# When is rank-1 perturbation to a positive operator still positive?

Let $$A : \mathcal{H} \to \mathcal{H}$$ and $$B : \mathcal{H} \to \mathcal{H}$$ be trace-class (hence compact) Hermitian operators on a separable Hilbert space. Assume that $$A$$ is strictly positive and that $$B$$ is positive and rank-1. I'm interested in conditions when $$A - \epsilon B \ge 0$$ for some strictly positive $$\epsilon \in \mathbb{R}$$ (as usual, $$>$$ and $$\ge$$ for operators refers to being positive (semi)definite).

If $$A$$ is finite dimensional, then $$A - \epsilon B \ge 0$$ for some $$\epsilon > 0$$ always. This is because the smallest eigenvalue of $$A - \epsilon B$$ obeys $$\lambda_\min >0$$ for $$\epsilon=0$$ and varies continuously with $$\epsilon$$.

If $$A$$ is infinite dimensional and $$B=\vert \phi\rangle\langle\phi\vert$$ for some eigenvector $$\vert \phi\rangle$$ of $$A$$ (with corresponding eigenvalue $$\lambda>0$$), then it is clear that $$A - \epsilon B\ge 0$$ for $$\epsilon\le \lambda$$.

What about when $$A$$ is infinite dimensional and $$B$$ does not have the form of $$\vert \phi\rangle\langle\phi\vert$$?

• It is not always true in finite dimensions, if $A$ can be positive semidefinite. You'd need all of its eigenvalues to be strictly positive. In infinite dimensions you need $\phi$ to belong to the spectral projection $\chi_{[\epsilon,\infty)}(A)$, assuming $\|\phi\|=1$. – Nik Weaver Apr 3 '20 at 17:00
• Sorry, I should have been more explicit. $A$ is strictly positive. – Artemy Apr 3 '20 at 17:24

In "On Majorization, Factorization, and Range Inclusion of Operators on Hilbert Space (1966)", R. G. Douglas proved the following result (Theorem 1 in the paper):

Theorem. Let $$C$$ and $$D$$ be bounded linear operators on a real or complex Hilbert space $$\mathcal{H}$$; then the following are equivalent:

(i) $$C\mathcal{H} \subseteq D \mathcal{H}$$.

(ii) There exists a number $$\lambda \in [0,\infty)$$ such that $$CC^* \le \lambda^2 DD^*$$.

(iii) There exists a bounded linear operator $$E$$ on $$\mathcal{H}$$ such that $$C = DE$$.

Now, if you choose $$C$$ in the theorem as the positive square root $$\sqrt{B}$$ of $$B$$ and $$D$$ in the theorem as the positive square root $$\sqrt{A}$$ of $$A$$, you can characterize the property you are interested in by means of a range condition.

More precisely:

Corollary. Write your rank-$$1$$ operator $$B$$ as $$B = \alpha \vert \phi\rangle\langle\phi\vert$$ for a number $$\alpha > 0$$ and a vector $$\vert \phi\rangle \in \mathcal{H}$$ of norm $$1$$ (not necessarily an eigenvector of $$A$$). Then the following are equivalent:

(i) There exists $$\varepsilon > 0$$ such that $$A \ge \varepsilon B$$.

(ii) $$\vert \phi\rangle$$ is an element of the range of $$\sqrt{A}$$.