Sign up ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Given a self-adjoint operator $\hat{T}$ on a Hilbert space $\mathcal{H}$, and assuming it has a basis of eigenvectors $\{\phi_n\}$ such that $\hat{T}\phi_n=\lambda_n\phi_n$, one can consider the subspace $$V=\textrm{span}\{\phi_n:\lambda_n>0\},$$ inside of which $\hat{T}$ is positive definite. My question is, does there exist a way of defining this subspace invariantly in terms of $\hat{T}$ without making reference to its eigenvectors?

(Some motivation: given the hydrogen-atom hamiltonian in quantum mechanics, $\hat{H}=-\frac{\hbar^2}{2m}\nabla^2 -\frac{e^2}{r}$, the subspace where $\hat{H}<0$ is the linear span of all bound states, which correspond to elliptical orbits in the analogous classical problem.)

share|cite|improve this question
When you say $span$, do you mean the algebraic span or its closure? The closure could be characterized as the orthogonal complement of the kernel of $T+|T|$, where $|T|=\sqrt{T^2}$. – doug Feb 25 '12 at 16:51
You could also look at the spectral projection corresponding to the right half spectral plane. – Aaron Hoffman Feb 25 '12 at 16:55
Xogn: I mean the closure of the algebraic span. Do you have references for this characterization? Also, by $|T|$, do you mean $\sqrt{T^2}$ or $\sqrt{T^\ast T}$? – Emilio Pisanty Feb 26 '12 at 13:29
Sorry, stupid question; $T$ is self-adjoint. – Emilio Pisanty Feb 26 '12 at 13:47

Your Answer


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

Browse other questions tagged or ask your own question.