It's well-know that if $\mathbf{X}\in \mathbb{R}^{n \times k}$ then dim(range($\mathbf{X}\theta$))=rank$(\mathbf{X})$ for $\theta \in \mathbb{R}^k.$
1) I was wondering to know dim(range($\mathbf{X}\mathbf{A}$)) for $\mathbf{A}\in \mathbb{R}^{k \times d}$. I guess that the answer is rank$(\mathbf{X}).d$
Because we can expand the multiplication as sum of $k$ parts which are independent of each other and each part has dimension of $d$. Note that we exploit this fact that elements of matrix $A$ can be any arbitrary numbers in $\mathbb{R}$.
2) My main question is about the comparison between dim(range($\mathbf{X}\mathbf{A}\mathbf{B}$)) for $\mathbf{A}\in \mathbb{R}^{k \times d}$ and $\mathbf{B}\in \mathbb{R}^{d \times p}$ as well as dim(range($\mathbf{X}\mathbf{M}$)) for $\mathbf{M}\in \mathbb{R}^{k \times p}$.
I think that dim(range($\mathbf{X}\mathbf{M}$))=rank$(\mathbf{X}).p$ but dim(range($\mathbf{X}\mathbf{A}\mathbf{B}$))$<$rank$(\mathbf{X}).p$ Since rank($\mathbf{A}\mathbf{B}$)$<$min(k,d,p)