Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I have a question regarding operator theory and would be glad if someone could help. I have a linear operator $A$ that is non-self-adjoint, unbounded and is densely defined in a Hilbert space $H$. I know how the spectrum of this operator looks like. However, not all points of the spectrum are eigenvalues. Given a point in the spectrum $\lambda_j$ (in my case $\lambda_j \in \mathbb{R}$), is there a way to check, if this point is an eigenvalue in the mathematical meaning of this word, i.e., that there exist a function $\varphi_j \in H$ such that $A\varphi_j = \lambda_j\varphi_j$?

Of course, the operator $A$ is given explicitly, i.e., I know how the corresponding linear prescription looks like.

share|improve this question

1 Answer 1

If the only data you have available is the spectrum, then the answer is no. You can have two operators with the same spectrum, where a certain point is an eigenvalue for one operator and not an eigenvalue for the other.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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