Prove that $T$ is the orthonal projection on the range of the right shift

Question

Consider the following operator on $\ell_{\mathbb{N}^*}^2(\mathbb{C})$ defined as:

$$ T(x_1,x_2,\cdots)=(0,x_2,x_3,\cdots).$$

I want prove that $$\sigma(T)=\{0,1\}.$$

I have the following idea: let us consider the right shift $$S_r(x_1,x_2,\cdots)=(0,x_1,x_2,x_3,\cdots).$$ If I show that $T$ is the orthonal projection on the range of $S_r$, then clearly $$\sigma(T)=\{0,1\}.$$

Because, it is well known that the spectrum of an Orthogonal Projection Operator is equal to $\{0,1\}$.

Note that $$T=S_rT_\ell,$$ where $$S_\ell(x_1,x_2,\cdots)=(x_2,x_3,\cdots).$$

Answer 1

I cannot see why you want to involve the shift at all.

You have $T^2-T=0$, so $$\{0\}=\{\lambda^2-\lambda:\ \lambda\in\sigma(T)\}.$$So $\sigma(T)=\{0,1\}$.

For a way to prove this without any theory, look at this.

Answer 2

First of all, let $e_1 = \left( 1, 0, \dots \right) $ be the first element of the canonical basis for $\ell^2(\mathbb{C})$ and $W= e_1^{\bot}=\langle e_i\rangle_{i\geq 2})$ (here $\langle \bullet \rangle$ identifies the linear span).

Both $\langle e_1\rangle$ and $W$ are $T$-invariant, so $T-\lambda Id$ is invertible iff $T-\lambda Id |_{\langle e_1 \rangle} $ and $T-\lambda Id|_W$ are both invertible.

$T-\lambda Id |_{\langle e_1 \rangle} \equiv -\lambda Id\ $ so it is invertible iff $\lambda \neq 0$.

$T-\lambda Id|_W \equiv \left(1-\lambda\right)Id\ $ so it is invertible iff $\lambda \neq 1$.