Let $k$ be a field. Let $A,B\in k^{m\times n}$ and $$C\_i=\pmatrix{A&B&&&\\&A&B&&\\&&\ddots&\ddots&\\&&&A&B}\in k^{im\times(i+1)n}.$$ Then how does the rank of $C\_i$ change with $i$? I conjucture that, for $i$ large enough, there exist non-negative integers $d,s$ such that the following equality holds: $$\mathrm{rank}(C\_i)=id+s.$$
**Edit:** Ilya Bogdanov gave a positive answer to the above conjucture. If we let $$C\_i=\pmatrix{A\_1&A\_2&\cdots&A\_s&&\\&A\_1&A\_2&\cdots&A\_s&&\\&&\ddots&\ddots&\ddots&\ddots\\&&&A\_1&A\_2&\cdots&A\_s}\in k^{im\times(i+s-1)n},$$ where $A\_1,\dots,A\_s\in k^{m\times n}$. Is the conjucture still valid?



How does the rank of $C_i$ change with $i$?
Your conjecture is true. Let us apply Gauss' elimination method with some changes. 


First, apply it to the first $(A\; B)$, among others, you obtain several rows containing nonzero values only in the $B$-area, with ones at some positions; let the fragments of these rows in the $B$-area span some subspace $U\_1$. 


Now you may vanish the elements of the next $A$ in these positions and apply Gauss' method to the rest of the second $(A\; B)$. You again obtain some rows nonzero only at the (second) $B$-area, and they will span some subspace $U\_2$ containing $U\_1$. 


Acting so, in each next $A$ you will vanish (non-strictly) more positions than in the precious one, so you get a chain of subspaces $U\_1\subseteq U\_2\subseteq \dots$. After it has been stabilized, everything will work in the same way, so the rank of obtaines matrix will gain a constant increment at each step.


