Timeline for Expected size of determinant of $AA^T$ for random circulant and Toeplitz matrices
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
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| Apr 26 at 14:52 | comment | added | Sean Eberhard | Any chance of getting further details about the lower bound? It seems surprising. Is it some sort of saddle-point analysis? | |
| Apr 26 at 14:51 | comment | added | Sean Eberhard | Hi, Padraig Ó Catháin and I were discussing this question and answer this week. This argument is very interesting. However, for any $(0,1)$-matrix one actually has the following variant of the Hadamard bound: $\det A \le (n+1)^{(n+1)/2} / 2^n$. This follows from $\det A = \det \begin{pmatrix} 1 & 1 \\ 0 & A\end{pmatrix} = 2^{-n} \det \begin{pmatrix} 1 & 1 \\ -1 & 2A - 1 \end{pmatrix}$ followed by the Hadamard bound for an $(n+1) \times (n+1)$ matrix. Therefore your upper bound $\lim f_{\mathrm{cir+}}$ can be improved to $0$. | |
| Jan 31, 2016 at 19:19 | vote | accept | Simd | ||
| Jan 29, 2016 at 17:06 | history | edited | Dierk Bormann | CC BY-SA 3.0 |
added 6806 characters in body
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| Jan 12, 2016 at 10:44 | comment | added | Simd | A result for circulant matrices would be really great. | |
| Jan 12, 2016 at 10:00 | review | Late answers | |||
| Jan 12, 2016 at 10:08 | |||||
| Jan 12, 2016 at 9:45 | review | First posts | |||
| Jan 12, 2016 at 10:04 | |||||
| Jan 12, 2016 at 9:43 | history | answered | Dierk Bormann | CC BY-SA 3.0 |