Commutators and the identity matrix – Miklos Schweitzer 2012 Problem 6
Suppose are complex matrices such that and , where is the usual commutator. Prove that is the identity matrix.
Miklos Schweitzer 2012 Problem 6
Proof: Let’s try to use the Hadamard formula because it involves both the commutators and the exponential:
If we replace by we get
Take in the above formula. We get
If we choose we get from where we get
If we denote , the last relation translates to and because we can obtain the same results by making a cyclic permutation we have . (these won’t be of any use in the sequel)
For we obtain which implies that , namely which means that .
If we apply the Hadamard like formula to and we obtain in the same way that which is equivalent to .
Now . In the same way we have .
So we have reached (from here the problem is solved using only simple group theory). In particular we have . Since is an exponential matrix, it is invertible, and therefore , which translates to and finishes the problem.
Note: It can be proved that if , where is a group and then where is the identity element of the group .

January 8, 2013 at 6:23 pmMiklos Schweitzer 2012 Problems « Beni Bogoşel's blog