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The normalized graph Laplacian fits the relationship $L=D^{-1/2}(D-A)D^{-1/2}=I-D^{-1/2}AD^{-1/2}$, where $I$ is the identity matrix, $D^{-1/2}$ is the diagonal matrix with $D(i,i)=\frac{-1}{\sqrt{n_{i}}}$, and $A$ is the $n \times n$ adjacency matrix. The Laplacian then takes the form:

$L(i,j)=1$ if $i=j$, and $\frac{-1}{\sqrt{n_{i}n_{j}}}$ if $i\neq j$.

Why, for $i \neq j$, is it not $\frac{-1}{ n_{i}n_{j}}$?

Thanks for any help.

Zach Langley
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user607205
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  • I'm not sure that there is a good reason other than "it works nicely". Both are valid descriptions of some linear operator. Some notes that develop theory using this definition of the Laplacian are here. – Joppy Oct 23 '18 at 02:08
  • OK, thanks. Can you point towards somewhere that shows how we get this operator from the relationship above? (Chung's book glosses over this.) – user607205 Oct 23 '18 at 14:25

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