Suppose $A$ is a positive definite matrix such that $$I \preceq A \preceq 1.01I.$$ Is it possible that $$\sum_{i=1}^n A_{1i}$$ can be arbitrarily large?
Thanks, Jack
Suppose $A$ is a positive definite matrix such that $$I \preceq A \preceq 1.01I.$$ Is it possible that $$\sum_{i=1}^n A_{1i}$$ can be arbitrarily large? Thanks, Jack 


Outline of one possible answer (but am doing this in a rush so have not written anything down to check). Consider the matrix $\Gamma$ which has 0 in topleft corner, 1 in rest of top row and rest of 1st column, and 0 in all other entries. (Note to hovering LaTeX hawks: please don't feel the need to enclose all those numbers in dollars.) Clearly $\Gamma$ is hermitian and a hasty calculation seems to show its eigenvalues are $\pm\sqrt{n1}$ and 0. Take $A= I + 0.005 (I+ (n1)^{1/2}\Gamma )$. This ought to satisfy your sandwich condition but it is clear that the sum of entries in 1st column will be ${\rm O}(\sqrt{n}$). 

