Timeline for Dimension (manifold) of matrices with exact $r$ positive and $r$ negative eigenvalues
Current License: CC BY-SA 4.0
7 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Aug 17, 2018 at 14:06 | vote | accept | Alpha001 | ||
| Aug 17, 2018 at 14:06 | vote | accept | Alpha001 | ||
| Aug 17, 2018 at 14:06 | |||||
| Jul 8, 2018 at 0:53 | comment | added | Peter Shor | @Alpha001: Hermitian matrices are not always symmetric matrices. The dimensionality of the space of symmetric matrices is 6. For Hermitian matrices, it's 9. | |
| Jul 5, 2018 at 15:03 | comment | added | Alpha001 | @IgorRivin : Look at the example $n=3$ and $r=2$ then your formula results in a dimension of $2nr-r^2 = 8$ but the space of symmetric matricies is only of dimension $\frac{n(n+1)}{2}=6$? Since the manifold is subset of the space of symmetric/hermitian matricies something went wrong? | |
| Jul 4, 2018 at 10:21 | comment | added | Alpha001 | But this is has the same dimension like in the case of $M_{2r}$? Could this be since we did not assume that the matricies in $M_{2r}$ are hermitian. | |
| Jul 4, 2018 at 1:22 | comment | added | Christian Remling | This is a nice simple approach, but it's not exactly what the OP is asking. Also, I'm not sure I agree with the dimension count of the $U$'s, for example the unit vectors (or $n\times 1$ matrices $U$ as above) have real dimension $2n-1$, not $2n-2$. | |
| Jul 3, 2018 at 23:53 | history | answered | Igor Rivin | CC BY-SA 4.0 |