Normal Operator: Empty Spectrum

Question

Given a Hilbert space $\mathcal{H}$.

For normal operators: $$N^*N=NN^*:\quad\sigma(N)\neq\varnothing$$

How can I check this?

Answer 1

Every bounded linear operator on a complex Banach space has non-empty spectrum $\sigma(B)$. A simple way to argue this is to assume the contrary, and conclude that $(B-\lambda I)^{-1}$ is an entire function of $\lambda$ which vanishes at $\infty$; then Liouville's theorem gives the contradiction that $(B-\lambda I)^{-1}\equiv 0$ for all $\lambda\in\mathbb{C}$.

If $N$ is unbounded normal and has empty spectrum, then $N^{-1}$ is bounded and has spectrum $\sigma(N^{-1})\subseteq\{0\}$ because, for $\lambda\ne 0$, $$ (N^{-1}-\lambda I) =(I-\lambda N)N^{-1}=\lambda(\frac{1}{\lambda}I-N)N^{-1} $$ has bounded inverse $$ \frac{1}{\lambda}N\left(\frac{1}{\lambda}I-N\right)^{-1}. $$ Spectral radius and norm are the same for a bounded normal operator, which leads to the contradiction that $N^{-1}=0$.

Answer 2

Note that the spectrum of any operator can't be empty. I think what you mean is that the operator might not have an eigenvalue- an example is given here.

Answer 3

I decided to make a separate answer for the Numerical Range question. They're different. Yes, the closure of the numerical range is the same as the closed convex hull of the spectrum.

Numerical Range: Suppose $\lambda_{1},\lambda_{2} \in \sigma(A)$ with $\lambda_{1}\ne \lambda_{2}$. Using the spectral theorem, you can find sequences $\{ e_{1,n}\}_{n=1}^{\infty}$ and $\{ e_{2,n}\}_{n=1}^{\infty}$ of unit vectors in the domain of $A$ such that $e_{1,n},Ae_{1,n}$ are orthogonal to $e_{2,n},Ae_{2,n}$, and such that $$ \lim_{n}(Ae_{j,n},e_{j,n})=\lambda_{j},\;\;\; j =1,2. $$ Then, for $\alpha \in [0,1]$, let $e_{n}=\alpha e_{1,n}+\sqrt{1-\alpha^{2}}e_{2,n}$, which is a unit vector, and $$ \begin{align} (Ae_{n},e_{n}) & = \alpha^{2}(Ae_{1,n},e_{1,n})+(1-\alpha^{2})(Ae_{2,n},e_{2,n}), \\ \lim_{n}(Ae_{n},e_{n}) & = \alpha^{2}\lambda_{1}+(1-\alpha^{2})\lambda_{2}. \end{align} $$ So the closure of the numerical range $\mathscr{N}(A)$ includes the closed convex hull of the spectrum. Conversely, suppose $x \in \mathcal{D}(A)$ is a unit vector. Let $E$ be the spectral resolution of the identity for $A$. Then $\mu_{x}(S)=\|E(S)x\|^{2}$ is a probability measure and $$ (Ax,x) = \int_{\sigma} \lambda d\mu_{x}(\lambda). $$ So, the numerical range $\mathscr{N}(A)$ is contained in the closed convex hull of the spectrum.