Timeline for How close to uniform are Perron-Frobenius eigenvectors?
Current License: CC BY-SA 4.0
12 events
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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 |
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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 |