I’m reading a paper and they make a claim about the eigenvector of a graph Laplacian that seems wrong and I wanted to check with the larger community if I’m missing something obvious.
Let $G = (V,E,W)$ be a weighted undirected graph with $|V|=n$ and $W$ being the weighted adjacency matrix if $u$ is the vector of all ones then the degree vector is $(d_1, d_2, \dots, d_n)= d=Wu$. Denoting by $\sqrt{d}$ the element wise square root $\sqrt{d}=(\sqrt{d_1},\dots, \sqrt{d_n})$ then we can normalize the adjacency matrix to get $\tilde{W} = \operatorname{diag}(\sqrt{d})W\operatorname{diag}(\sqrt{d})$.
The claim is that the leading eigenvector associated to the largest eigenvalue is
$$v = \frac{\sqrt{d}}{\|\sqrt{d}\|_2} = \frac{d}{\|d\|_1}$$
My issue is the equality
$$\frac{\sqrt{d}}{\|\sqrt{d}\|_2} = \frac{d}{\|d\|_1}$$
This seems like it is obviously false. If we have the degree vector $d=(2,3,2,1)$ then $$\frac{\sqrt{d}}{\|\sqrt{d}\|_2} = \frac{1}{\sqrt{7}}(\sqrt{2},\sqrt{3},\sqrt{2},1)$$ and $$\frac{d}{\|d\|_1} = \frac{1}{7}(2,3,2,1)$$ which are not equal after clearing radicals $\frac{\sqrt{2}\sqrt{7}}{7}\not=\frac{2}{7}$.
Is this an error by the authors or am I missing something?