Example of normal operator on infinite-dimensional Hilbert spaces 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.

 A: 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.
A: 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 $S$.
A: 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.
