TITLE: Consequences when the commutator is a scalar multiple of the identity matrix
QUESTION [5 upvotes]: I just stumbled over the question below. As to the first, I could easily find out the answer (D) by invoking the commutation relation. But I don't figure out how to solve other two. Could anybody give a hint?
Here's the question:   
The commutator
$[B,A] \equiv BA-AB = \lambda I$
of two $n\times n$ matrices $A$ and $B$ is proportional to the identity matrix $I$.
Choose the only answer for each question.
(1) $[B,A^n]=?\;$ (A) $0\;$ (B) $n\;$ (C) $n\lambda A\;$ (D) $n\lambda A^{n-1}\;$ (E) $n\lambda A^n$
(2) $e^{A+B}=?\;$ (A) $e^{A}e^{B}\;$ (B) $e^{B}e^{A}\;$ (C) $e^{A}e^{B}e^{{1\over 2}[B,A]}\;$ (D) $e^{A}e^{B}e^{[B,A]}$
(3) $e^{A}e^{B}=?\;$ (A) $e^{A+B}\;$ (B) $e^{B}e^{A}\;$ (C) $e^{A}e^{B}e^{[B,A]}\;$ (D) $e^{A}e^{B}e^{-[B,A]}$

REPLY [3 votes]: Since this has floated to the front page anyway, let me supply a more elementary answer than the current one.
I assume you work over either the reals or the complex numbers since the question mentions $e^A$.
It is a standard exercise to show that then $AB$ and $BA$ have the same trace, so the trace of $AB - BA$ is $0$, which forces $\lambda = 0$ and hence $AB = BA$.
This immediately gives the answer to all the questions (and in fact means that most of the possible answers are identical).