Timeline for Rank inequality for spectral measures
Current License: CC BY-SA 4.0
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jan 17, 2022 at 18:52 | comment | added | Fei Cao | @ChristianRemling Yes I really want a essentially different argument than the argument based on Cauchy's interlacing theorem, if your approach more or less used the interlacing result that is probably not the approach I want to see... | |
| Jan 17, 2022 at 18:41 | comment | added | Christian Remling | @ShannonStarr: Of course, in another sense you cannot escape the interlacing principle because the statement the OP wants proved is essentially a rephrasing of it. | |
| Jan 17, 2022 at 18:36 | comment | added | Christian Remling | @FeiCao: If $V,W$ are subspaces with $\dim V>\dim W+k$, then $\dim V\ominus W>k$. | |
| Jan 17, 2022 at 18:34 | comment | added | Christian Remling | @ShannonStarr: I just said "min-max" for ease of reference. What I need here is that if the quadratic form is $\le x$ on a $d$-dimensional subspace, then there are at least $d$ eigenvalues in $(-\infty,x]$. This is a very easy exercise. | |
| Jan 17, 2022 at 15:48 | comment | added | Shannon Starr | The min-max principle is how you prove the interlacing theorem. | |
| Jan 17, 2022 at 1:36 | comment | added | Fei Cao | May I know why "$B$ had $>m+k$ eigenvalues there, then we could find $>k$ orthogonal vectors $R(E_B(-\infty,x])$ (with $E$ denoting the spectral projection) that are also orthogonal to (the $m$-dimensional space) $R(E_A(-\infty, x])$"? It will be better if you can explain your notations in more detail as well. | |
| Jan 16, 2022 at 23:30 | history | undeleted | Christian Remling | ||
| Jan 16, 2022 at 23:28 | history | deleted | Christian Remling | via Vote | |
| Jan 16, 2022 at 23:26 | history | answered | Christian Remling | CC BY-SA 4.0 |