A normal operator is self-adjoint over $\mathbb{C}$ or $\mathbb{R}$?

Question

A normal operator is self-adjoint over $\mathbb{C}$ or $\mathbb{R}$?

I know that if it's self-adjoint, then it is indeed normal, but is the converse true?

Answer 1

No, for example unitary operators are also normal.

Answer 2

Hint: for $(a_n)\subset\mathbb{C}$ a bounded sequence that does not consist of only real numbers, consider the diagonal operator $D:\ell^2\to\ell^2$ given by $$(x_n)_{n=1}^\infty\mapsto(a_nx_n)_{n=1}^\infty$$ Compute the adjoint of $D$ (answer: it is the diagonal operator with diagonal sequence $(\bar{a_n})$. Observe that it is normal, since $D^*D=DD^*=$ the diagonal operator with diagonal sequence $(|a_n|)$.

Observe that since $(a_n)$ does not consist only of real numbers, we have $D\neq D^*$, so $D$ is not self-adjoint.