Let $A$ and $B$ be two matrices of arbitrary shape where the number of columns of $A$ is the same as the number of rows of $B$.
Is it always true that
$$\textbf{rref}[A]\cdot \textbf{rref}[B] = \textbf{rref}[AB]$$
and if it is not true all the time, what conditions guarantee that it will be?
I came across this as a claim by a student while grading and am unable to think of an appropriate proof or counterexample.
In the case that $\ker(A)=\{0\}$ and $\ker(B)=\{0\}$, it will be true that $\ker(AB)=\{0\}$. This would imply that the columns of $A$ are linearly independent as well as the same for columns of $B$ and $AB$. This would imply that it each individually rowreduces to an identity above some number of zero-rows and the identity should hold, but what of other cases where a kernel is nontrivial?
How about this for a counterexample:
$$A = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \end{pmatrix}, \quad B = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 1 \end{pmatrix}, \quad AB = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix}$$
Then $$rref[A] = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \end{pmatrix}, \quad rref[B] = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{pmatrix}$$ so $$rref[A] \cdot rref[B] = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{pmatrix} \neq rref[AB] = AB$$