Timeline for Determinant of the random matrix $X^2+Y^2$
Current License: CC BY-SA 4.0
24 events
| when toggle format | what | by | license | comment | |
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| Dec 4, 2023 at 11:30 | history | edited | YCor | CC BY-SA 4.0 |
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| Dec 4, 2023 at 11:13 | history | edited | loup blanc | CC BY-SA 4.0 |
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| Oct 20, 2023 at 21:56 | answer | added | Dan Piponi | timeline score: 5 | |
| Oct 20, 2023 at 16:38 | comment | added | Michael Hardy | The probability distribution of this determinant seems to have tails so fat that the sample mean probabily has a much larger variance than the sample median. | |
| Oct 20, 2023 at 15:39 | answer | added | Mykola Pochekai | timeline score: 19 | |
| Oct 20, 2023 at 9:58 | comment | added | loup blanc | @Abdelmalek Abdesselam ; thanks ; great proof. | |
| Oct 18, 2023 at 21:58 | comment | added | Dan Piponi | Going back to my example of the sum of $m$ squares of $n\times n$ matrices. By an argument related to the classical matchings-via-determinants one, each non-zero term contributing to the determinant corresponds to a permutation where each cycle is given 1 of m colors and it contributes precisely 1. Eg. for n=3, m=2 we have 1 cycle conjugate to (1)(2)(3) contributing 2^3, 3 cycles conjugate to (1)(23) contributing 2^2 and 2 cycles conjugate to (123) contributing 2^1=8+3*4+2*2=24=(3+1)! | |
| Oct 18, 2023 at 20:29 | answer | added | Abdelmalek Abdesselam | timeline score: 21 | |
| Oct 18, 2023 at 20:14 | comment | added | Denis Serre | Another instance of non-commutative probabilities ... :) | |
| Oct 18, 2023 at 20:02 | comment | added | Carlo Beenakker | @DanPiponi -- FWIW, my attempt at an answer, which I was not able to complete for arbitrary $n$, does directly prove your equation for $n=2$ and arbitrary $m$ [when the determinant equals $m(m+1)$] | |
| Oct 18, 2023 at 17:32 | comment | added | Dan Piponi | More generally I guess $E[\det(X_1^2+X_2^2+\ldots+X_m^2)]=(m+n-1)!/(m-1)!$ for independent matrices $X_i$ of the same type. | |
| Oct 18, 2023 at 16:00 | comment | added | loup blanc | @YCor, yes you are right. | |
| Oct 18, 2023 at 15:57 | history | edited | loup blanc | CC BY-SA 4.0 |
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| Oct 18, 2023 at 13:22 | comment | added | YCor | The tag gaussian already exists, no need to create normal-law. Unless you mean something else that Gaussian/Normal distribution (but what?). | |
| Oct 18, 2023 at 13:21 | history | edited | YCor |
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| Oct 18, 2023 at 13:06 | answer | added | Carlo Beenakker | timeline score: 7 | |
| Oct 18, 2023 at 11:45 | comment | added | Timothy Chow | Wayback Machine link to Moo K. Chung's paper: On Expected Gaussian Random Determinants. | |
| Oct 18, 2023 at 8:19 | comment | added | Carlo Beenakker | @TimothyBudd --- yes, apologies, $\mathbb{E}[\det(X+X^\top)]=(-1)^m (2m)!/m!$ (this is stated correctly by Chung, it's my mistake) | |
| Oct 18, 2023 at 7:24 | comment | added | Timothy Budd | @CarloBeenakker I suppose in your formula $Y$ is supposed to be the transpose of $X$? If $X$ and $Y$ are independent, the determinant vanishes. | |
| Oct 17, 2023 at 22:58 | history | edited | YCor | CC BY-SA 4.0 |
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| Oct 17, 2023 at 21:22 | comment | added | Joseph Van Name | I formulated and verified a conjecture similar to Conjecture 1 for quaternionic matrices. If $A$ is an $m\times n$-quaternionic matrix, then let $\chi_A$ be the corresponding $2m\times 2n$-complex matrix. Then whenever $A$ is a random $n\times n$-quaternionic matrices, we experimentally have $E(\det(\chi_A))=n!$. Here, the entries $v$ in $A$ are Gaussian with independent with mean $0$ and where if $\alpha$ is a unit quaternion, then $\text{Var}(\text{Re}(\alpha v))=1/4$. | |
| Oct 17, 2023 at 20:50 | history | edited | Carlo Beenakker | CC BY-SA 4.0 |
somewhat more informative title
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| Oct 17, 2023 at 20:48 | comment | added | Carlo Beenakker | interesting conjecture; a similar identity, $\mathbb{E}[\det(X+Y)]=(-1)^m (2m)!/m!$ for $n=2m$, was proven by M.K. Chung | |
| Oct 17, 2023 at 19:31 | history | asked | loup blanc | CC BY-SA 4.0 |