Is there a compact normal injective operator on a infinite dimensional Hilbert space? 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)&lt;∞?
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.
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