My question is : Does there exist a bounded linear operator $T:\ell^{2}(\mathbb{N})\rightarrow\ell^{2}(\mathbb{N})$ which is normal but not self-adjoint?
Just to be clear, if $H$ is an Hilbert space, then a bounded linear operator $T:H\rightarrow H$ is said to be normal if $TT^{*}=T^{*}T$, or equivalently if $\Vert T(x)\Vert =\Vert T^{*}(x)\Vert $ for every $x\in H$.
It is easy to see that the operator $S:\ell^{2}(\mathbb{Z})\rightarrow\ell^{2}(\mathbb{Z}):(x_n)_{n\in\mathbb{Z}}\mapsto (x_{n-1})_{n\in\mathbb{Z}}$ works for $\ell^{2}(\mathbb{Z})$ (it is normal without being self-adjoint), but I'm unable to find a map that would work for $\ell^{2}(\mathbb{N})$.
Does anyone have any idea?
Best Answer
You could take $T$ to be multiplication with some complex number, $\lambda \in \Bbb{C}$. Then $T^*$ would be multiplication with the complex conjugate $\bar{\lambda}$. Hence if $\lambda \not \in \Bbb{R}$ then $T \neq T^*$ and clearly $TT^* = T^*T$.