If nullity$(AB)=0$, then nullity$(B)=0$. I know my question is essentially the same as this one: Show that $nullity(B)\leq nullity(AB)$
but I don't understand this one.
Here's what I know: 


*

*The rank-nullity theorem gives us that rank$(AB)=n$ (assuming $AB$ is $m\times n$).

*Every finite dimensional matrix $Z_{m\times n}$ with rank $k$ has a non-unique rank factorization as $X_{m\times k}\times Y_{k\times n}$.

*rank$(M)=$rank$(M^T)$.


I'm not sure how to compose these facts (or others) into a working proof of this fact.
 A: Hint:
Prove the contrapositive:  if $\ker B\ne  \{0\}$, then $\ker AB\ne \{0\}$.
Variant: interpret this assertion in terms of the linear maps associated with the matrices:
As a linear map is injective if and only its kernel is $\{0\}$, it comes down to the following well-known result on maps  between sets:

If the composition  $g\circ f$ of two maps is one-to-one, the first map (in the order of composition) $f$ is one-to-one.

A: The facts that you give aren't really all that useful for this proof.  It's easier to just use the definition of nullity (the dimension of the null space of the matrix).
Sketch: 


*

*$\operatorname{Nullity}(AB)=0$ means that the dimension of the null space of $AB$ is $0$.  In other words, the only vector such that $AB\vec{v}=\vec{0}$ is $\vec{v}=\vec{0}$.

*Suppose that there is some vector $\vec{w}$ so that $B\vec{w}=\vec{0}$.  Then, $AB\vec{w}=A\vec{0}=\vec{0}$.  So, by the first point $\vec{w}=\vec{0}$.
This shows that the nullity of $B$ is $0$ since only the zero vector satisfies $B\vec{w}=\vec{0}$.
A: You can do this using the rank-nullity theorem, but it's a long way.
Suppose that $A$ is a $m\times p$ matrix and $B$ is a $p\times n$ matrix.  Then $AB$ is an $m\times n$ matrix.  By the rank-nullity theorem, it must be that 
$$
\operatorname{rank}(AB)+\operatorname{Nullity}(AB)=m.
$$
You are given that $\operatorname{Nullity}(AB)=0$, so this implies that 
$$
\operatorname{rank}(AB)=m.
$$
On the other hand, by the rank-nullity theorem applied to $B$, we have that 
$$
\operatorname{rank}(B)+\operatorname{Nullity}(B)=m.
$$
We also know that $\operatorname{rank}(AB)=\operatorname{rank}(B^TA^T)$.  Since $\operatorname{rank}(B^TA^T)$ is the dimension of the column space of $B^TA^T$, we know that the column space of $B^TA^T$ is spanned by vectors of the form
$$
a_{i,1}(B^T)_1+a_{i,2}(B^T)_2+\dots+a_{i,p}(B^T)_p.
$$
Since this is a linear combination of the vectors $(B^T)_i$, these vectors are in the column space of $B^T$, so the column space of $B^TA^T$ is a subspace of the column space of $B^T$.  Therefore, 
$$
\operatorname{rank}(AB)=\operatorname{rank}(B^TA^T)\leq\operatorname{rank}(B^T)=\operatorname{rank}(B).
$$
Therefore, $\operatorname{rank}(B)\geq m$, but, since $B$ has $m$ columns, $\operatorname{rank}(B)\leq m$.  Therefore, $\operatorname{rank}(B)=m$ and so $\operatorname{nullity}(B)=0$. 
