Commutant of two bounded linear operators Let $E$ be an infinite-dimensional complex Hilbert space and $A,B\in  \mathcal{L}(E)$.

Why it is impossible to find $c\in \mathbb{C}^*$ such that $[A,B]=cI$?

 A: First of all, suppose that 
$[A, B] = I; \tag 1$
this implies $B \ne 0$; it also implies
$[A, B^n] = nB^{n - 1}, \; B^{n - 1} \ne 0, \tag 2$
which may be seen as follows:
$AB - BA = [A, B] = I \Longrightarrow AB^2 - BAB = B \ne 0; \tag 3$
since
$AB =  BA + I, \tag 4$
we have
$AB^2 - B^2A - B = AB^2 - B(BA + I) = AB^2 - BAB = B \ne 0, \tag 5$
or
$AB^2 - B^2A = 2B \ne 0; \tag 6$
we may now proceed by induction:  if
$AB^k - B^kA = kB^{k - 1} \ne 0, \tag 7$
$AB^{k + 1} - B^k AB = kB^k \ne 0, \tag 8$
since if $B^k = 0$, (7) would imply $B^{k - 1} = 0$; thus, again using $AB = BA +I$,
$AB^{k + 1} - B^{k + 1}A - B^k = AB^{k + 1} - B^k(BA + I) = AB^{k + 1} - B^kAB  = kB^k \ne 0, \tag 9$
$AB^{k + 1} - B^{k + 1}A = (k + 1)B^k \ne 0; \tag{10}$
we therefore see that
$AB^n - B^nA = n B^{n - 1} \ne 0, \tag{11}$
holding for all positive integers $n$; taking norms yields
$n\Vert B^{n - 1} \Vert = \Vert n B^{n - 1} \Vert = \Vert AB^n - B^ nA \Vert$
$\le \Vert AB^n \Vert + \Vert B^nA \Vert \le 2 \Vert A \Vert \Vert B^n \Vert \le 2 \Vert A \Vert \Vert B \Vert \Vert B^{n - 1} \Vert \ne 0; \tag{12}$ 
since $\Vert B^{n - 1} \Vert \ne 0$ we may divide it out to find
$n \le 2 \Vert A \Vert \Vert B \Vert, \tag{13}$
a contradiction since $n$ may be arbitrarily large.  Thus we cannot have (1), and this precludes
$[A, B] = cI, \; 0 \ne c \in \Bbb C,\tag{14}$
since it may be written
$[A, c^{-1}B] = I, \tag{15}$
thus reducing (14) to the case (1).
