5
$\begingroup$

The Schur norm of a matrix $A$ is defined to be $\|A\|_S=\max\{\|A\circ X\|: \|X\|\leq 1\}$, where $\|\cdot \|$ is the operator norm of a matrix, i.e., the largest singular value.

Let $a_1,\ldots, a_m, b_1,\ldots, b_n$ be positive reals.Let $A$ be an $m\times n$ matrix defined to be $A_{i,j}=(a_i-b_j)/(a_i+b_j)$.

My question is how to compute $\|A\|_S$. Is it upper bounded by an absolute constant independent of $m, n$?

$\endgroup$
3
  • $\begingroup$ Is it upper bounded by an absolute constant independent of $m,n$? You cannot get that much unless you have some extra restrictions on $a,b$. Let $m=n$ and let $a_j=b_j$ be a fast increasing sequence. Then your multiplier is essentially $-1$ if $i<j$, $0$ if $i=j$ and $1$ if $i>j$, which has norm about $\log n$ $\endgroup$
    – fedja
    Jun 9, 2019 at 18:47
  • $\begingroup$ Isn't it just 0 matrix if $a_i=b_j$? $\endgroup$ Jun 9, 2019 at 21:17
  • $\begingroup$ Not $a_i=b_j$ but $a_j=b_j$ (not a_i=b_j but a_j=b_j). Sorry for the small font in comments. $\endgroup$
    – fedja
    Jun 9, 2019 at 22:54

0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Browse other questions tagged or ask your own question.