A linear operator $T$ on a separable Hilbert space $H$ is said to be a weighted shift operator if there is some orthogonal basis $\{e_n\}_n$ and weight sequence $\{w_n\}_n$ such that $$Te_n=w_n e_{n+1}, \forall n\in N. $$ Its adjoint operator is given by $$T^* e_n=w_{n-1} e_{n-1} \quad (n\geq 1), \quad T^*e_0 =0$$
In the paper by Allen Shields where he discussed on the the spectrum of a weighted shift, the following theorem was stated:
Theorem: Let $T$ be a weighted shift then the eigenvalues of $T^*$ are simple.
Proof: Let $0\neq \lambda\in \sqcap_0 (T^*)$ with $f=\sum_{n\geq 0} \alpha_n e_n$ as a corresponding eigenvector. From $T^* f = \lambda f$ we have $$\sum_{n\geq 1} \alpha_n w_{n-1} e_{n-1}=\sum_{n\geq 0} \lambda \alpha_n e_n.$$ And so, $\alpha_{n+1} w_n=\lambda \alpha_n \text{ for all }n \geq 0$. Therefore $$\alpha_n=\frac{\alpha_0 \lambda^{n}}{w_0 w_1 ...w_{n-1}} \text{ for all } n\geq 1.$$
From this we see that the eigenvalues are simple.
I know that an eigenvalue is simple if it is of geometric multiplicity $1$. How does this the conclusion relate. thank you
The formula shows that the eigenvector is uniquely determined up to a scalar multiple by the eigenvalue. In other words, the eigenspace is $1$-dimensional. This is the definition of "simple eigenvalue" that the author appears to be using. (Note that there is some ambiguity when one says "multiplicity $1$" since there are two notions of multiplicity which need not coincide.)