# Approximation theoretic question about operator norm

Let $\|M\|:=\sup_{u:\|u\|=1}\|Mu\|$ be the operator norm induced by the Euclidean distance.

Suppose $A$ is a $k\times k$ symmetric matrix with $A_{ij}>0$ for all $i,j$ and $\sum_{i,j} A_{ij} = 1.$ Let $A_i$ be the sum of all elements in row $i$ (or column $i$) of $A.$ Let $B$ be a matrix with entries $$B_{ij} = \frac{A_{ij}^2}{A_i A_j}~.$$ We may show $\|B\|\in \left[1/k,1\right].$ Let $B^\prime$ be identical to $B$ except that its row $k$ and column $k$ are zeroed out.

I want to show a quantitative version of the following qualitative conjecture:

If $A_i$ is large, say $\sim \Theta\left(1/k\right)$ for all $i$ except $i=k,$ for which $A_k\sim o\left(1/k\right),$ then $\frac{\|B^\prime\|}{\|B\|}$ is very close to 1, and approaches $1$ as $A_k$ approaches zero.

Note that the entry $B_{kk}$ need not be small, although yes, $B_{ik}, B_{ki}$ for $i\neq k$ will be small.

Note also that I am interested in the ratio of the two norms being close to 1, not their difference being close to 0. Since each norm may be individually as small as $\frac 1k,$ it is the ratio that is of interest to me.

Concretely, suppose $$A_i = \begin{cases} \frac{1-\epsilon}{k-1}, & i=1,2,\ldots,k-1~, \\ \epsilon, & i = k~. \end{cases}$$ Can we show $$\frac{\|B^\prime\|}{\|B\|} = 1 + O(\epsilon^2)~?$$

Including here the simple argument for why $\|B\|\in\left[1/k,1\right]:$

Note that $B$ is similar to $C$ where $C_{ij} = \frac{A_{ij}^2}{A_j^2}$ and then observe that $\sum_{j=1}^k C_{ij} \in\left[1/k,1\right]$ for each $i.$ By Perron-Frobenius theory, the spectral radius of $C$ lies in $\left[1/k,1\right]$ and $B$ is symmetric, so its spectral norm equals its spectral radius.

I discovered that my conjecture is incorrect. Here is a simple counterexample.

First, let's allow $A_{ij}=0$ for some $i,j.$ Let

$$A_{ij} = \begin{cases}\frac{1-\epsilon}{(k-1)^2} & 1\leq i,j\leq k-1 \\ \epsilon & i=j=k \\ 0 & \mbox{else}. \end{cases}$$

Then, $\|B\| = 1$ since $B_{kk} = 1,$ and we know that $\|B\|\in[1/k,1].$ However, $\|B^\prime\| = \frac{1}{k-1}.$

To get a counterexample with all $A_{ij}$ strictly positive, let

$$A_{ij} = \begin{cases}\frac{1-\epsilon}{(k-1)^2} & 1\leq i,j\leq k-1 \\ \epsilon-(2k-2)\delta & i=j=k \\ \delta & \mbox{else} \end{cases},$$

and choose $\delta$ much much smaller than $\frac{\epsilon}{k}.$