no, because addition and scalar multiplication do not impose any mixing on the matrix elements. Consider the 5 x 5 matrix with a in the first 4 positions on the diagonal, c is the last position, b at positions (4,5) and (5,4) and zeros elsewhere. Taking (a, b, c) = (1, 0, 1), (1, 0, 0), (0, 1, 0), (0, 0, 1), (0, 0, 0) gives matrices of rank 5, 4, 2, 1, 0 respectively. If a = 0, then all the entries in rows 1, 2 and 3 are zero, so the rank is at most 2. If a is non-zero, then there is certainly a 4 x 4 unit submatrix, so the rank is at least 4. Thus no member of the set has rank 3. If we add two columns of zeros to every member of the set, then we get a counter-example for the 5 x 7 case.