While I was learning about operators I stumbled apon the following defintion.
Definition:
Let $T : H_1 \rightarrow H_2$ be a linear map.
There exists an operator $T^* H_2 \rightarrow H_1$ such that
$\langle Tx,y \rangle_{H_2}=\langle x,T^*y \rangle_{H_1}$ (1)
The operator $T^*$ is called the adjoint operator (regarding $T$).
I wanted to try some (simple looking exercise) to get a feeling for this topic.
Exercise:
The two-sided right shift operator $T: l^2(\mathbb{Z}) \rightarrow l^2(\mathbb{Z})$
, where $l^2(\mathbb{Z}):=${$(…,x_{-n},x_{-n+1},…,x_{-1},x_0,x_1,…,x_{n},…) \text{sucht that} x_k \in \mathbb{K} \text{and} \sum_{\infty}^{\infty}|x_k|^2 < \infty $}.
The operator $T$ is defined as $ T(x_n)_n = (x_{n-1})_n $.
Determine $T^*$, and find out if $T$ is a unitary operator (i.e. $T^*T=TT^*$)
My approach:
First, I wanted to understand how $T$ "acts".
So let $(x_n)_{n \in \mathbb{Z}}$ be in $l^2(\mathbb{Z})$, $(x_n)_{n \in \mathbb{Z}}=(…,x_{-n},…,x_{-1},x_0,x_1,…,x_n,…)$
than $T(x_n)_{n \in \mathbb{Z}}=(x_{n-1})_{n \in \mathbb{Z}}=(…,x_{-n-1},…,x_{-2},x_{-1},x_0,…,x_{n-1},…)$
just shifts the sequence to the right (okay, I see where the name for this operator comes from)
I am not really sure how to use (1) to determine $ T^* $, by guessing I would say that $T^*(x_n)_{n \in \mathbb{Z}}=(x_{n+1})_{n \in \mathbb{Z}}$.
Then it would be easy to see that $T^*T=id$ and $TT^*=id$ and thus $T$ is unitary.
Question: Is my answer correct? And although it was possible to solve this by guessing, I would like to know how to solve it by using the definition of adjoint operator. Would be thankful if someone could show me.
Best Answer
Observe that $$\langle T^*Tx,x\rangle =\|Tx\|^2=\|x\|^2=\langle x,x\rangle $$ Hence $T^*T=I.$ Moreover $T$ is invertible and $T^{-1}(x_n)=(x_{n+1}).$ Thus $T^*=T^{-1}$ therefore $T$ is a unitary operator.
Summarizing an invertible isometry is always a unitary operator.