I want to prove the following statement:
Let $H$ be a self-adjoint operator in a Hilbert space $\mathcal{H}$. Suppose that $H^2 = H^3$. Prove that $H = H^2$.
I tried to prove that proposition and I couldn't.
It seems to be very easy but I think that I need some trick to prove that proposition that I haven't noticed.
I need help.
On
A proof using spectral theory in the case where $\mathcal{H}$ is complex:
Observe that if $T$ a self adjoint operator on $\mathcal{H}$, then $T$ is bounded (Hellinger- Toeplitz) and for the spectral radius $r(T)$ of $T$ we have
$$ r(T)=||T||.$$
Let us denote the spectrum of $T$ by $ \sigma(T)$
If $H$ is self adjoint, then the operators $H^2-H$ and $H^3-H^2$ are also self adjoint.
Now let $\lambda \in \sigma(H^2-H)$. The spectral mapping theorem gives that there is $\mu \in \sigma(H)$ such that $\lambda=\mu^2- \mu$.
We get (use again the spectral mapping theorem)
$$ \lambda \mu=\mu^3-\mu^2 \in \sigma(H^3-H^2)=\sigma(0)=\{0\}.$$
Thus $ \lambda \mu =0$. Together with $\lambda=\mu^2- \mu$ we derive that $\lambda=0$. Therefore we have
$$ \sigma(H^2-H)=\{0\},$$
hence $||H^2-H||=r(H^2-H)=0$, which gives $H^2=H$.
Since $H$ is self adjoint, note that $$\left<H^2x-Hx,H^2x-Hx\right>=\left<H^2x,H^2x\right>-\left<Hx,H^2x\right>-\left<H^2x,Hx\right>+\left<Hx,Hx\right>\\=\left<H^4x,x\right>-\left<2H^3x,x\right>+\left<H^2x,x\right>=\left<(H^4-2H^3+H^2)x,x\right>.$$ But, $H^3=H^2$, so $H^4=H^3$, therefore $H^4-2H^3+H^2=0$, showing that $$\left<H^2x-Hx,H^2x-Hx\right>=0$$ for all $x$. So, $H^2x=Hx$ for all $x$, therefore $H^2=H$.