Let $H$ be a complex, infinite dimensional, separable complex Hilbert space with orthonormal basis $(e_n)_n$.
Let $T$ be the unilateral forward shift with $Te_n= e_{n+1}$.
We say that $T \in \mathcal{B(H)}$ is essentially normal if the image $\pi(T)$ of $T$ under the canonical quotient map $\pi: \mathcal{B(H)} \rightarrow {\mathcal{B(H)}}/{\mathcal{K(H)}}$ in a sense that $\pi(T)\pi(T^*)= \pi(T^*)\pi(T)$
Show that $T$ is essentially normal.
What I've tried:
Since the quotient map is linear, I want to show that
$\pi(T)\pi(T^*) – \pi(T^*) \pi(T)=\pi(TT^*) – \pi(T^*T)= \pi(TT^*- T^*T)=0$
But not I'm confused because isn't that equivalent to showing that $T$ is normal?
i.e. $TT^* – T^*T = 0$.
Also, I'm not sure how to find $TT^* – T^*T$.
Any help will be appreciated!
Thank you in advance!
Best Answer
Hints: since $\langle T^*e_n,e_m\rangle=\delta_{n(m+1)}$ show that $\langle T^*e_n,x\rangle=\langle e_{n-1},x\rangle$ and therefore that $T^*e_n=e_{n-1}$. Then, $T^*T-TT^*=I-TT^*.$ which you can show by direct calculation, to be projection on the rank one subspace spanned by $e_0.$