Eigenvalues of an operator in an infinite-dimensional Hilbert space

functional-analysishilbert-spacesisometryoperator-theoryspectral-theory

This is from Cheney's Analysis for Applied Mathematics:

Let $X$ be a Hilbert space having a countable orthonormal base $\{u_1, u_2, \dots\}$. Define an operator $A$ by the equation
$$Ax = \sum_{n=1}^\infty\langle x, u_n \rangle u_{n+1}$$
What are the eigenvalues of $A$?
Is $A$ compact?
Is $A$ Hermitian?
What is the norm of $A$?

Here are my thoughts on this:
From the orthonormal basis theorem, for each $x$ in the Hilbert space $X$, we have $x =\sum \langle x,u_i \rangle u_i$. Let $a_i :=\langle x,u_i\rangle$. Then $Ax = \sum a_n u_{n+1}.$ Solving for eigenvalues, $Ax = \lambda x$, yields $$\sum_{n=1}^{\infty}a_n u_{n+1}=\sum_{n=1}^{\infty}\lambda a_n u_n.$$ But I cannot go further from here to find the eigenvectors and corresponding eigenvalues. This is specifically difficult to intuitively imagine, as each component of $x$ in the $n$-th direction is projected by $A$ to the $(n+1)$-th one.

Am I on the right track so far? I'd appreciate any help on this part and the rest of the problem (whether $A$ is compact and Hermitian).

Best Answer

Summarising answer, to avoid the question dangling in the unanswered queue, made community:

  • $A$ has no eigenvalue.
    $A$'s spectrum is the closed unit disk: $\,\sigma(A)=\big\{\lambda\in\mathbb C\,\big\vert\, |\lambda|\leqslant 1\big\}$
  • $A$ is the Right shift operator, thus an isometry, and in particular non-compact.
  • Its adjoint $A^*$ is a partial isometry, so $A$ is not Hermitian.
  • $\|A\|=1\,$ follows from the preceding.

This subject is well represented on this site$-$do a search in the header line, by looking up "shift operator spectrum" for instance.

Related Question