I know that it has been proved on this site that $\operatorname{rank}(A + B) ≤ \operatorname{rank}(A) + \operatorname{rank}(B)$.
What I still want now is an application of this theorem or the method of proving this theorem to show that $\operatorname{rank}(L + J) =1 + \operatorname{rank}(L)$.
L is a Laplacian matrix of a simple graph (V,E), and J is a square matrix whose entries are ones and has the same dimension of L.
I am sure this is not a duplicate.
You know that $rk(L+J)\leq 1+rk(J)$. We want to prove the equality. Write $$L=D-A,$$ where $A$ is the adjacency matrix and $D$ the diagonal matrix whose diagonal is given by the degrees of the vertices.
Moreover define the hyperplane
$$H=\left\{v\in \mathbb R^n|\sum v_i=0 \right\}.$$
Observe that $$H=\ker(J)$$ I would do it in three steps
1) $\ker(L+J)\cap H \subseteq \ker (L)$.
2) $\ker(L+J)\subseteq H$ (Hence $\ker(L+J)\subseteq \ker(L)$).
3) $\exists u\in \ker(L)\smallsetminus \ker(L+J)$.
Proof of 1: If $v\in\ker(L+J)\cap H$, then $$0=(L+J)v=Lv.$$
Proof of 2: If $v\in \ker(L+J)$ then $$0=v^t(L+J)v=v^tLv+\left(\sum v_i\right)^2$$ Since $L$ is symmetric positive-semidefinitive, it must be that both $$v^tLv=\left(\sum v_i\right)=0,$$ therefore $v\in H$.
Proof of 3: Just pick $v=(1,1,\ldots,1)^T$. We have $Lv=0$ and $(L+J)v=nv$.
CONCLUSION: $$\dim \ker(L+J)+1\leq \dim \ker(L)$$ and so $$rk(L+J)\geq 1+rk(L)$$