An Example of Matrices $A$, $B$ such that $\mathrm{rref}(AB)\neq \mathrm{rref}(A) \mathrm{rref}(B)$

Problems and solutions in Linear Algebra

Problem 569

For an $m\times n$ matrix $A$, we denote by $\mathrm{rref}(A)$ the matrix in reduced row echelon form that is row equivalent to $A$.
For example, consider the matrix $A=\begin{bmatrix}
1 & 1 & 1 \\
0 &2 &2
\end{bmatrix}$
Then we have
\[A=\begin{bmatrix}
1 & 1 & 1 \\
0 &2 &2
\end{bmatrix}
\xrightarrow{\frac{1}{2}R_2}
\begin{bmatrix}
1 & 1 & 1 \\
0 &1 & 1
\end{bmatrix}
\xrightarrow{R_1-R_2}
\begin{bmatrix}
1 & 0 & 0 \\
0 &1 &1
\end{bmatrix}\] and the last matrix is in reduced row echelon form.
Hence $\mathrm{rref}(A)=\begin{bmatrix}
1 & 0 & 0 \\
0 &1 &1
\end{bmatrix}$.

Find an example of matrices $A$ and $B$ such that
\[\mathrm{rref}(AB)\neq \mathrm{rref}(A) \mathrm{rref}(B).\]

 
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Proof.

Let
\[A=\begin{bmatrix}
0 & 1\\
0& 0
\end{bmatrix} \text{ and } B=\begin{bmatrix}
0 & 0\\
1& 0
\end{bmatrix}.\] Then $A$ is already in reduced row echelon from.
So we have $\mathrm{rref}(A)=A$.

Applying the elementary row operations, we obtain
\[B=\begin{bmatrix}
0 & 0\\
1& 0
\end{bmatrix} \xrightarrow{R_1\leftrightarrow R_2}\begin{bmatrix}
1 & 0\\
0& 0
\end{bmatrix}.\] As the last matrix is in reduced row echelon from, we have $\mathrm{rref}(B)=\begin{bmatrix}
1 & 0\\
0& 0
\end{bmatrix}$.
Therefore, we see that
\[\mathrm{rref}(A) \mathrm{rref}(B)=\begin{bmatrix}
0 & 1\\
0& 0
\end{bmatrix}
\begin{bmatrix}
1 & 0\\
0& 0
\end{bmatrix}=\begin{bmatrix}
0 & 0\\
0& 0
\end{bmatrix}.\]


The product of $A$ and $B$ is
\[AB=\begin{bmatrix}
0 & 1\\
0& 0
\end{bmatrix}
\begin{bmatrix}
0 & 0\\
1& 0
\end{bmatrix}=\begin{bmatrix}
1 & 0\\
0& 0
\end{bmatrix}.\] It follows that $\mathrm{rref}(AB)=\begin{bmatrix}
1 & 0\\
0& 0
\end{bmatrix}$.


In summary, we have
\[\mathrm{rref}(AB)=\begin{bmatrix}
1 & 0\\
0& 0
\end{bmatrix} \neq
\begin{bmatrix}
0 & 0\\
0& 0
\end{bmatrix}
=\mathrm{rref}(A) \mathrm{rref}(B),\] as required.


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