Eigenvector of a complete graph Laplacian

Question

Can somebody help me prove why $v=\begin{bmatrix} 1 \dots 1\end{bmatrix}^T$ is the eigenvector of every complete graph Laplacian matrix?

Thanks!

Answer 1

Recall that the definition of a graph Laplacian is $L=D-A$, where $D$ is the degree matrix and $A$ is the adjacency matrix.

Now $Lv$ is a vector whose $i^\text{th}$ component is the sum of entries in the $i^\text{th}$ row in $L$. But the sum of entries in the $i^\text{th}$ row of $D$ is $\deg(v_i)$ by definition, as is the sum of entries of the $i^\text{th}$ row of $A$, because this counts all vertices that $v_i$ is adjacent to, (with multiplicity of edges if the graph is not simple).

So these cancel out, leaving $0$, and hence $Av = 0v$.