## Eigenvalues of directed Laplacian matrix $L$ and $DL$, where $D$ is a diagonal matrix with positive entries

I have a weighted Laplacian matrix $L$ of a strongly connected directed graph and a diagonal matrix $D$ with positive entries. Since the graph is directed, $L$ is non-symmetric real. Further, since the graph is strongly connected, $L$ has a simple zero eigenvalue and all its nonzero eigenvalues have positive real part. Is it possible to establish a relation e.g., a bound, between the eigenvalues of $L$ and those of the product $DL$?

Thanks a lot!

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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 $0,a+b$. – Delio Mugnolo Mar 3 at 17:10
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? – Delio Mugnolo Mar 3 at 17:57
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 $\sigma(DL)\leq max_i(D_{ii})\sigma(L)$. – FS Mar 3 at 18:12
I see. So my example would indeed fit your scheme. – Delio Mugnolo Mar 3 at 18:31
Yes, absolutely. – FS Mar 3 at 19:47
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UPDT: Let $G$ be a weighted undirected graph with Laplacian matrix $L$. Let $D$ be a positive diagonal matrix. Let $d=min(diag(D))$ and let $\Delta$ be the maximum diagonal entry of $L$. Let $i$ be the weighted isoperimetric number of $G$. Then: $$\lambda_{2}(DL) \geq d (\Delta-\sqrt{\Delta^{2}-i^{2}})$$