Proving that $AB-BA=cI$ for nontrivial $c \in \mathbb{C}$ I have a homework question I can`t solve:
Let $X$ be a normed linear space, $A,B \in B(X)$. Show that there exists no nontrivial $c \in \mathbb{C} $ such that $AB-BA=cI$.
Thanks alot already guys! I seem to have a professor who gives impossible homework problems.
The hint he gave: $A^{n}B-BA^{n}=ncA^{n-1}$, $n \in \mathbb{N}$
 A: The finite dimensional  case:
If $X$ is finite dimensional, say $m=\dim X$, then from the equality $AB-BA=cI$ we conclude, on taking traces that $cm=c{\rm Tr}(I_m)= {\rm Tr}(AB-BA)=
 {\rm Tr}(AB)-{\rm Tr}(BA)=0$.$~~$ This implies that $c=0$.
The infinite dimensional case:
Here we prove by induction that for all $n\geq 1$ we have $A^nB-BA^n=ncA^{n-1}$. Indeed, this is true for $n=1$ by assumption,  and assuming it is true for $n$ we get
$$\eqalign{
A^{n+1}B-BA^{n+1}&=A^n(cI+BA)-BA^{n+1}\cr&=cA^n+(ncA^{n-1}+BA^n)A-BA^{n+1}=(n+1)cA^n
}
$$
Now, using the fact that we have bounded operators we conclude that, for every $n$ we have
$$
(n+1)|c|\Vert{A^n\Vert}\leq \Vert{A^n\Vert}(\Vert{AB\Vert}+\Vert{BA\Vert})
$$
Now suppose that $c\ne 0$, then for $n+1>\frac{1}{|c|}\left(\Vert AB\Vert +\Vert BA\Vert \right)$ the preceding inequality implies that $\Vert{A^n}\Vert=0$, or $A^n=0$. So,
let us define $n_0$ to be the smallest $n$ such that $A^n=0$:
$$
n_0=\min\{n:A^n=0\}
$$
$n_0$ exists because of the preceding discussion, Now the formula
$$n_0cA^{n_0-1}=A^{n_0}B-BA^{n_0}=0$$ yields a contradiction. This shows that we must have $c=0$ and we are done.$\qquad\square$
Remark.
If we do not assume the boundedness of the operators involved the result might not be true. Consider $E=\Bbb{R}[X]$ the normed space of real polynomials with $\Vert P\Vert=\sup_{t\in[0,1]}|P(t)|$, and let
$$
A:E\to E, A(P)=P',\quad B:E\to E, B(P)=XP.
$$
Then clearly $(AB-BA)(P)=(XP)'-XP'=P$ that is $AB-BA=I$. But in this case $A$ is not bounded.
