$V$ be a vector space ; $f,g \in \mathcal L (V)$ ; $f\circ g-g \circ f=I$ ; then is the set $\{g^n: n\ge 0\}$ linearly independent? Let $f,g$ be linear operators on a vector space $V$ such that $f\circ g-g \circ f=I$ , where $I$ is the identity operator on $V$ ; then is it true that the set $\{g^n: n\ge 0\}$ is linearly independent ? ( here $g^0:=I)$  
All I know is that under the given conditions ; if the grouond field of $V$ has characteristic $0$ then $V$ cannot be finite dimensional and that $f \circ g^n - g^n \circ f=ng^{n-1} , \forall n \in \mathbb N$  . Please help . Thanks in advance 
 A: You have laid the groundwork.  Let me generalize things slightly, because the techniques have come up for me repeatedly.
Let us write $[f,g]=f\circ g -g\circ f$ for the commutator of $f$ and $g$.  It is not hard to show that $[f,-]$ is a derivation, that is, $[f,gh]=[f,g]h+g[f,h]$ (where the multiplication is function composition, although this will hold more generally in any associative ring).  For any derivation $\partial$, if $\partial(g)$ commutes with $g$, then we have by induction that $\partial(g^n)=ng^{n-1}\partial(g)$, and hence for any polynomial $P$, that $\partial(P(g))=P'(g)\partial(g)$, which is simply the chain rule.
Now, assume that the set $\{g^n \mid n\geq 0 \}$ were not linearly independent.  Then from a linear dependence we would have a polynomial relation that $g$ satisfied.  Let $P$ be the minimum degree polynomial relation, so that $P(g)=0$.  Then $[f,P(g)]=P'(g)=0$, but since $P'$ has lower degree than $P$, either P' is the zero polynomial (impossible in characteristic zero) or we have contradicted minimality.  Therefore, no minimal polynomial can exist, and we must have that the powers of $g$ are linearly independent.  
