Let $A_1$ be an $m \times n_1$ matrix, $B_1$ an $m \times n_1$ matrix and $A_2$ an $m \times n_2$ matrix such that,
- $\text{Rank}(A_1)=\text{Rank}(B_1)=n_1$,
- $\text{Rank}(A_1:A_2)=n_1+n_2$, where $(A_1:A_2)$ is the matrix that concatenates $A_1$ and $A_2$.
Show that if $\text{Rank}(A_1:B_1)=n_1$ then $\text{Rank}(B_1:A_2)=n_1+n_2$.
Edit 1: The original question didn't make sense so I thought I'd modify it (rather than deleting and posting a new question. Not sure if this is the right thing to do or if it's good to have my original question). I also think I know the solution and will post later if no-one wants to post an answer.
Edit 2: Here's more context. The problem I'm working on has the assumptions:
- $\text{Rank}(A_1)=\text{Rank}(B_1)=n_1$,
- $\text{Rank}(A_1:A_2)=\text{Rank}(B_1:B_2)=n_1+n_2$
I need to have $\text{Rank}(A_1:B_2)=n_1+n_2$ but I can only impose conditions (preferably rank conditions) on the relationships between $(A_1,A_2)$, $(B_1,B_2)$, $(A_1,B_1)$ and $(A_2,B_2)$... obviously I would like the additional conditions to be as "mild" / general as possible.
The one I thought about was the additional assumption $\text{Rank}(A_1:B_1)=n_1$ but I'm now having problems proving that this additional condition implies that $\text{Rank}(A_1:B_2)=n_1+n_2$.
I'll show that with the assumption $\text{Rank}(A_1:B_1)=n_1$ then $\text{Rank}(B_1:A_2)=n_1+n_2$. From $\text{Rank}(A_1:B_1)=n_1$ and $\text{Rank}(A_1)=\text{Rank}(B_1)=n_1$, we can express each column of $B_1$ as $$ B_{1i}=\sum_{j=1}^{n_1} a_{ij} A_{1j} \qquad (*) $$ so $$ B_1 = \begin{pmatrix} a_{11} & \ldots & a_{1n_1} \\ \vdots & \ddots & \vdots \\ a_{n_11} & \ldots & a_{n_1n_1} \end{pmatrix} A_1 $$ Similarly we can express $A_1$ in terms of $B_1$ so we must have that the above matrix is invertible.
Now in order for $\text{Rank}(B_1:A_2)=n_1+n_2$ we need to show that the only solution to the following equation is $b_i=c_i=0$ for all $i$. $$ \sum_{i=1}^{n_1} b_{i} B_{1i} + \sum_{i=1}^{n_2} c_{i} A_{2i}=0 $$ Substituting in $(*)$, $$ \sum_{j=1}^{n_1} \sum_{i=1}^{n_1} b_{i} a_{ij} A_{1j} + \sum_{i=1}^{n_2} c_{i} A_{2i}=0 $$ Now we know that $\text{Rank}(A_1:A_2)=n_1+n_2$ so, $$ \sum_{i=1}^{n_1} b_{i} a_{ij}=0 \qquad j=1,\ldots n_1 $$ Hence, $$ \begin{pmatrix} a_{11} & \ldots & a_{1n_1} \\ \vdots & \ddots & \vdots \\ a_{n_11} & \ldots & a_{n_1n_1} \end{pmatrix} \begin{pmatrix} b_1 \\ \vdots \\ b_n \end{pmatrix} =0 $$ Now because the above matrix is invertible we must have $b=0$ so $\text{Rank}(A_1:B_2)=n_1+n_2$.