Timeline for Explicit expression for determinant
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Aug 30, 2011 at 8:22 | comment | added | Dan Fox | I've not deleted my misguided comment only so that other comments remain understandable. | |
| Aug 30, 2011 at 8:19 | comment | added | Dan Fox | Noam Elkies is right, I misread the equation. Sorry for creating confusion. | |
| Aug 30, 2011 at 7:35 | comment | added | fabee | Hmm, I was afraid so. Thanks for your thoughts anyway. | |
| Aug 29, 2011 at 19:18 | comment | added | Noam D. Elkies | Sorry, I (and apparently Dan) must have misread $w_{ij}$ as $w_i w_j$. As it stands, the matrix seems to be so general that there'll be nothing simpler than using general determinant formulas. | |
| Aug 29, 2011 at 18:08 | comment | added | Robert Israel | It's true that $J y = \sigma^2 y$, but I don't see anything useful about the action on vectors annihilated by $y_i$. | |
| Aug 29, 2011 at 14:07 | comment | added | fabee | I am not sure whether I can follow your answers. If my matrix had the form $\sigma^2I+J_1$ with rank one $J_1$ I would use Sylvester's determinant theorem and be done. However, in my case both matrices can have full rank: the first one is a diagonal matrix and the second is the elementwise product of an outer product with the matrix $W$. | |
| Aug 29, 2011 at 13:02 | comment | added | Noam D. Elkies | Dan Fox's technique works because the matrix is $\sigma^2 I + J_1$ where $I$ is the identity matrix and $J_1$ is of rank 1: in general if $M,M_1$ are square matrices of the same size and $M_1$ has rank 1 then $\det(M+xM_1)$ is a polynomial in $x$ of degree at most 1. | |
| Aug 29, 2011 at 12:53 | comment | added | Dan Fox | The eigenvalues can be found explicitly. Consider separately how $J_{ij}$ acts on the vector $y^{i}$ Euclidean dual to $y_{i}$ and how it acts on space of vectors annihilated by $y_{i}$. | |
| Aug 29, 2011 at 12:41 | history | asked | fabee | CC BY-SA 3.0 |