Timeline for Growth of symmetric positive definite integral matrices.
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| May 15, 2011 at 7:57 | comment | added | Did | Roland: I finally understood your point. But it yields the crude bound $1/4$ (to choose $a$) times $1/4$ (to choose $c$) times $1/2$ (to choose $b$), that is, $c_2\ge1/32$ (which is true). | |
| May 4, 2011 at 17:48 | comment | added | Robert Israel | For a symmetric 2 x 2 matrix with entries uniformly distributed in $[-1,1]$ the probability of being positive definite is $\frac{1}{8} \int_0^1 da \int_0^1 dc \int_{-\sqrt{ac}}^{\sqrt{ac}} db = \frac{1}{9}$. | |
| May 4, 2011 at 17:07 | comment | added | Roland Bacher | The above computation (for $c_2\geq 1/4$) uses $\alpha_2(2N)$. | |
| May 4, 2011 at 15:49 | comment | added | Did | I thought $|a|$ and $|c|$ were a priori at most $N$? And that the conditions for positive definiteness were $a\ge0$, $c\ge0$ and $b^2\le ac$. | |
| May 4, 2011 at 15:31 | history | edited | Roland Bacher | CC BY-SA 3.0 |
added 383 characters in body; edited body
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| May 4, 2011 at 15:13 | comment | added | Roland Bacher | In fact, $c_2=2\left(\int_0^1 t^{3/2}dt\right)^2=\frac{8}{25}$. | |
| May 4, 2011 at 14:48 | comment | added | Roland Bacher | $c2\geq 1/4$ by considering $\left(\begin{array}{cc}a&b\\b&c\end{array}\right)$ with $a,c\in\lbrace N+1,\dots,2N\rbrace$ and $b\in \lbrace −N,…,N\rbrace$. | |
| May 4, 2011 at 14:43 | comment | added | Did | Roland: nice question, as usual. $c_2=2/9$? | |
| May 4, 2011 at 14:28 | history | asked | Roland Bacher | CC BY-SA 3.0 |