# $2$-norm distance between square roots of matrices

Suppose two square real matrices $$A$$ and $$B$$ are close in the Schatten 1-norm, i.e. $$\|A-B\|_1=\varepsilon$$. Can this be used to put a bound on the Schatten 2-norm distance between their square roots. Namely, is there something of the form $$\|\sqrt{A}-\sqrt{B}\|_2\leq f(\varepsilon)$$? It is important that $$f(\varepsilon)$$ be independent of the dimension of the matrices. One can assume that these are symmetric, positive definite matrices.

I have a proof for the above statement when $$A$$ and $$B$$ are taken to be simultaneously diagonal. However, I was wondering if there is a more general proof.

• How do you make sense of $\sqrt{A}$ when $A$ is not symmetric $\ge 0$? – YCor Jul 31 '19 at 14:33
• @YemonChoi It should be true as stated. I have an idea how to get it (though it'll take me some time to check the details) but, if indeed true, it should be a textbook stuff, so even if I'm not mistaken, I'll wait a bit before sharing my home-made computations in case somebody has a good reference. – fedja Jul 31 '19 at 18:18
• @DavidRoberts: This arose in a quantum information context, in trying to prove that trace distance between density matrices being small implies fidelity of their canonical purifications is large. – Pratik Rath Aug 1 '19 at 7:16
• @fedja Thanks for the help. I found a reference where something equivalent to this is proved. Look at theorem 1 of arxiv.org/pdf/1207.1197.pdf – Pratik Rath Aug 1 '19 at 7:18
• PratikRath @fedja Indeed I am now extremely embarrassed for not recognising this result straightaway; not because I claim I could easily come up with the proof (unlike fedja) but because this result is mentioned and used a lot in areas related to my research. It is (assuming positivity etc as in YCor's edit) the Powers-Stormer inequality c.f. mathoverflow.net/questions/198610/… – Yemon Choi Aug 1 '19 at 13:28

Theorem (Powers–Størmer, 1970, Lemma 4.1; link) Let $$S$$ and $$T$$ be positive Hilbert-Schmidt operators on a Hilbert space. Then $$\Vert S - T \Vert_2^2 \leq \Vert S^2-T^2\Vert_1$$, where $$\Vert \quad\Vert_p$$ denotes the Schatten $$p$$-norm.
The starting idea of the proof is to work in an ONB with respect to which $$R=S-T$$ is diagonal (which is possible by the spectral theorem for compact self-adjoint operators). One then observes that, putting $$Q=S+T$$, we have $$S^2-T^2 = (RQ+QR)/2$$ and then one exploits the fact that $$Q\geq \pm R$$.
I don't have the answer to your question, but I can give you the following: $$\|\sqrt A-\sqrt B\|_\infty\le\sqrt{\|A-B\|_\infty\,}\,,$$ where the $$\infty$$-Schatten norm is nothing but the operator norm.