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Jun 8, 2022 at 8:19 answer added Anurag Sahay timeline score: 1
Feb 16, 2022 at 3:41 comment added Will Sawin @IosifPinelis Sure, see my answer.
Feb 16, 2022 at 3:39 answer added Will Sawin timeline score: 4
Feb 16, 2022 at 3:22 comment added Iosif Pinelis @WillSawin : Can you elaborate on your comment, especially on $\approx\sqrt 2^n$?
Feb 15, 2022 at 12:21 comment added Will Sawin I believe the adjacency matrix of a depth $n$ binary tree gives a symmetric counterexample, as $\max_i v_i / \min_i v_i$ should be close to $\sqrt{2}^n$ but the row sums are bounded between $1$ and $3$.
Feb 15, 2022 at 7:06 comment added H A Helfgott The operator norm. But don't trust that bound! A friend tekks me he just found a counterexample (not symmetric I think). My "proof" in the shower was most likely wrong (but then I may have forgotten a crucial detail).
Feb 14, 2022 at 23:42 comment added Will Sawin (1) I think $A$ symmetric is necessary - i.e. it does help. Take a path, so the adjacency matrix has nonzero entries only on the diagonal, just above, and just below, and multiply all the entries just above the diagonal by $2$ and just below the diagonal by $1/2$. We have now multiplied the ratio between the first and last entry by $2^{n-1}$. (2) what is $|A|$ in the displayed equation?
Feb 14, 2022 at 22:47 history edited H A Helfgott CC BY-SA 4.0
added 102 characters in body
Feb 14, 2022 at 17:51 comment added H A Helfgott Oh, it can be useful, but I was assuming people would have a battery of results better than something I thought of in the shower.
Feb 14, 2022 at 17:47 comment added Iosif Pinelis What do you mean by "useful bounds", and why is the "easy bound" not useful?
Feb 14, 2022 at 17:37 history edited YCor
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Feb 14, 2022 at 17:08 history asked H A Helfgott CC BY-SA 4.0