2
$\begingroup$

I need an example of a ring consisting of 2 by 2 matrices where $a^3=a$ with $a$ belonging to this ring. If someone can list the elements I would be satisfied.

What I'm trying to get at it is conceptualize why a ring $R$ is always commmuative when $a^3=a$. I know of one such example and that is the factor ring $\mathbb{Z}/3\mathbb{Z}.$ Does anyone know how to prove this statement mathematically as well as giving me an example of a ring of 2 by 2 matrices?

$\endgroup$
  • $\begingroup$ It looks like you want a proof of the theorem attributed to Jacobson (too long to prove here, I think) and an example of a 2x2 matrix $a$ which satisfies $a^3=a$? Is this correct? $\endgroup$ – rschwieb Aug 21 '12 at 16:18
  • 3
    $\begingroup$ You’re not going to get a $2\times2$ matrix ring over $\Bbb R$ in which every element satisfies $a^3=a$, if that’s what you want. $\endgroup$ – Brian M. Scott Aug 21 '12 at 16:19
  • $\begingroup$ If your matrix is invertible it satisfies $a^2=I$ or alternatively $a=a^{-1}$. If not, then it is singular ... the options are limited. $\endgroup$ – Mark Bennet Aug 21 '12 at 16:21
  • $\begingroup$ Yes, but I want a non-trivial example of a 2x2 matrix if possible. I don't know of a matrix $a$ where $a^2$ gives the identity matrix. An intuitive proof would be nice but not necessary $\endgroup$ – Student Aug 21 '12 at 16:21
  • $\begingroup$ The only non-trivial example is the one with the identity matrix. Brian points out that I can't get a 2x2 matrix if that is the case then I won't press any further $\endgroup$ – Student Aug 21 '12 at 16:26
2
$\begingroup$

There is a theorem due to Jacobson that says if for every $a\in R$ there exists an $n\in\mathbb{N}$ such that $a^n=a$, then $R$ is commutative. (See this, or this for example).

Obviously the identity matrix cubed is itself... is this the sort of thing you're looking for?!

In general matrix rings are going to have a lot of idempotent elements $e$ such that $e^2=e$, and for all of those $e^3=e$ as well.

For an example where $a^2\neq a$, you could use $\begin{bmatrix}0&1\\1&0\end{bmatrix}$.

$\endgroup$
  • $\begingroup$ I didn't see that, thanks but is there a nontrivial example? Sorry if I'm putting you on the spot someone else can respond $\endgroup$ – Student Aug 21 '12 at 16:15
  • 1
    $\begingroup$ @Shaniqua I listed 3 examples with decreasing triviality. $\endgroup$ – rschwieb Aug 21 '12 at 16:19

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.