Timeline for Smallest eigenvalues of block Kronecker product
Current License: CC BY-SA 4.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Feb 18, 2019 at 18:25 | comment | added | JKay | now I see it! thank you again | |
| Feb 18, 2019 at 18:21 | comment | added | LeechLattice | As $L$ has size $2n^2×n^2$, $LL^T$ has rank at most $n^2$. But $LL^T$ is $2n^2$ dimensional, so it must have nontrivial null space. It follows that the least eigenvalue is $0$. | |
| Feb 18, 2019 at 18:17 | comment | added | JKay | hm. for $d=64$ it was $-0.000000000073413774739704132619464336358199$ in vpa mode. but our matrix is symmetric so probably it would be zero as you said @Bullet51 :) thanks for your suggestion! | |
| Feb 18, 2019 at 18:07 | comment | added | LeechLattice | It could be a numerical error. What about trying in vpa mode? | |
| Feb 18, 2019 at 18:07 | history | edited | JKay | CC BY-SA 4.0 |
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| Feb 18, 2019 at 18:05 | comment | added | JKay | @Bullet51 No. I tried with small $d$ in matlab. The eigenvalue is small but strictly positive (around $10^{-6}$) and tends to increase as $d$ increase. Since in my problem $d$ is $512$ or even $1024$, I think that this value is not that small | |
| Feb 18, 2019 at 17:58 | comment | added | LeechLattice | Zero? | |
| Feb 18, 2019 at 17:28 | history | asked | JKay | CC BY-SA 4.0 |