Let $\mathbb F$ be a field, and let $r_1, r_2, s_1, s_2$ be positive integers. Consider the matrix $$X:=\left[\begin{array}{cc} A & B \\ C & D \end{array} \right],$$ where $A \in \mathbb F^{r_1 \times s_1}$, $B\in \mathbb F^{r_1\times s_2}$, $C \in \mathbb F^{r_2\times s_1}$ and $D \in \mathbb F^{r_2\times s_2}$.
Q1: Is the following inequality true? $$rank [X] \geq rank \left[\begin{array}{c} B \\ D \end{array} \right]+\max \left\{rank[A\;\; B]- rank[B], rank[C\;\;D]-rank[D] \right\}$$ If so, when does the equality hold?
Q2: Is there some nice formula for $rank[X]$ depending only on the blocks?
Thanks. Alessandro
Edit: The answer to the first question is YES. Is there an elementary proof of it? I can prove it using using results on completions of partial matrices, but I believe there should be an easy way to prove it.
For the equality I suspect that it holds if and only if $$rank \left[\begin{array}{c} A \\ C \end{array} \right]=\max \left\{rank[A],rank[C] \right\},$$ but I'm not sure.
You can arrange row reduction of a block matrix $\left[ \matrix{A\cr B\cr} \right]$ so that you get a set of $\text{rank} \left[\matrix{A\cr B\cr}\right] - \text{rank}(A)$ rows of $B$ that are linearly independent of the rows of $A$, i.e. the only linear combination of these rows of $B$ that is in the row space of $A$ has all coefficients $0$. Taking transposes and interchanging $A$ and $B$, we have a similar result for columns: there is a set of $\text{rank}[A\ B] - \text{rank}(B)$ columns of $A$ that are linearly independent of the columns of $B$. Now for any $C$ and $D$, the corresponding columns of $\left[\matrix{A\cr C\cr}\right]$ are linearly independent of the columns of $\left[\matrix{B\cr D\cr}\right]$.
Thus $\text{rank}\left[\matrix{B\cr D\cr}\right]$ linearly independent columns of $\left[\matrix{B\cr D\cr}\right]$ together with those $\text{rank}[A\ B] - \text{rank}(B)$ columns of $\left[\matrix{A\cr C\cr}\right]$ form a linearly independent set, which says $$\text{rank}(X) \ge \text{rank}\left[\matrix{B\cr D\cr}\right] + \text{rank}[A\ B] - \text{rank}(B)$$ Similarly for $\text{rank}\left[\matrix{B\cr D\cr}\right] + \text{rank}[C\ D] - \text{rank}(D)$.
EDIT: Your suspicion is incorrect in both directions. Try
$$ \eqalign{A = \pmatrix{0 & 1\cr 0 & 1\cr},\ & B = \pmatrix{0 & 1\cr 1 & 1\cr}\cr C = \pmatrix{0 & 1\cr 1 & 1\cr},\ & D = \pmatrix{0 & 0\cr 1 & 1\cr}}$$ Then $X$ has rank $4$, while $\text{rank}\left[\matrix{B\cr D\cr}\right] = 2$, $\max(\text{rank}[A\ B] - \text{rank}(B), \text{rank}[C\ D] - \text{rank}(D)) = 1$, and $\text{rank} \left[ \matrix{A\cr C\cr} \right] = \max\{\text{rank}[A],\text{rank}[C]\} = 2$.
In the other direction, try $$ \eqalign{A = \pmatrix{0 & 0\cr 1 & 1\cr}, \ &B = \pmatrix{1 & 1\cr 0 & 0\cr}\cr C = \pmatrix{0 & 1\cr 0 & 1\cr}, \ &D = \pmatrix{1 & 0\cr 1 & 0\cr}}$$ Here $\text{rank}(X) = 3$, $\text{rank}\left[\matrix{B\cr D\cr}\right] = 2$, $\max(\text{rank}[A\ B] - \text{rank}(B), \text{rank}[C\ D] - \text{rank}(D)) = 1$, but $\text{rank} \left[ \matrix{A\cr C\cr} \right] =2 \ne \max\{\text{rank}[A],\text{rank}[C]\} = 1$.