How to prove $\operatorname{rank}{(AB)} = \operatorname{rank}{A}$?

Question

How to prove $\operatorname{rank}{(AB)} = \operatorname{rank}{A}$ given that $A$ is any $m \times n$ matrix, and $B$ is is any $n\times k $ matrix where $\operatorname{rank}{(B)}=n$ ?

It is a property of rank, written on Wikipedia.

EDIT

I'm sorry there is a typo, it should be $\operatorname{rank}{(AB)} = \operatorname{rank}{A}$ but I wrote $\operatorname{rank}{(AB)} \le \operatorname{rank}{A}$

Answer 1

Let $V \to_B W \to_A U$ be the composition of two linear maps $A \circ B:V \to U$, otherwise known as $AB$ (multiplying matrices.) Notice that $\mathrm{Rank}$ is just the dimension of the image. From this, notice that $\ker B \subset \ker A \circ B$, since if $B(v)=0$, then certainly $A \circ B(v)=0$. It follows that $\dim \ker B \leq \dim \ker A \circ B$, and rank nullity implies the result.

Answer 2

If you view the matrices as linear maps, then the matrix product correspond to the composition $A \circ B$. Let $V = \mathrm{img}(B)$ be a subspace of $\mathbb{K}^n$. Then the rank of $AB$ is equal to the rank of $A$ restricted to $V$ $$ \mathrm{rank}{(AB)} = \mathrm{rank}{(A|_V)} $$

But since $\mathrm{rank}{(B)} = n$, then $V = \mathbb{K}^n = \mathrm{dom}{A}$. Hence, $A|_V = A$ and you have your result.

edit: Sorry I misread the property you wanted to prove. Well the reasonning is similar to prove $\mathrm{rank}{(AB)} \leq \mathrm{rank}{(A)}$ for any $B$.

Answer 3

Quite simple: if $K$ is the base field, interpret matrix $A$ as the matrix of a linear map $u\colon K^n\longrightarrow K^m$, matrix $B$ as the matrix of a linear map $v\colon K^k\longrightarrow K^n$ and $AB$ as the matrix of the composition $u\circ v\colon n K^k\longrightarrow K^m$.

The rank of $A$ is the rank of $u$, i.e. $\;\DeclareMathOperator{\im}{Im}\dim(\im u)$, the rank of $B$ is $\;\dim(\im v)$ and the rank of $AB$, $\;\dim(\im(u\circ v))$. Now $$\dim(\im(u\circ v))\le \dim(\im u)\quad\text{since}\enspace \im(u\circ v\subset \im u.$$

Answer 4

Since $\text{rank}(B) = n$ it means that $\text{min}(n,k) = n$.

$$ \text{rank}(AB) = \text{min}(\text{rank}(A), n) = \text{min}(\text{min}(m,n), n). $$

$$ \text{rank}(AB) \leq \text{rank}(A)\\ $$ This can be re-written as $$ \text{min}(\text{min}(m, n),n) \leq \text{min}(m,n), $$

This can be true for any case 1) $m<n$, 2)$m = n$, 3) $m>n$. But I think this is not inequality but equal, $\text{rank}(AB) = \text{rank}(A)$