upper bounds on a certain matrix norm - MathOverflow most recent 30 from http://mathoverflow.net2013-05-22T03:35:02Zhttp://mathoverflow.net/feeds/question/97579http://www.creativecommons.org/licenses/by-nc/2.5/rdfhttp://mathoverflow.net/questions/97579/upper-bounds-on-a-certain-matrix-normupper bounds on a certain matrix normFelix Goldberg2012-05-21T17:43:26Z2012-05-21T23:20:51Z
<p>Is there some simple upper bound on $||(B^{-1}+A^{-1})^{-1}||$, where $A,B$ are $n \times n$ symmetric matrices?</p>
http://mathoverflow.net/questions/97579/upper-bounds-on-a-certain-matrix-norm/97587#97587Answer by Federico Poloni for upper bounds on a certain matrix normFederico Poloni2012-05-21T18:36:23Z2012-05-21T18:36:23Z<p>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.</p>
http://mathoverflow.net/questions/97579/upper-bounds-on-a-certain-matrix-norm/97588#97588Answer by Will Sawin for upper bounds on a certain matrix normWill Sawin2012-05-21T18:42:52Z2012-05-21T19:16:50Z<p>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.</p>
<p>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:</p>
<p>$x^T( A^{-1}+B^{-1})^{-1} x= x^T A^{-1} x+ x^T B^{-1} x\geq 1/||A||+1/||B||$</p>
<p>So the spectral norm of the harmonic sum is bounded by the harmonic sum of the spectral norms!</p>
http://mathoverflow.net/questions/97579/upper-bounds-on-a-certain-matrix-norm/97592#97592Answer by S. Sra for upper bounds on a certain matrix normS. Sra2012-05-21T19:05:38Z2012-05-21T19:05:38Z<p>To expand on my first comment, if $A, B > 0$ are symmetric positive definite matrices. Then, it is known that</p>
<p>$$\left(\frac{A^{-1}+B^{-1}}{2}\right)^{-1} \le A\sharp B \le \frac{A+B}{2},$$
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. </p>
<p>These operator inequalities are of course, <strong>stronger</strong> than corresponding norm inequalities (based on unitarily invariant norms).</p>
<p>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.</p>