Example of normal operator on infinite-dimensional Hilbert spaces

Question

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.

Answer 1

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 $S$.

Answer 2

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.

Answer 3

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.