I think the fact that this problem is about operators over finite-dimensional vector space is irrelevant. This statement can be proved about any associative algebra over a ring.
I'm studying for a PhD preliminary exam, and it seems I have some holes in my linear algebra knowledge. I need help with the following problem:
Let A and B be operators V -> V, where V is a finite-dimensional vector space over the complex numbers. Let p be any polynomial such that p(AB) = 0. Then (a) Prove that that q(x) = xp(x) satisfies q(BA) = 0.
Interestingly, I have shown that this problem is equivalent to proving that the minimal polynomial of AB is equal to the minimal polynomial of BA. Therefore, a direct solution to this problem would be to directly show that q(BA) = 0 whereas, given what I've already shown, it would suffice to simply prove that the minimal polynomials of BA and AB are equal. I am interested to a solution of either kind (ideally both!)
Thanks!
I think the fact that this problem is about operators over finite-dimensional vector space is irrelevant. This statement can be proved about any associative algebra over a ring.
suppose p(x) = c_{0} + c_{1}x + ...+ c_{m}x^{m}, so that
q(x) = c_{0}x + c_{1}x^{2} +....+ c_{m}x^{m+1}.
then q(BA) = c_{0}BA + c_{1}(BA)^{2} +...+ c_{m}(BA)^{m+1}.
note that (BA)^{k} = B(AB)^{k-1}A, so we have:
q(BA) = B(c_{0}I + c_{1}(AB) +.... + c_{m}(AB)^{m})A
= B(p(AB))A = B(0)A = 0.
your assertion that AB and BA must have the same minimal polynomial is false. suppose A =
[1 -1]
[1 -1], and that B =
[0 1]
[0 1]
then AB = 0, while BA = A.
clearly, the minimal polynomial for AB is x, while the minimal polynomial for BA is x^{2}. note this agrees with the result above.