Suppose $A$ is a positive definite matrix such that$$ I \preceq A \preceq 1.01I.$$ Is it possible that $\sum\limits_{i=1}^n A_{1i}$ can be arbitrarily large?
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1$\begingroup$ Your definition seems to say, that the matrix A is diagonal. What do you actually mean? $\endgroup$– Gottfried HelmsCommented Jul 16, 2013 at 7:56
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$\begingroup$ @GottfriedHelms I read this as the usual partial order on the set of positive semi-definite matrices, so that one can take any PSD matrix X and put A = I + X / (100 r(X)) $\endgroup$– Yemon ChoiCommented Jul 16, 2013 at 8:00
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1$\begingroup$ Is $n$ fixed? Meaning, are you asking whether there is a uniform bound for $\sum_{i =1}^n A_{1i}$ independent of $n$ (which I assume is the dimension of the matrix), or are you asking whether for each $n$ there is a bound which may depend on $n$? $\endgroup$– Willie WongCommented Jul 16, 2013 at 8:29
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1 Answer
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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 top-left 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 $\pm\sqrt{n-1}$ and 0.
Take $A= I + 0.005 (I+ (n-1)^{-1/2}\Gamma )$. This ought to satisfy your sandwich condition but it is clear that the sum of entries in 1st column will be ${\rm O}(\sqrt{n}$).
answered Jul 16, 2013 at 8:23