# Find the Hilbert-Adjoint Operator $T^*$

I have started operator theory in Functional Analysis and have got stuck here:

If $$(e_n)_n$$is a total orthonormal sequence in a separable Hilbert Space $$H$$ and the right shift operator $$T:H\to H$$ is defined by $$T(e_n)=e_{n+1}$$ for $$n=1,2\dots$$ .

Show that $$T$$ is not normal.

Step 1:I need to find $$T^*$$ i.e. the Hilbert -Adjoint Operator first.I am unable to find the operator $$T^*$$.

Step 2: Need to show that $$TT^*\neq T^*T$$.

You have $\langle T^*e_n,e_k\rangle=\langle e_n,Te_k\rangle=\langle e_n,e_{k+1}\rangle=\delta_{n,k+1}=\delta_{n-1,k}$, which implies $T^*e_n=e_{n-1}$ if $n\geq 2$, and $T^*e_1=0$. Then compare $T^*Te_1$ with $TT^*e_1$.
Define for all $\xi \in H = l_2(\mathbb{N})$ the operator $(S\xi)(n) := \xi(n+1)$.
We then calculate for $\xi,\eta \in H$ on the one hand $$\langle \xi, S \eta \rangle = \sum_{n = 1}^\infty \xi(n)\, S \eta(n) = \sum_{n = 1}^\infty \xi(n)\, \eta(n + 1) = 0 \cdot \eta(1) + \sum_{n = 2}^\infty \xi(n - 1)\, \eta(n)\\ = T\xi(1)\, \eta(1) + \sum_{n = 2}^\infty T\xi(n)\, \eta(n) = \sum_{n = 1}^\infty T\xi(n) \, \eta(n) = \langle T \xi, \eta \rangle.$$ On the other hand, we have $$\langle \xi, T^* \eta \rangle = \langle T \xi, \eta \rangle.$$
Uniqueness of the adjoint then gives $S = T^*$.