Show that C is nilpotent?
Let A and B be nxn matrices and C = AB-BA. Show that if C commutes with A and B, then C ^ p = 0 for some positive integer p. Anonymous dude: It is not clear to me that A and B must share an eigenvector based only on the condition that C commutes with A and B. Take C = 0 such that commutes with all A and B (if we disregard that C is in this case AB-BA). Or have I misunderstood? Anonymous dude: I'm always a little suspicious. 
It seems that you are using that A and B commute. It is not clear (if true). The only problem, indicates that both C commutes with A and B. (Note also that the problem is trivial to solve if A and B commute.) A great way to solve the problem is to calculate the trace of C ^ 2.
I'm not sure I can do on any what body, but here is an argument over the real numbers or complex. Let t be an eigenvalue of C and V is the corresponding eigenvector. The assumption that C commutes with A and B implies that v is an eigenvector of A and B. It is a fact of general matrices on the shuttle. Given that, suppose that Av = Bv = BC and BV. Then tv = Cv = ABV – BAV = ABV – ABV = 0. So we need that T = 0. In other words, all eigenvalues of C are 0. But this is equivalent the statement that C is nilpotent. EDIT: What I wrote is not quite right, and you're right to be suspicious. But I think I can fix it. Suppose P and Q are nxn matrices commuting. It is not true, as I wrote, that any eigenvector of P is a vector own Q, the example P = 0, Q any nonzero matrix serves cons-example. However, the rest of my proof does not need to be correct. Everything I need is that for every t eigenvalue of P there exists a vector v in the T-eigenspace of P, which is an eigenvector of Q. If, for example, we assume that P and Q are diagonalizable, then it follows from well-known fact that commuting diagonalizable matrices are simultaneously diagonalizable. I will try to indicate how to prove the more general statement. Then, we show that the T-eigenspace W of P contains an eigenvector for Q. Take a vector v in W, we that PV = tv. This means QPV = TQV, but on the other hand QPV = PQV, so we PQV = TQV meaning that Qv in W. This shows that Q stabilizes W, it is ie the linear transformation Q '= Q restricted to W is a linear transformation of the West itself. But any linear transformation of a complex vector space to itself has an intrinsic value by the fundamental theorem of algebra (which is why my proof works only for matrices with Real or complex entries), and there is an eigenvector associated with each eigenvalue. Q So "and thus Q is an eigenvector in W, as desired.
Dahon Matrix
