Consequences when the commutator is a scalar multiple of the identity matrix 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) $e^{A}e^{B}e^{-[B,A]}$
 A: Given the commutation relation $AB-BA=\lambda I$, it is always possible to assume that $A$ is invertible because $(A-\mu I)B-B(A-\mu I)=\lambda I$ holds for all scalars $\mu$. So, without loss of generality, assume $A$ is invertible.
Establish 1(D) by induction: By assumption, $[B,A]=\lambda A^{0}$. Suppose $[B,A^{k}]=\lambda A^{k-1}$ for some $k \ge 1$. Then
$$
\begin{align}
   [B,A^{k+1}]= BA^{k+1}-A^{k+1}B & =(BA^{k}-A^{k}B)A+A^{k}BA-A^{k+1}B \\
     & = [B,A^{k}]A+A^{k}[B,A] \\
     & = \lambda k A^{k}+A^{k}(\lambda I)=\lambda (k+1) A^{k}.
\end{align}
$$
So, by induction, $[B,A^{k}]=\lambda k A^{k-1}$ for all $k \ge 1$.
It is a well-known result that $[B,A]=I$ cannot hold for bounded linear operators on a normed linear space, which certainly includes matrices. To see why, let $\|M\|=\sum_{j,k}|m_{j,k}|$ for an $n\times n$ matrix $M$. Then $\|M_1+M_2\|\le \|M_1\|+\|M_2\|$ and $\|M_1 M_2\|\le \|M_1\|\|M_2\|$ for any two such matrices $M_1$, $M_2$. Assuming the commutation relation for $A$ and $B$, it follows that
$$
    k|\lambda|\|A^{k-1}\| \le \|A^{k}B\|+\|BA^{k}\| \le 2\|A\|^{k}\|B\|\le 2\|A^{k-1}\|A\|\|B\|,\;\;\; k \ge 1.
$$
Either $A^{k-1}=0$ for some $k \ge 1$, which cannot happen because $A$ is invertible, or
$$
          k|\lambda| \le \|A\|\|B\|,\;\;\; k=1,2,3,\cdots,
$$
forces $\lambda=0$, which means $AB=BA$. That makes everything else straightforward.
A: 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).
