Rank of a block-triagonal matrix

Question

Given a matrix $C=\left [ \begin{matrix} A & 0 \\ B & A \end{matrix} \right ]$, where rank(A+B)=rank(B), and rank(B)>rank(A), does rank(C)=rank(A)+rank(B) hold?

A,B are Laplacian matrices.

Answer 1

First, if $A$ is a $M$-matrix then it is invertible and $\mathrm{rank}(B) > \mathrm{rank}(A)$ is impossible.

Here is a counter-example, if $A = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}$ and $B = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$ then:

• $B$ is a $M$-matrix;
• $\mathrm{rank}(A+B)=\mathrm{rank}(B)=2$;
• $ 2 = \mathrm{rank}(B) > \mathrm{rank}(A) = 1$;
• $ \mathrm{rank}(C) = \mathrm{rank}  \begin{pmatrix} 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 1 \\ 0 & 1 & 0 & 0 \end{pmatrix} = 2 < \mathrm{rank}(A) + \mathrm{rank}(B)$.