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?
$AB \neq 0$ but $BA=0$
4 Answers
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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0@NieldeBeaudrap If you can get your hands on the papers I mentioned, I'm sure you'll find something useful along these lines :) – 2012-10-18
What about $B = \begin{pmatrix}0 & 1 \\ 0 & 0 \end{pmatrix} $ and $A = \begin{pmatrix}1 & 0 \\ 0 & 0 \end{pmatrix}$
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}. $
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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0Any $n \times n$ 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 $X = v u^T$. – 2012-10-18