$AB \neq 0$ but $BA=0$ Do there exists to matrices or objects such that $AB \neq 0$ but $BA=0$?
Another way to ask this question is if there exists objects or matrices $A$ and $B$ such that...
$[A,B]=AB$
where $[ \, , \, ]$ is the commutator $[A,B]=AB-BA.$
If such matrices do not exist, what does that imply about the algebra that the elements are in?
 A: What about $$B = \begin{pmatrix}0 & 1 \\ 0 & 0 \end{pmatrix} $$ and $$A = \begin{pmatrix}1 & 0 \\ 0 & 0 \end{pmatrix}$$
A: Sure - here's an example, with matrices taking entries in the field $\mathbb{Z}_{2}$:
$$
\begin{bmatrix}
1 & 1\\
0 & 0\\
\end{bmatrix}
\begin{bmatrix}
1 & 0\\
1 & 0
\end{bmatrix}
=
\begin{bmatrix}
0 & 0\\
0 & 0
\end{bmatrix},
$$
but
$$
\begin{bmatrix}
1 & 0\\
1 & 0\\
\end{bmatrix}
\begin{bmatrix}
1 & 1\\
0 & 0
\end{bmatrix}
=
\begin{bmatrix}
1 & 1\\
1 & 1
\end{bmatrix}.
$$
A: As the other answer show: there are uncountably many pairs of matrices $(A,B)$ such that $AB = 0$ while $BA \neq 0$. If you think of square matrices as linear transformations then it is obvious why this should be so: in $AB$, we can think of the product as recording the image of each of the columns of $B$ under the linear transformation $A$, likewise with $BA$.
A nice question to ask is the following: Given a fixed matrix $A \in \text{Mat}_n\mathbb{R},$ what is the following:
$$\widetilde{A} := \{ X \in \text{Mat}_n\mathbb{R} : AX = 0 \ \wedge \ XA \neq 0\} \, ?$$
Clearly, if $X \in \widetilde{A}$ then $\lambda X \in \widetilde{A}$ for all $\lambda \neq 0.$ Interestingly, this space is not a vector space because the zero matrix $0 \notin \widetilde{A}$ and $X,Y \in \widetilde{A}$ does not imply that $X+Y \in \widetilde{A}.$
A: Noncommutative algebra is filled with examples of this.
For example, take $A=\begin{bmatrix}0&1\\0&0\end{bmatrix}$ and $B=\begin{bmatrix}0&0\\0&1\end{bmatrix}$.
You have $AB\neq 0$ but $BA=0$.
Rings in which $ab=0$ implies $ba=0$ are called reversible rings. That is a particularly strong condition, and is pretty interesting to study. I highly recommend Greg Marks' paper: Reversible and symmetric rings (2002), and P.M. Cohn's paper Reversible rings (1999).
This would imply that $AB=0$ does not necessarily imply $[A,B]=0$.
