Dear all, I have checked by computer that for any complex matrix , there we have However, I could not prove it mathematically [without tedious calculation].

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- Nov 3rd 2012, 04:40 PMxinglongdadaA(BC-CB)^2=(BC-CB)^2A, for any 2 by 2 complex matrix?
Dear all, I have checked by computer that for any complex matrix , there we have However, I could not prove it mathematically [without tedious calculation].

- Nov 3rd 2012, 05:40 PMchiroRe: A(BC-CB)^2=(BC-CB)^2A, for any 2 by 2 complex matrix?
Hey xinglongdada.

Try expanding all the matrices out and use the fact that (A+B)(C+D) = AC + AD + BC + BD for matrices if they are well defined (dimension wise that is) and if you get zero no matter what, then you're done. - Nov 4th 2012, 12:41 PMDevenoRe: A(BC-CB)^2=(BC-CB)^2A, for any 2 by 2 complex matrix?
it's not true.

for a counter-example, let:

.

then:

.

so:

.

now:

,

but:

.

your "computer check" apparently didn't check some "simple matrices". - Nov 4th 2012, 03:12 PMxinglongdadaRe: A(BC-CB)^2=(BC-CB)^2A, for any 2 by 2 complex matrix?
your calculation of BC-CB is wrong....

- Nov 4th 2012, 05:47 PMDevenoRe: A(BC-CB)^2=(BC-CB)^2A, for any 2 by 2 complex matrix?
*sigh*, this happens to me, more often than i'd like to admit. yes, it should be:

this means that:

and:

but these 2 still aren't equal.

*******

in general, unless B and C commute, BC - CB will be non-zero, and unless BC - CB is of nilpotency degree 2, (BC - CB)^{2}will also be non-zero.

in fact, unless (BC - CB)^{2}is a scalar multiple of the identity, there will always be SOME matrix A that doesn't commute with it.

so your "theorem" would only be true if (BC - CB)^{2}= I or 0, which implies BC - CB is either 0, or invertible.

so all we have to do is find matrices B and C so that BC - CB is singular. this isn't hard to do (and is probably what i SHOULD have done). - Nov 4th 2012, 06:01 PMtopsquarkRe: A(BC-CB)^2=(BC-CB)^2A, for any 2 by 2 complex matrix?
- Nov 5th 2012, 08:39 AMDevenoRe: A(BC-CB)^2=(BC-CB)^2A, for any 2 by 2 complex matrix?
omg. i am...wrong. just....wrong. how embarrassing.

it turns out the theorem IS true, and i can prove it.

it suffices (from a previous post) to prove that (BC - CB)^{2}is a scalar multiple of the identity (such matrices lie in the center of the ring of 2x2 matrices, and thus will commute with any A).

let:

.

then:

that is, BC - CB is of the form:

hence (BC - CB)^{2}is of the form:

so sorry :( - Nov 5th 2012, 12:26 PMxinglongdadaRe: A(BC-CB)^2=(BC-CB)^2A, for any 2 by 2 complex matrix?
Thank you very much. This proof is nice.