vector orthogonal to range of $T - \lambda I$

Question

Suppose that $T$ denotes the unilateral right shift operator on $\ell^2(\mathbb{N})$. My argument here $$\| T \| = \sup_{\| f \|=1} \| Tf \| = \sup_{\| f \| =1} \| T \left( \sum_{k=1}^{\infty} \langle e_k, f \rangle e_k \right) \| = \sup_{\| f \| =1} \| \sum_{k=1}^{\infty} \langle e_k, f \rangle e_{k+1} \| = 1,$$ show that the norm of $T$ is 1. I want to show that there is a vector $v \in \ell^2(\mathbb{N})$ orthogonal to the range of $T - \lambda I$ if and only if $\left| \lambda \right| <1$. I've played around with inner products for a few days and haven't made any progress. Note that this is a self-study question.

Cheers.

Answer 1

If $\langle (T - \lambda I)x, v \rangle = 0$ for all $x$ then $\langle x, (T - \lambda I)^\ast v \rangle = 0$ for all $x$. This is equivalent to the statement that $v$ is an eigenvector of the adjoint $T^\ast$ with eigenvalue $\bar{\lambda}$.

You have to do three things:

1. compute the adjoint $T^\ast$ (you can write it down expliclitly).

2. Show that every $\mu$ with $|\mu| < 1$ is an eigenvalue of $T^\ast$ by finding an eigenvector (you can write it down explicitly, too).

3. Show that every $\mu$ with $|\mu| \geq 1$ is not an eigenvalue of $T^\ast$.