 '16 at 14:49
  • No. You already assumed that $A,B$ are symmetric (as part of positive definite). Please read up on "operator monotone functions." –  Jul 05 '16 at 16:45