If Ker(A)=$\{\vec{0}\}$ and Ker(B)=$\{\vec{0}\}$ Ker(AB)=? I need to demonstrate that Ker(AB)=${\vec{0}}$, just like Ker(A) and Ker(B). I have a not-so-elegant way of showing that (I think): A and B are n * p and p * m, and it has to be the case that n $\geq$ p $\geq$ m, and rref's of A and B are either equal to the identity matrix, or are of form I and row(s) of 0's on the bottom. So, product of rref's of A and B will equal a matrix that is either I or I with row(s) of 0's on the bottom, and is n * m where n $\geq$ m, so it has more equations than variables. Multiplication of matrices of the aforementioned forms can only yield a matrix of the aforementioned form (I or I with row(s) of zeros), so it must also have Ker=${\vec{0}}$. Is there a more simple (though not too advanced, as we are barely mid-way through the course) way of showing that this is true? Also, while I know that Ker(A)=Ker(rref(A)) for any matrix A, I am not sure if I can say that Ker(rref(A) * rref(B))=Ker(AB). Is this statement true?
Thanks!
 A: Composition of one-to-one functions is one-to-one.
Alternatively and directly: let $\mathbf{v}\in\mathrm{Ker}(AB)$. Then $A(B(\mathbf{v}))=\mathbf{0}$, so $B(\mathbf{v})\in\mathrm{ker}(A)=\{\mathbf{0}\}$. Therefore, $B(\mathbf{v})=\mathbf{0}$, so $\mathbf{v}\in\mathrm{ker}(B) = \{\mathbf{0}\}$. Therefore, $\mathbf{v}=\mathbf{0}$.
This proves that $\mathrm{ker}(AB)\subseteq\{\mathbf{0}\}$, which suffices. 
No need to muck about with sizes of matrices.
Added: (as per request in the comment) I assume that "rref" means "row reduced echelon form". It is true that $\mathrm{Ker}(A) = \mathrm{Ker}(\mathrm{rref}(A))$, because we obtain $\mathrm{rref}(A)$ be left-multiplying $A$ by invertible elementary matrices, and this does not change the nullspace (equivalently, elementary row operations do not change the nullspace). However, it is false that $\mathrm{Ker}(AB) = \mathrm{Ker}(\mathrm{rref}(A)\mathrm{rref}(B))$; if you think about it, you'll see that the problem is that those invertible matrices will now also be between $A$ and $B$. Here is an explicit example: take
$$A = \left(\begin{array}{cc}
1 & 0 \\
0 & 0
\end{array}\right),\qquad B = \left(\begin{array}{cc}
0 & 0\\
1 & 0
\end{array}\right).$$
Then the reduced row echelon form of $B$ is $A$, so $\mathrm{rred}(A)\mathrm{rred}(B) = AA = A$. However, $AB$ is the zero matrix. So $\mathrm{Ker}(AB) = \mathbb{R}^2$, but $\mathrm{ker}(\mathrm{rred}(A)\mathrm{rred}(B)) = \mathrm{ker}(A) = \{(0,y)\mid y\in\mathbb{R}\}\neq \mathrm{ker}(AB)$. 
The reason it does not work is that if $C$ is invertible, then $\mathrm{ker}(CA) = \mathrm{ker}(A)$ (use the same argument as above). But $\mathrm{ker}(AC) = C^{-1}(\mathrm{ker}(A))$, so this may change the kernel. Since $\mathrm{rred}(A) = CA$ for some invertible matrix $C$, and $\mathrm{rred}(B) = DB$ for some invertible matrix $D$, you get $\mathrm{rred}(A)\mathrm{rred}(B) = CADB$, so that $D$ in the middle can mess you up.
A: Hint: Assume that $\text{Kernel}(AB)$ is non-trivial. Prove by contradiction by making use of the fact that the nullspace of $A$ and $B$ are trivial.
A: You can use the rank nullity theorem.
