5
$\begingroup$

In this post all my matrices will be $\mathbb R^{N\times N}$ symmetric positive semi-definite (psd), but I am also interested in the Hermitian case. In particular the square root $A^{\frac 12}$ of a psd matrix $A$ is defined unambigusouly via the spectral theorem. Also, I use the conventional Frobenius scalar product and norm $$ <A,B>:=Tr(A^tB), \qquad |A|^2:=<A,A> $$

Question: is the folowing inequality true $$ |A^{\frac 12}-B^{\frac 12}|^2\leq C_N |A-B|\quad ??? $$ for all psd matrices $A,B$ and a positive constant $C_N$ depending on the dimension only.

For non-negative scalar number (i-e $N=1$) this amounts to asking whether $|\sqrt a-\sqrt b|^2\leq C|a-b|$, which of course is true due to $|\sqrt a-\sqrt b|^2=|\sqrt a-\sqrt b|\times |\sqrt a-\sqrt b|\leq |\sqrt a-\sqrt b| \times |\sqrt a+\sqrt b|=|a-b|$.

If $A$ and $B$ commute then by simultaneous diagonalisation we can assume that $A=diag(a_i)$ and $B=diag(b_i)$, hence from the scalar case $$ |A^\frac 12-B^\frac 12|^2 =\sum\limits_{i=1}^N |\sqrt a_i-\sqrt b_i|^2 \leq \sum\limits_{i=1}^N |a_i-b_i| \leq \sqrt N \left(\sum\limits_{i=1}^N |a_i-b_i|^2\right)^\frac 12=\sqrt N |A-B| $$

Some hidden convexity seems to be involved, but in the general (non diagonal) case I am embarrasingly not even sure that the statement holds true and I cannot even get started. Since I am pretty sure that this is either blatantly false, or otherwise well-known and referenced, I would like to avoid wasting more time reinventing the wheel than I already have.


This post and that post seem to be related but do not quite get me where I want (unless I missed something?)


Context: this question arises for technical purposes in a problem I'm currently working on, related to the Bures distance between psd matrices, defined as $$ d(A,B)=\min\limits_U |A^\frac 12-B^\frac 12U| $$ (the infimum runs over unitary matrices $UU^t=Id$)

$\endgroup$
7
$\begingroup$

The classical operator generalization of the scalar inequality $|\sqrt{a}-\sqrt{b}|^2 \leq |a-b|$ is the Powers-Størmer inequality, which involves two different norms : the trace norm $\|X\|_1 = \operatorname{Tr}|X|$ and the Froebenius norm $\|X\|_2 = (\operatorname{Tr}(X^* X))^{\frac 1 2}$, where $|X| = (X^* X)^{\frac 1 2}$ is the usual absolute value of matrices. It says that for all positive matrices $A,B$ (or operators on a Hilbert space), $$ \|\sqrt{A} - \sqrt{B}\|_2^2 \leq \|A-B\|_1.$$

It implies a positive answer to your question with $C_N = \sqrt{N}$, because by Hoelder's inequality, $\|A-B\|_1 \leq \sqrt{N} \|A-B\|_2$. The constant is optimal (take $A=\operatorname{Id},B=0$).

The Powers-Størmer surprisingly does not have a wikipedia page yet, but it probably appears in most textbooks on operator algebras or matrix analysis. The original reference is R. T. Powers, E. Størmer, Free states of canonical anticommutation relations, Commun. Math. Phys. 16, 1-33 (1970).

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Great, merci beaucoup Mikael! This $L^1$ version is even better than what I needed: I really was trying to control by $L^1$, but for some resaon I was convinced that $L^2$ should be used as an intermediate step. I guess I was wrong. Thank you again. $\endgroup$ – leo monsaingeon May 28 at 12:50
  • $\begingroup$ Content que ça puisse t'être utile ! $\endgroup$ – Mikael de la Salle May 29 at 11:46
  • $\begingroup$ Oh yes, merci encore, ça me tire une belle épine du pied! By the way, since you appear to be the expert here, and mostly out of curiosity: Have you ever heard of the Bures distance in the infinite-dimensional/operator-theoretical framework? It seems to be quite popular in information theory these days, and apparently plays a role in quantum mechanics, so I was just wondering... (Although I must admit I didn't particularly search the literature for this) $\endgroup$ – leo monsaingeon May 29 at 12:03
  • $\begingroup$ @leomonsaingeon : I do not remember having heard of the Bures distance before your question (but I am no expert in Quantum information theory). But its definition makes perfect sense for infinite dimensional spaces (and more generally for any tracial von Neumann algebra), and I guess that most properties known for finite matrices remain true in this setting. $\endgroup$ – Mikael de la Salle May 29 at 12:35

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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