0
$\begingroup$

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

$\endgroup$
1
$\begingroup$

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.)

$\endgroup$
  • $\begingroup$ I meant the geometric multiplicity. Thanks . $\endgroup$ – stackuser Jun 19 '19 at 5:19
  • $\begingroup$ The geometric multiplicity is defined to be the dimension of the eigenspace. Thus, an eigenvalue is simple iff it has a $1$-dimensional eigenspace, which is precisely the condition I described in my answer above. It seems you don't yet feel the question has been answered, and if so, please let me know what part of my answer you do not follow, and I will happily clarify. $\endgroup$ – pre-kidney Jun 19 '19 at 5:23
  • $\begingroup$ I totally align with your answer. I was at fault for not specifying the multiplicity type. $\endgroup$ – stackuser Jun 19 '19 at 5:29
  • $\begingroup$ Ok, in that case it will be least confusing if you mark the question as answered. $\endgroup$ – pre-kidney Jun 19 '19 at 5:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.