Timeline for Is there an improvement for the Schur-Horn inequalities for positive semi-definite matrices?
Current License: CC BY-SA 3.0
15 events
| when toggle format | what | by | license | comment | |
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| Nov 26, 2014 at 20:42 | comment | added | user6818 | @AllenKnutson hmmm..so I asked a separate question about that - mathoverflow.net/questions/188172/… | |
| Nov 26, 2014 at 20:26 | comment | added | Allen Knutson | If your matrix is special then it's unlikely that Schur-Horn gives the best possible results. | |
| Nov 26, 2014 at 20:15 | comment | added | user6818 | @AllenKnutson If $A$ is a $\pm 1$ signed graph Laplacian or an adjacency matrix ? | |
| Nov 26, 2014 at 14:46 | comment | added | Allen Knutson | Consider the case $A$ real diagonal, and realize, "No." | |
| Nov 25, 2014 at 22:35 | comment | added | user6818 | especially if p.s.d nature is assumed.. | |
| Nov 25, 2014 at 22:10 | comment | added | user6818 | @AllenKnutson Okay - but does Schur-Horn say anything stronger about $A_{ii}$ than this? Can it be constrained any further than $[\lambda_{min},\lambda_{max}]$? (...for something I was thinking of I have been wondering if there is a constrain which is $O(1/dim(A))$...does something like this exist?..) | |
| Nov 25, 2014 at 21:22 | comment | added | Allen Knutson | Let $\vec v = \sum_j c_j \vec e_j$, where $\vec v$ is the $i$th vector in your orthonormal basis, and the $(e_j)$ are an orthonormal eigenbasis (with eigenvalues $(\lambda_i))$, and $\sum_i |c_i|^2=1$. Then $A_{ii} = \langle \vec v,A \vec v\rangle = \sum_i |c_i|^2 \lambda_i \in [\lambda_{min},\lambda_{max}]$. It's a little weird to credit Schur and Horn with this. | |
| Nov 25, 2014 at 18:34 | comment | added | Student | @AllenKnutson Yes :P But this is not a trivial statement - right? The inequality for $A_{ii}$ that I wrote above comes because of the Schur-Horn inequality - right? | |
| Nov 25, 2014 at 18:10 | comment | added | Allen Knutson | If you're asking what I think you're asking, the answer is "Very obviously yes," enough so that I wonder if I'm misunderstanding the question. | |
| Nov 25, 2014 at 17:34 | comment | added | Student | @AllenKnutson About my second comment - what I mean is this - if $A$ is any Hermitian matrix then can I always say that $ (min-eigenvalue) \leq A_{ii} \leq (max-eigenvalue)$ ? (for any choice of basis) ? | |
| Nov 25, 2014 at 14:19 | comment | added | Allen Knutson | By "improvement", I assumed that the question was "if we impose more conditions, do we get a better result?" i.e. that the matrices were still Hermitian. Then my dumb trick says "no, we don't". As to the second comment: any orthonormal basis will work, giving you the same permutahedron. | |
| Nov 25, 2014 at 7:26 | comment | added | Student | @AllenKnutson In the Schur-Horn inequalities can the diagonal entries be defined in any basis? (..because I am a bit confused by the statement here in equation 9 - terrytao.wordpress.com/2010/01/12/… - where Terence Tao seems to want to choose the "standard basis"..) | |
| Nov 25, 2014 at 6:17 | comment | added | Student | @AllenKnutson I am not getting you - are you saying that any semi-definite matrix is also Hermitian and decomposable in some particular way such that the Schur-Horn can be used on the parts individually? Can you kindly elaborate? | |
| Nov 25, 2014 at 2:49 | comment | added | Allen Knutson | For the first one, can't you just add a multiple of the identity to embed the usual problem into the semidefinite one, i.e. they're the same level of difficulty? | |
| Nov 24, 2014 at 22:33 | history | asked | user6818 | CC BY-SA 3.0 |