Timeline for What it is the volume of the unit ball section of the cone of positive definite matrices?
Current License: CC BY-SA 3.0
14 events
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| May 3, 2014 at 20:32 | vote | accept | Felix Goldberg | ||
| May 2, 2014 at 16:35 | comment | added | Yoav Kallus | @RobertBryant: Sure, but since they only specified the operator norm later, I thought they might be interested in other cases as well. | |
| May 2, 2014 at 15:16 | comment | added | Robert Bryant | @YoavKallus: The question you reference uses a different norm on the space of symmetric $n$-by-$n$ matrices, so that discussion really isn't relevant to this question with the operator norm, as specified by the OP. | |
| May 2, 2014 at 11:44 | answer | added | Robert Bryant | timeline score: 15 | |
| May 1, 2014 at 16:26 | comment | added | Yoav Kallus | Have you taken a look at the following question? Looks relevant: mathoverflow.net/questions/118481/… | |
| May 1, 2014 at 15:06 | history | edited | Felix Goldberg | CC BY-SA 3.0 |
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| May 1, 2014 at 15:06 | comment | added | Felix Goldberg | @Suvrit I want the operator norm, aka $2$-norm. | |
| May 1, 2014 at 15:01 | comment | added | alvarezpaiva | Sometime ago I took a look at the paper "The volume of a certain set of matrices" by H. Jack and A. M. Macbeath. Maybe it will be useful. journals.cambridge.org/action/… | |
| Apr 27, 2014 at 14:10 | comment | added | Suvrit | I'm presuming the cone is of "real symmetric positive definite" matrices, and the unit sphere that is meant is of the operator norm or perhaps Frobenius norm --- but needs to be clarified. | |
| Apr 27, 2014 at 12:44 | comment | added | Robert Bryant | Actually, do you mean to ask this in the symmetric $n$-by-$n$ matrices? Otherwise, I don't know what 'positive definite' means. Also, what (positive definite) norm are you assuming on the symmetric $n$-by-$n$ matrices? One of the ones that is invariant under $\mathrm{SO}(n)$? There is a $2$-parameter family of those, and the answer will depend on which one you choose. There is no such norm that is invariant under the full group $\mathrm{GL}(n,\mathbb{R})$. | |
| Apr 27, 2014 at 8:21 | comment | added | Felix Goldberg | @RobertIsrael Sorry, fixed that. <blush> | |
| Apr 27, 2014 at 8:21 | history | edited | Felix Goldberg | CC BY-SA 3.0 |
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| Apr 27, 2014 at 8:15 | comment | added | Robert Israel | Please clarify your question. The symmetric $n \times n$ real matrices form a linear space of dimension $n(n+1)/2$. It's certainly not a subset of ${\mathbb R}^n$. | |
| Apr 27, 2014 at 7:54 | history | asked | Felix Goldberg | CC BY-SA 3.0 |