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Is there some simple upper bound on $||(B^{-1}+A^{-1})^{-1}||$, where $A,B$ are $n \times n$ symmetric matrices?

flag	asked May 21 2012 at 17:43 Felix Goldberg 2,407●5●12

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What norm are you denoting by $||\cdot ||$? – Noah Stein May 21 2012 at 18:22

Of course, for posdef matrices suchs bounds follow trivially using the operator harmonic, geometric, arithmetic mean inequalities. So I guess, the broader class of symmetric or Hermitian matrices in your question is deliberate.... – suVRit May 21 2012 at 18:41

My conjecture is that for any symmetric norm, your lhs $\le \| |a^{-1}b^{-1}|^{-1/2} \|$, where $|x|$ denotes the operator absolute value. Corresponding to this, one can then derive (i guess) an arithmetic-mean-style upper bound. – suVRit May 21 2012 at 18:51

(I threw in a factor of $2$ into your lhs to replace it by the more symmetric harmonic mean, while writing out the above inequality) – suVRit May 21 2012 at 18:51

Federico Poloni · Accepted Answer · 2012-05-21 18:36:23Z UTC

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You can use the surprising identity $(A^{-1}+B^{-1})^{-1}=A(A+B)^{-1}B$, and take the norms of the three factors separately.

answered May 21 2012 at 18:36

Federico Poloni
4,449●1●12●25

	Depending on the norm you are using, this might not give a very tight bound. Suppose $A=B$ and has one eigenvalue $\epsilon$ and the rest $1$. Then the spectral norm of $A$ is $1$, the spectral norm of $(A+B)^{-1})=1/(2\epsilon$, and the spectral norm of $B$ is $1$, while the actual spectral norm of $(A^{-1}+B^{-1})^{-1}$ is $1/2$. But if $\epsilon <<1$, then $1/(2\epsilon)<<1/2$. – Will Sawin May 21 2012 at 18:45
	I'm afraid this throws me back to the Woodbury formula, which I was coming from, but it's still a nice answer! – Felix Goldberg May 21 2012 at 19:41

suVRit · Answer 2 · 2012-05-21 19:05:38Z UTC

To expand on my first comment, if $A, B > 0$ are symmetric positive definite matrices. Then, it is known that

$$\left(\frac{A^{-1}+B^{-1}}{2}\right)^{-1} \le A\sharp B \le \frac{A+B}{2},$$ where the inequalities are in the Löwner partial order, and $A\sharp B := A^{1/2}(A^{1/2}BA^{1/2})^{-1/2}A^{1/2}$ denotes the matrix geometric mean.

These operator inequalities are of course, stronger than corresponding norm inequalities (based on unitarily invariant norms).

For the case where you don't have positive matrices, I think the conjecture mentioned in my second argument can be expanded into a proof --- maybe if I get time, I'll try to expand that.

Will Sawin · Answer 3 · 2012-05-21 19:16:50Z UTC

Probably not unless $A$ and $B$ are positive-definite, since if $B$ is very close to $-A$ then $B^{-1}+A^{-1}$ is very small and so its inverse is very large. In fact, depending on the norm, they probably need to be close only on one shared or almost-shared eigenvector.

For spectral norm of positive-definite matrices, we have a nice answer. The highest eigenvalue of $(A^{-1}+B^{-1})^{-1}$ is the lowest eigenvalue of $A^{-1}+B^{-1}$, which one can find by minimizing $x^T(A^{-1}+B^{-1})x$ with respect to $x^Tx=1$. But the minimum for $A^{-1}$ is its lowest eigenvalue, $1/||A||$, and the minimum for $B^{-1}$ is similarly $1/||B||$. Thus:

$x^T( A^{-1}+B^{-1})^{-1} x= x^T A^{-1} x+ x^T B^{-1} x\geq 1/||A||+1/||B||$

So the spectral norm of the harmonic sum is bounded by the harmonic sum of the spectral norms!

upper bounds on a certain matrix norm

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