TITLE: Rank product of matrix compared to individual matrices.
QUESTION [18 upvotes]: Possible Duplicate:
How to prove $\text{Rank}(AB)\leq \min(\text{Rank}(A), \text{Rank}(B))$? 


If $A$ is an $m\times n$ matrix and $B$ is a $n \times r$ matrix, 
  prove that the rank of matrix $AB$ is at most $\mathrm{rank}(A)$.

I asked a similar question earlier phrased incorrectly. The above is closer to the actual question generalised.

REPLY [7 votes]: $$\mathcal{C}(A) = \{Ax | x \in \mathbb{R}^n\}$$
$$\mathcal{C}(AB) = \{ABy | y \in \mathbb{R}^r\}$$
Notice how $By$ is an n by 1 vector. 
Assume to the contrary that the rank of the latter is strictly greater than the first. Then there exists some $y \in \mathbb{R}^m$ that is in $\mathcal{C}(AB)$,  but not $\mathcal{C}(A)$. This implies $\exists z \in \mathbb{R}^r$ such that 
$$
ABz = y.
$$ 
But since $Bz$ is an n by 1 vector, this $y$ will also necessarily lie in $\mathcal{C}(A)$. This is a contradiction.