Here's now an idea how I could show it -- but actually I think, this is not totally correct
Let's look at the rows of (A B). Since all rows of A are linearly independent, we need a linear combination of n+1 rows of M such that we obtain a vector
v_2 is unequal 0 at all entries since the rows of B are linearly independent of the rows of A. (IS THIS TRUE?)
There exist further r-1 rows of B which are linearly independent of (0 v_2).
Hence rank(A B) = 2n.
Is this proof correct?? I am not happy with this proof at all...