Normality of unilateral forward shift

Question

Let $H$ be an infinite dimensional, separable complex Hilbert space with orthonormal basis $(e_n)_n$.

Let $T$ be the unilateral forward shift with $Te_n= e_{n+1}$.

Let $\mathcal{K}(H)$ denote the set of compact operators acting on $H$.

1. Show $T$ is essentially normal.

2. Show that there doesn't exist a normal $N \in B(H), K\in \mathcal{K}(H)$ such that $T= N+K$.

Things I've done:

1. To show that $T$ is essentially normal, we want to show that $\pi(T)\pi(T^*) = \pi(T^*)\pi(T) $, where $\pi$ is the canonical quotient map $\pi: B(H) \rightarrow \mathcal{L}(H)$, and $\mathcal{L}(H)$ is the Calkin algebra.

From here I'm not sure how to continue.

2. We will proceed by contradiction.

Suppose there exists a normal $N \in B(H), K\in \mathcal{K}(H)$ such that $T= N+K$.

Then we have $ e_{n+1}=T(e_n)= N(e_n)+ K(e_n)$. From here, I don't know how to continue to have a contradiction.

I also know that if $N$ is normal and $K$ is compact, then $N+K$ is essentially normal.

Any help or ideas will be appreciated!

Thank you!

Answer 1

$T$ is Fredholm with index $-1$. So, if $T = N+K$, then $N = T-K$ would be Fredholm with index $-1$. But a normal Fredholm operator always has index zero (because $H = \ker N\oplus \operatorname{ran}N$). Contradiction!