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

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

How can I check this?


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 $N^{-1}=0$.

  • $\begingroup$ Nice proof - thanks :) so but then my question is the closure of the numerical range always the convex hull of the spectrum of a selfadjoint operator even for unbounded ones? $\endgroup$ – C-Star-W-Star Aug 28 '14 at 11:35
  • $\begingroup$ @Freeze_S : I added information for you about numerical range, but I assume the spectral theorem. $\endgroup$ – Disintegrating By Parts Aug 28 '14 at 16:43
  • $\begingroup$ Shouldn't the last line read: "Spectral radius and norm are the same for an unbounded selfadjoint operator, ..." Assuming a bounded operator would relate to the former paragraph, or? Besides your answer greatly applies to normal operators, too. Do you mind replacing selfadjoint by normal? (I would adapt the question then.) $\endgroup$ – C-Star-W-Star Jun 23 '15 at 4:55
  • $\begingroup$ @Freeze_S : No, spectral radius and norm are the same for a bounded selfadjoint operator. Norm doesn't exist for an unbounded operator. So I reduce the unbounded case to the bounded by knowing that some point is in the resolvent. $\endgroup$ – Disintegrating By Parts Jun 23 '15 at 4:59
  • 1
    $\begingroup$ @AshvinSwaminathan : If $N$ is unbounded with empty spectrum, then, $N^{-1}$ is bounded because, by assumption, $0$ is in the resolvent set. $\endgroup$ – Disintegrating By Parts Nov 26 '17 at 17:05

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.


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.

  • 3
    $\begingroup$ Side note: For general unbounded operators, the spectrum can be empty, but for self adjoint operators (even unbounded), the spectrum is always nonempty. $\endgroup$ – PhoemueX Aug 28 '14 at 5:20
  • $\begingroup$ @PhoemueX: Do you have some good reference on these issues for further reading? $\endgroup$ – C-Star-W-Star Aug 28 '14 at 11:32
  • $\begingroup$ @voldemort: I mean spectrum of unbounded operator. (The bounded case is clear.) $\endgroup$ – C-Star-W-Star Jul 11 '15 at 22:54
  • 1
    $\begingroup$ @C-Star-W-Star Let $A=\frac{d}{dx}$ on the dense domain $\mathcal{D}(A)\subset L^2[0,1]$ consisting of all absolutely continuous $f\in L^2$ such that $f'\in L^2$ and $f(0)=0$. $(A-\lambda I)f=g$ is uniquely solved for $f\in\mathcal{D}(A)$ with $f=e^{\lambda x}\int_{0}^{x}e^{-\lambda u}f(u)du$, and this defines a bounded operator on $L^2$ for all $\lambda\in\mathbb{C}$. $\endgroup$ – Disintegrating By Parts Aug 28 '19 at 21:42
  • $\begingroup$ @DisintegratingByParts: And it has necessarily empty spectrum since it is not closed? $\endgroup$ – C-Star-W-Star Aug 31 '19 at 14:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.