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.



Approximation theoretic question about operator norm
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 $\frac{\epsilon}{k}.$


