Welcome to MathOverflow

Question

3

2

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 2,383●4●12

2

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.... – S. Sra 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. – S. Sra 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) – S. Sra May 21 2012 at 18:51

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

4

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
4,400●1●12●24

	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

Will Sawin · Answer 2 · 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!

S. Sra · Answer 3 · 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.

upper bounds on a certain matrix norm

Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)

3 Answers

You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you.