Self adjoint operator, proof for $f^{k}$ Let $f$ be an endomorphism of the vector space $V$. $f$ is alto self adjoint. Prove that, if $f^{k}(v)=0$ for $k\ge2$, then $f(v)=0$.
Let choose $v \in V$, then we get the following:
$$
0 = <f^{2}(v),v> = <f(v),f(v)>
$$
Then we get that $f(v) = 0$
Is it correct for $k=2$? How can I generalize it?
Thanks.
 A: Note that if $f$ is self-adjoint then for all $n$, $f^n$ is self-adjoint. Let's prove the result by induction on $k$. For $k=2$, we have $\langle fv, fv\rangle = \langle f^2 v, v \rangle = 0$, so $fv = 0$ for all $v$. Suppose that this is true up to $k$ and suppose that $f \in \text{end}(V)$ such that $f^{k+1} =0$. Then, $f^{2k} =0$ since $2k \ge k+1$ for $k \ge 2$. We have $\langle f^k v, f^k v \rangle = \langle f^{2k} v , v\rangle = 0$, so $f^k = 0$ and so by induction hypothesis, $f =0$.
A: Your proof for $k=2$ looks good, and the general case can be proved in a similar way using induction on $k$: the base case is $k=2$, which you have already proved. 
Now suppose that $k>2$, and $f^jv=0$ implies that $fv=0$ for all $2\leq j\leq k-1$, and suppose that $f^kv=0$.
If $k=2l$ is even then we have
$$ 0=\langle f^{2l}v,v\rangle=\langle f^lv,f^lv\rangle $$
Therefore $f^lv=0$, and by the induction hypothesis it follows that $fv=0$.
Finally, if $k=2l+1$ is odd then $f^kv=0$ implies that $f^{k+1}v=0$, and since $k+1$ is even we can apply the same argument as above to conclude that $fv=0$.
