Let $G$ be a simple graph with adjacency matrix $A(G)$. Let $v$ be a cut-vertex of $G$. Let $G\_1, G\_2,\dots, G\_k$ be the connected components of the induced graph $G-v$ ( the subgraph resulting after the removal of $v$ and its incident edges from $G$). Assume that the rank of $A(G\_1)$, denote by $r(G\_1)$ satisfies $r(G\_1+v)=r(G\_1)+2.$ Then prove the following :
$$r(G)=\sum\_{i=1}^kr(G\_i)+2=r(G-v)+2.$$
This is given in Theorem 2.3 ([here](https://www.sciencedirect.com/science/article/pii/S0024379511004897)), but I am not getting the proof.
I am not able to understand the following. Assume that there are only 2 connected components in $G-v$, let these are $G\_1, G\_2$. Then 
$$A(G)=\begin{bmatrix} 0 & A(v,G\_1) & A(v,G\_2) \\ A(v,G\_1)^T & A(G\_1) & O \\ A(v,G\_2)^T & O & A(G\_2)\end{bmatrix}.$$ I understood that under given hypothesis, the vector $[0 \ A(v,G\_1)]$ is independent of relevant vectors in the matrix $[A(v,G\_1)^T \ A(G\_1)]$. But how does this lead to that the first row of $A(G)$ is also independent of the relevant rows of the matrix $[A(v,G\_2)^T \ O \ A(G\_2)]$.



Rank of the connected components
The proof given in the article that you mention relies on the fact that when you stack a row vector $X$ at the top of a matrix $M$, the rank increases by at most $1$ and does so if and only if $X$ is linearly independent of the rows of $M$. The same goes for column vectors stuck to the left (or anywhere) of $M$.


Hence if you have an $n\times n$ matrix $M$ and a submatrix $N$ of size $(n-1)\times (n-1)$, the rank of $M$ is at most $\operatorname{rk}(N)+2$ and this bound is achieved if and only if the $1\times (n-1)$ row and the $n\times 1$ column that you successively added to $N$ are independent of the relevant sets of rows and columns.


Besides this, they use the rank-nullity theorem and I can't see any other tricky points.


**UPDATE**



> 
> how does this lead to that the first row of $A(G)$ is also independent of the relevant rows of the matrix $[A(v,G\_2)^T \ O \ A(G\_2)]$.
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That is because of the zeroes underneath $A(G\_1)$, any writing of $A(v,G)$ as a linear combination of the other rows of $A(G)$ would give you in particular $A(v,G\_1)$ as a linear combination of the rows of $A(G\_1)$. There is nothing here.


