Commutant of two bounded linear operators

Question

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 $[A,B]=cI$?

Answer 1

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).