# Schur norm of weighted Cauchy matrix

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 $$m, n$$?

• 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 $\log n$ Jun 9 '19 at 18:47
• Isn't it just 0 matrix if $a_i=b_j$? Jun 9 '19 at 21:17
• 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. Jun 9 '19 at 22:54