# $AB \neq 0$ 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?

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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 $[A,B]=0$.

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Obvious examples of reversible rings include commutative rings and rings without nilpotent elements. – rschwieb Oct 18 '12 at 15:37
What can be said about algebras where $AB = BA$, for instance over commutative rings? – Niel de Beaudrap Oct 18 '12 at 15:44
@NieldeBeaudrap Well, if you read my comment, you see I mentioned that commutative rings are obviously reversible. They also have the property that $[A,B]=0$ for all $A, B$, too. Algebras are rings, unless you specifically have nonassociative algebras in mind. Do you? – rschwieb Oct 18 '12 at 16:11
@rscwieb: Sigh. Yes, obviously, algebras are rings; and obviously an algebra which happens to be commutative satisfy $AB = 0 \iff BA = 0$. I had hoped for something more. Are you telling me that there is nothing more known about when $AB = 0 \iff BA = 0$, for algebras as a particular case of rings? That in this case, the only thing we know about algebras are things which hold for rings in general? – Niel de Beaudrap Oct 18 '12 at 16:16
@NieldeBeaudrap Since this definition only deals with the ring multiplication of the algebra, I can't forsee why having additional algebra structure would make any difference. In fact, I can't think of a single substantial thing where being an algebra makes a difference on a result for rings. But if you have a good example, I would love to hear it :) – rschwieb Oct 18 '12 at 16:39

What about $$B = \begin{pmatrix}0 & 1 \\ 0 & 0 \end{pmatrix}$$ and $$A = \begin{pmatrix}1 & 0 \\ 0 & 0 \end{pmatrix}$$

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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

$$\begin{bmatrix} 1 & 0\\ 1 & 0\\ \end{bmatrix} \begin{bmatrix} 1 & 1\\ 0 & 0 \end{bmatrix} = \begin{bmatrix} 1 & 1\\ 1 & 1 \end{bmatrix}.$$

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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 $X+Y \in \widetilde{A}.$

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@JasonDeVito Quite right! Thanks for that. I've changed it. – Fly by Night Oct 18 '12 at 16:01
$AX = 0$ iff ${\cal R}(X) \subseteq {\cal N}(A)$ (i.e. the column space of $X$ is contained in the null space of $A$). So $\tilde{A}$ consists of those $X$ such that ${\cal R}(X) \subseteq {\cal N}(A)$ but not ${\cal R}(A) \subseteq {\cal N}(X)$. – Robert Israel Oct 18 '12 at 18:26
Any $n \times n$ matrix $A$ whose rank is strictly between $0$ and $n$ has $\tilde{A}$ nonempty: just let $u$ and $v$ be nonzero vectors in ${\cal R}(A)$ and ${\cal N}(A)$ respectively, and take $X = v u^T$. – Robert Israel Oct 18 '12 at 18:32