Rank and the column space of matrix product If $A$ and $B$ be matrices for which the product $AB$ makes sense,
and the rank of $AB$ equals the rank of $A$. Is the column
space of $AB$ equals the column space of $A$?
 A: Clearly ${\cal R} (AB) \subset {\cal R} A$, and we are given $\dim {\cal R} (AB) = \dim  {\cal R} A$.
Suppose we have two finite dimensional subspaces $S_1 \subset S_2$ and there exists some $w \in S_2 \setminus S_1$, then we must have $\dim S_1 < \dim S_2$. 
To see this, suppose $v_1,...,v_r$ is a basis for $ S_1$, then I claim that $w,v_1,...,v_r$ are linearly independent in $S_2$. Suppose $\sum_i \alpha_i v_i +\beta w = 0$. Then we have $\beta w = \sum_i (-\alpha_i) v_i$. If $\beta \neq 0$, we would have $w \in S_1$, which would be a contradiction, so $\beta = 0$. It follows that $\alpha_i = 0$ since they form a basis for $S_1$.
Hence $\dim S_2 \ge \dim S_1 +1$.
So, if $S_1 \subset S_2$, and $\dim S_1 = \dim S_2$, then we must have $S_1 = S_2$.
If we let $S_1 = {\cal R} (AB)$, and $ S_2 = {\cal R} A$, we get the desired result.
A: Obviously, column space of $AB$ is a subspace of the column space of $A$. Since the $AB$ and $A$ have the same rank, i.e., their column spaces have the same dimension, what do we conclude?
