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].
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].
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.
*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).
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