Let $\mathcal{B}(F)$ the algebra of all bounded linear operators on an infinite-dimensional complex Hilbert space $F$.

Definition Let $A\in\mathcal{B}(F)$. $A$ is said to be normal if $$AA^*=A^*A.$$

All self-adjoint operators are normal.

I look for an example of normal operator on an infinite-dimensional complex Hilbert space $F$ which is not self-adjoint.

  • 2
    $\begingroup$ Do you know an example on $\mathbb{C}^n$ for some $n$? Because then you can just embed $\mathbb{C}^n$ into $\ell^2(\mathbb{C})$ in the obvious way and use the map on $\ell^2$ induced by this example. $\endgroup$ Mar 26 '18 at 15:48
  • $\begingroup$ Surely Dr. Google can help you. $\endgroup$
    – user223391
    Mar 26 '18 at 15:48
  • $\begingroup$ @Schüler Have you made any effort to solve this problem yourself? It's clear from the speed of your reply that you didn't think about my hint and (as Zachary Selk notes) you could definitely google this question. If you aren't willing to put effort into solving the problem, I am certainly not willing to put effort into writing a solution for you. $\endgroup$ Mar 26 '18 at 15:56

A unitary operator is not generally selfadjoint. For example, consider $T : L^2[-\pi,\pi]\rightarrow L^2[-\pi,\pi]$ defined by $$ (Tf)(\theta) = e^{i\theta}f(\theta) $$ The spectrum of $T$ is the unit circle in the complex plane. And $T^*T=TT^*=I$ because $(T^*f)(\theta) = e^{-i\theta}f(\theta)$.

The fundamental building blocks for selfadjoint and normal operators are multiplications on some $L^2$ space. So they are similar in many ways, and the Spectral Theorem applies to both.

For another example: Let $\mu$ be a finite Lebesgue measure on the complex plane with support $S\subset \mathbb{C}$, where $S$ is a bounded set. Let $T : L^2(S,\mu)\rightarrow L^2(S,\mu)$ be defined by $(Tf)(z)=zf(z)$. Then $T$ is a bounded normal operator with spectrum $S$.

  • $\begingroup$ I hope to get an exemple of normal operator which is neither self adjoint nor unitary $\endgroup$
    – Schüler
    Mar 27 '18 at 10:33
  • $\begingroup$ @Schüler Let $(Tf)(z) = zf(z)$, where $H=L^2(\Omega)$ and $\Omega$ is a bounded open region of the complex plane. Then $T$ is normal, and has spectrum equal to the closure of $\Omega$. $\endgroup$ Mar 27 '18 at 14:43
  • $\begingroup$ @Schuler : I added a brief example at the end of my post. $\endgroup$ Mar 27 '18 at 19:59

Consider $\ell^2(\mathbb C) = \{ (z_n)_{n\in\mathbb Z} : \sum_{n\in Z}|z_n|^2<\infty\}$ and the right-shift operator $T:\ell^2(\mathbb C)\to\ell^2(\mathbb C)$ which satisfies $T((z_n)_{n\in\mathbb Z}) = (z_{n+1})_{n\in\mathbb Z}$. Then for $z,w\in\ell^2(\mathbb C)$, $$ \langle Tz,w\rangle = \sum_{n\in\mathbb Z} z_{n+1}\overline w_n = \sum_{n\in\mathbb Z} z_n\overline w_{n-1} = \langle z,T^\star w\rangle, $$ that is, $T^\star$ is the left-shift operator which maps $(z_n)_{n\in\mathbb N}$ to $(z_{n-1})_{n\in\mathbb N}$. Clearly $T\ne T^\star$, but by symmetry $$ TT^{\star}z = T^\star Tz = z, $$ so that $TT^\star = T^\star T=I$. Indeed, $T$ is a unitary (and hence normal) operator which is not self-adjoint.

  • $\begingroup$ There is an error, indeed $$TT^{*}(0,1,0,\dots)=0$$ and $$T^{*}T(0,1,0,\dots)=(0,1,0,\dots).$$ $\endgroup$
    – A. Bag
    May 3 at 0:27

Soit $$ A= \begin{bmatrix} 1 & 0 \\ 0 & i \end{bmatrix}.$$ Then $A$ is normal and $i \in\sigma(A)$. So $ \sigma(A)\nsubseteq \mathbb R $, and hence $A$ is not self-adjoint.


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.