I was told that for any finite dimensional Hilbert space, $[A,B]=i$ is impossible, while it is possible for an infinite dimensional Hilbert space. How can I show it is impossible for finite dimensional?
Edit: I just realized it's trivial with by taking trace. But is it possible to show without taking trace?
The trace is the best way, but just for fun:
Suppose $AB - BA = I$ and $v$ is an eigenvector of $BA$ corresponding to eigenvalue $\lambda$. Then, $$ABv - BAv = v \implies ABv = (\lambda + 1)v \implies BABv = (\lambda + 1)Bv.$$ If $\lambda \neq -1$, then we have $Bv \neq 0$, hence $Bv$ is an eigenvector of $BA$ corresponding to eigenvalue $\lambda + 1$. On the other hand, if $\lambda = -1$, then we must have $ABv = 0$. Since $v \neq 0$, this means that $A$ or $B$ is not invertible, and hence neither is $BA$, which means $\lambda + 1 = 0$ is an eigenvalue of $BA$.
In either case, if $\lambda$ is an eigenvalue of $BA$, then so is $\lambda + 1$. Hence, there are an infinite number of eigenvalues of $BA$!