# 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!

• 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 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? Mar 3, 2013 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 $\sigma(DL)\leq max_i(D_{ii})\sigma(L)$. Mar 3, 2013 at 18:12
• I see. So my example would indeed fit your scheme. Mar 3, 2013 at 18:31
• Yes, absolutely. Mar 3, 2013 at 19:47

UPDT: Let $$G$$ be a weighted undirected graph with Laplacian matrix $$L$$. Let $$D$$ be a positive diagonal matrix. Let $$d=\min(\operatorname{diag}(D))$$ and let $$\Delta$$ be the maximum diagonal entry of $$L$$. Let $$i$$ be the weighted isoperimetric number of $$G$$. Then: $$\lambda_{2}(DL) \geq d \left(\Delta-\sqrt{\Delta^{2}-i^{2}}\right)$$