Let $G$ be a non-directed graph with non negative weights. Prove that the multiplicity of the eigenvalue $0$ of $L_s$ is the same as the number of convex components $A_1,\dots, A_k$ of the graph. And the subspace associated to the eigenvalue $0$ is generated by the vectors $D^{1/2}1_{A_i}$.
Any hins on how to start the problem?
Essentially, this comes down to the following linear algebraic identity. Let $f, g$ be two functions $V \to \mathbb R$ satisfying $g = D^{1/2} f$. Then $$\langle g, L_sg\rangle = \langle f, L f\rangle = \sum_{uv \in E} (f(u) - f(v))^2.$$
This identity is first used as a proof that all the eigenvalues of $L$ (or $L_s$) are nonnegative: the right-hand side of the identity is nonnegative by inspection, so the left must be, also.
Moreover, if $g$ is an eigenvector of the eigenvalue $0$, then $\langle g, L_sg\rangle = 0$, and this identity characterizes the vectors $f$ for which this can occur.