Timeline for Eigenvalues of directed Laplacian matrix $L$ and $DL$, where $D$ is a diagonal matrix with positive entries
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
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| Mar 4, 2013 at 0:21 | answer | added | Felix Goldberg | timeline score: 1 | |
| Mar 3, 2013 at 20:17 | comment | added | Delio Mugnolo | If you were considering the adjacency matrix (and hence you could apply Perron-Frobenius' theorem, I would see how to work this out using Gelfand's theorem. But in this way... On the other hand, I do not see how to use strong connectedness. There is a result of Chung (btw: are you using the normalized Laplacian or the usual one?) that yields an upper bound on the diameter using some spectral information - but conversely? | |
| Mar 3, 2013 at 19:47 | comment | added | user31905 | Yes, absolutely. | |
| Mar 3, 2013 at 18:31 | comment | added | Delio Mugnolo | I see. So my example would indeed fit your scheme. | |
| Mar 3, 2013 at 18:12 | comment | added | user31905 | Hi Delio, thanks for the reply. Let $\sigma(L)$ respectively $\sigma(DL)$ denote the spectrum of $L$ respectively $DL$. Let $D_{ii}$ denote the positive diagonal entries of $D$. I am hoping for an estimate of the form $\sigma(DL)\leq max_i(D_{ii})\sigma(L)$. | |
| Mar 3, 2013 at 17:57 | comment | added | Delio Mugnolo | Sorry, I meant −1 on the off-diagonal entries of L, but my computations still hold. So, do you possibly want to make more precise what kind of estimates are you hoping for? | |
| Mar 3, 2013 at 17:10 | comment | added | Delio Mugnolo | hardly, it seems. Think of the Laplacian of the oriented 2-cycle, i.e., $L:=\begin{pmatrix}1 & 1\\ 1 & 1\end{pmatrix}$ and take any diagonal matrix $D:=\begin{pmatrix}a & 0\\ 0 & b\end{pmatrix}$. Then $L$ will always have eigenvalues $0,2$, but the eigenvalues of $D\cdot A$ are $0,a+b$. | |
| Mar 3, 2013 at 14:12 | history | edited | user31905 | CC BY-SA 3.0 |
added 5 characters in body
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| Mar 3, 2013 at 14:03 | history | asked | user31905 | CC BY-SA 3.0 |