$A$ is an $m\times n$ matrix and $B$ is an $n\times p$ matrix, show that $\text{rank}(AB)\le\text{rank}(A)$.

Question

The problem is asking a proof for $\text{rank}(AB)$ is smaller or equal to $\text{rank}(A)$. Given the conditions $A$ is an $m\times n$ matrix and $B$ is an $n\times p$ matrix. Any idea about the proof? Many thanks!

Answer 1

Let $w \in$ Range$(AB)$, then $w = AB(v)$ for some $v \in \mathbb{F}^p$, hence $w = A(B(v)) \in$ Image$(A) \implies $ Range$(AB) \subset$ Range$(A) \implies $ rank$(AB) \leq$ rank$(A)$ as needed.

Answer 2

The rank of a matrix is the dimension of its image. If the linear mapping with matrix $A$ maps $\mathbb{R}^n$ to $\mathbb{R}^m$, how can the composition of the lineair mappings B and A have an image with a dimension larger than m?