# 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.

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.
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 $A$.
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.
• This is not completely correct. Yes, the spectrum contains 0, but 0 need not to be an eigenvalue (e.g., laplacian with Dirichlet b.c.). The spectrum of an infinite dimensional operator does not contain only its eigenvalues. That would be the 'point spectrum', that is the set of values $\lambda$ such that $A-\lambda I$ is not injective. The spectrum also contains the 'continuous spectrum' (i.e.,$A-\lambda I$ is injective and has dense range) and the 'residual spectrum' (i.e., $A-\lambda I$ does not have a dense range but $(A-\lambda I)^{-1}$ is not bounded). – bartgol May 2 '17 at 3:57