Is there a compact normal injective operator on a infinite dimensional Hilbert space?

Question

Is there a compact, normal and injective operator $T$ in an infinite-dimensional Hilbert space $H$ such that $sp(T)$ is finite?

Thanks.

Related: Is there a injective compact operator TT such that #sp(T)<∞?

Answer 1

This is not possible. One way to show this is the following:

If $T$ is normal, and has finite spectrum, then each element of the spectrum is an eigenvalue (see theorem 12.29 of Rudin's Functional Analysis). If $T$ is also injective, then $0$ is not in the spectrum, because if it were it would be an eigenvalue. But then $T$ cannot be compact, because $H$ is infinite-dimensional.

Answer 2

No, the spectrum of a compact operator in an infinite dimensional Banach space contains $0$, so it is not injective since every element of the spectrum is an eigenvalue.

https://en.wikipedia.org/wiki/Spectral_theory_of_compact_operators#Statement

Answer 3

Suppose $A$ is normal with a finite spectrum $\{\lambda_1,\lambda_2,\cdots,\lambda_n\}$, then $$ A = \lambda_1 E_1 + \lambda_2 E_2 + \cdots \lambda_n E_n, $$ where $E_j$ are the orthogonal projections onto $\mathcal{N}(A-\lambda_j I)$. And, $$ I = E_1 + E_2 + \cdots + E_n, $$ which gives the orthogonal decomposition $$ H = \mathcal{N}(A-\lambda_1 I)\oplus\mathcal{N}(A-\lambda_2 I)\oplus\cdots\oplus\mathcal{N}(A-\lambda_n I). $$ Assuming that $A$ is compact and injective, then $\lambda_k \ne 0$ for all $k$, and one of these null spaces must be infinite-dimensional, which contradicts the compactness of $A$ because $\mbox{dim}\mathcal{N}(A-\lambda I) < \infty$ for all $\lambda \ne 0$ for any compact $A$.