For a symmetric real matrix $A$, denote by $\rho(A)$ the spectral radius of $A$, and by $\sigma(A)$ the largest absolute row sum of $A$; that is, $\sigma(A)=\max_i \sum_j |a_{ij}|$, where $a_{ij}$ are the elements of $A$. We thus have $\rho(A)\le\sigma(A)$.
A general question:
What can we say about $A$ given that $\sigma(A)\le K\rho(A)$ with a real $K>0$?
A pinpointed version (this is what I eventually need):
Given that $\sigma(A)\le K\rho(A)$, does there necessarily exist a unit-length vector $x=(x_1,\ldots,x_n)$ (where $n$ is the order of $A$) such that $\|Ax\|\ge 0.1\rho(A)$ and $(|x_1|+\dotsb+|x_n|)\max_i|x_i|\le 10K^{10}$?
(It is not difficult to see that a unit-length vector $x=(x_1,\ldots,x_n)$ with $\|Ax\|=\rho(A)$ and $(|x_1|+\dotsb+|x_n|)\max_i|x_i|\le 10K^{10}$ may fail to exist.)
An ideologically motivated restatement:
For a (not necessarily square) matrix $A\ne 0$, define the height of $A$ by $h(A):=\|A\|_1\|A\|_\infty/\|A\|^2$, where $\|A\|_1,\|A\|_\infty$, and $\|A\|$ are induced operator norms (see Wikipedia for the definitions). Notice, that $1\le h(A)\le\sqrt{mn}$, where $m$ and $n$ are the dimensions of $A$.
If $m=n$ and $A$ is symmetric, then $\|A\|_1=\|A\|_\infty=\sigma(A)$ and $\|A\|=\rho(A)$. If $A$ is actually a vector, then $$ h(A)=\frac{(|x_1|+\dotsb+|x_n|)\max_i|x_i|}{x_1^2+\dotsb+x_n^2},$$ where $x_1,\ldots x_n$ are the coordinates of $A$. My question can be equivalently restated as follows:
If $A$ is a symmetric real matrix with $h(A)\le K$, does there necessarily exist a vector $x$ with $\|Ax\|\ge 0.1\|A\|\|x\|$ and $h(x)\le 10K^{10}$?
The following special cases may be worth mentioning.
- if $A$ is diagonal, then the coordinate vector corresponding to the absolutely largest eigenvalue has height $1$; thus, denoting this vector by $x$, we have $\|Ax\|=\|A\|\|x\|$ and $h(x)=1$.
- if $K=1$, then $\rho(A)=\sigma(A)$. I have a complete characterization of matrices with this property, and it turns out that every principal eigenvector of such a matrix has height $1$. Hence, the assertion is true in this case, too.
- If $A$ is rank-$1$ then, in view of $A^t=A$, we have $A=\pm xx^t$ for some vector $x$. It is easy to see that in this case $\sigma(A)=h(x)\|x\|^2$, $\rho(A)=\|x\|^2$, and $x$ is a principal eigenvector of $A$. Thus, $\|Ax\|=\|A\|\|x\|$, and $h(x)=\sigma(A)/\rho(A)\le K$.
Update 23.04
I doubt this will result in a major breakthrough, but I noticed recently that my original requirement $|\langle Ax,x\rangle|\ge 0.1\rho(A)$ can be relaxed to $\|Ax\|\ge 0.1\rho(A)$, and I update the problem accordingly, for a good record-keeping