The problem
How does one show that $\text{null}(A) \subseteq \text{null}(BA)$? is it even true for all $A$ and $B$?
My understandings
Please correct me if I'm wrong.
By definition, null space of any matrix $M$, also $\text{null}(M)$, is the set of all of the answers for $Ax=0$. This equation has one obvious answer where $x=0$, that's also the same thing in saying that linear transformations keep/do not move the origin.
So, I can know that null space of matrices is at least zero.
Now, for arbitrary $m \times n$ matrices $A$ and $B$, we can say $\text{null}(A) > 0$ and $\text{null}(B) > 0$, so $\text{null}(BA) > 0$.
Where I need help
I suspect that I can say $BA$ is a linear transformation by $B$ on $A$.
Now, since $B$ might or might not nullify things that $A$ spanned, and since $A$ itself is can be thought of as a linear transformation on the identity matrix, and $A$ might or might not have nulled the space spanned by those unit vectors, $\text{null}(BA) \ge \text{null}(A)$, because at least everything that $A$ has nullified is still nullified in $BA$ because of multiplication by zero, and $B$ might or might not have nulled more stuff.
Let $x\in\ker A$ so $Ax=0$. Then $BAx=0$ and hence $x\in\ker BA$.
This works even for non-square matrices, as long as all the matrices/vectors have compatible dimensions.