The problem statement.
Let $(V,<,>)$ be a finite dimensional vector space equipped with an inner product, whose dimension is $n$, and let $f:V \to V$ be a nilpotent linear transformation such that $f \circ f^*=f^* \circ f$. Prove that $f=0$.
The attempt at a solution:
I know that there is $ k \in \mathbb N, k\leq n$: $f^k=0$. If I compose $f \circ f^*$ with itself $k-1$ times, then $(f \circ f^*)^k$. As $f \circ f^*=f^* \circ f$, one can easily verify $(f \circ f^*)^k=f^k \circ (f^*)^k=0 \circ (f^*)^k=0$
Then $(f \circ f^*)^k=(f^* \circ f)^k=0$
At this point I got stuck.
I suppose that at some point, if I could show that $f^* \circ f=0$, I must use the property that characterizes $f^*$, which is: $<f(v),w>=<v,f^*(w)>$ and try to conclude that $||f(v)||=0$ for all $v \in V$. I would appreciate suggestions to go in this direction or another idea to get to the solution.
Using $f^*f=0$ $$(f(x),f(x))=(f^*f(x),x)=0$$ so $f(x)=0$.
To show $f^*f=0$ let $n$ be minimal such that $(f^*f)^n=0$ then
$$((f^*f)^{n-1}(x), (f^*f)^{n-1}(x))= ((f^*f)^{2n-2}(x), x)$$ this follows from the fact that $f^{**}=f$ and $f$ commutes with $f^*$. Now if $n\geq 2$ then $2n-2 \geq n$ and $(f^*f)^{2n-2}=0$ so we have $((f^*f)^{n-1}(x), (f^*f)^{n-1}(x))=0$ for all $x$ and thus $(f^*f)^{n-1}=0$ contradicting the choice of $n$. So $n \leq 1$.