I'm working through a book section on linear bounded operators at the moment and there is a section where it is mentioned that there are examples of such operators $T:H \rightarrow H$, such that $\lambda$ is an Eigenvalue of $T$ but $\lambda^*$ is not an Eigenvalue of its adjoint operator $T^*$. $H$ is a Hilbert space (potentially infinite dimensional).
This is discussed in a section dealing with the distinctions between things such as point spectrum, continuous spectrum and residual spectrum, which leads me to believe such an example can only exist in an infinite dimensional setting and that my failure to find an example is due to intuitions based in finite dimensional linear algebra. My questions are as follows:
1) Why is the following not a proof that such a thing does not exist: $$\left[\langle T x | y\rangle = \langle \lambda x | y\rangle = \langle x | \lambda^* y\rangle \right] \land \left[ \langle T x | y\rangle = \langle x | T^* y\rangle \right]$$
2) Please share with me such an example if a proper one can be constructed, since the book I'm working with only mentions this in passing and doesn't give one.
3) There is a chance that I misunderstood the slightly unclear and hasty remark and the book is instead trying to tell me that there are operators $T^*$ with Eigenvalues $\lambda^*$, for which the operator $T$ does not have $\lambda$ as its eigenvalue. But would this not be equivalent? It seems to me that $T^{**} = T$ and $\lambda^{**}=\lambda$. If not, why not and can you give me an example of such objects?
Answers to any of these or all of these would be greatly appreciated.
Consider the left shift operator $T : \ell^2\to \ell^2$. This has an eigenvalue $\lambda = 0$. However, the adjoint $T^*$ is the right shift operator, which does not have $0$ as an eigenvalue.
Your attempted proof does not show that $T^*y = \lambda^*y$, merely that $\langle x, T^*y\rangle = \langle x, \lambda^*y\rangle$ for all eigenvectors $x$ of $T$ with eigenvalue $\lambda$. You could write the same proof attempt with the same $y$ and with any other eigenvalue $\lambda$, and clearly $y$ can't be an eigenvector with more than one eigenvalue.
The condition that you mentioned in 3) is equivalent, as taking the double adjoint of an operator on a Hilbert space produces the original operator. You could let $T$ be the right shift operator instead, and then you would have your condition.