Timeline for Minimum distance of a symmetric matrix to diagonal matrices
Current License: CC BY-SA 4.0
17 events
| when toggle format | what | by | license | comment | |
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| Oct 29, 2023 at 11:36 | vote | accept | Mostafa - Free Palestine | ||
| Oct 28, 2023 at 14:48 | history | edited | Michael Hardy | CC BY-SA 4.0 |
deleted 2 characters in body
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| Oct 26, 2023 at 19:26 | answer | added | Mostafa - Free Palestine | timeline score: 2 | |
| Dec 26, 2019 at 20:04 | comment | added | Mahdi - Free Palestine | Dear Mostafa, I think WLOG we could consider diagonal entries of $ A $ equal to zero. | |
| Dec 26, 2019 at 14:35 | comment | added | Mostafa - Free Palestine | @fedja I think you should also consider the absolute value of diagonal entries in the upper bound estimate, so I can't see that "the sum of the squares of the eigenvalues $\leq n(n-1)\max_{i\neq j} |a_{ij}|^2$" ... | |
| Dec 26, 2019 at 3:30 | comment | added | MTyson | @NathanielJohnston I see, thanks! I was thinking of the Schatten $2$-norm rather than the operator $2$-norm. | |
| Dec 26, 2019 at 3:06 | comment | added | Nathaniel Johnston | @MTyson - Does your argument suggest that the closest diagonal matrix to A is the diagonal of A? If so, that's not true (when $n \geq 3$). For what it's worth, we solved the problem of finding the closest matrix D when A has rank 1 in arXiv:1601.06269, but the formula is quite nasty (see Theorem 1). | |
| Dec 25, 2019 at 13:07 | comment | added | fedja | @MarkL.Stone To set up the "triviality level", the obvious lower bound is $n/2$ (consider all ones outside the diagonal) and the obvious upper bound is $\sqrt{\frac{n(n-1)}2}$ (take the distance from the off-diagonal part to multiples of identity and take into account that the sum of the squares of the eigenvalues is $\le n(n-1)\max|a_{ij}|^2$). So, roughly speaking, we are between $\frac n2$ and $\frac n{\sqrt 2}$ and I do not know which bound is closer to the truth... | |
| Dec 25, 2019 at 10:41 | comment | added | Mostafa - Free Palestine | @ChristianRemling Yes, my argument also works for $n>2$. I edited this remark in the question. | |
| Dec 25, 2019 at 10:38 | history | edited | Mostafa - Free Palestine | CC BY-SA 4.0 |
added 24 characters in body
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| S Dec 25, 2019 at 10:36 | history | suggested | Rodrigo de Azevedo | CC BY-SA 4.0 |
Added tag. Fixed typo. Minor improvements in wording and formatting.
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| Dec 25, 2019 at 10:12 | review | Suggested edits | |||
| S Dec 25, 2019 at 10:36 | |||||
| Dec 24, 2019 at 18:06 | comment | added | Christian Remling | I don't think there's a smaller constant for $n=1, 2$. | |
| Dec 24, 2019 at 13:47 | comment | added | Mostafa - Free Palestine | @Stone I don't know a smaller constant, but there is a smallest constant (compactness argument) and I know that there is no matrix such that $d(A)/\max(|a_{ij}|)=n-1$. My proof for this fact is rather long and may be irrelevant to finding the best $C$, so I didn't write it here. | |
| Dec 24, 2019 at 13:38 | comment | added | Mark L. Stone | "But it can be shown that it's not the smallest constant." Please show us a smaller (the smallest) constant you know with proof. | |
| Dec 24, 2019 at 11:02 | history | edited | Mostafa - Free Palestine | CC BY-SA 4.0 |
edited title
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| Dec 24, 2019 at 10:56 | history | asked | Mostafa - Free Palestine | CC BY-SA 4.0 |