# A square root inequality for symmetric matrices?

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 unambiguously via the spectral theorem. Also, I use the conventional Frobenius scalar product and norm $$\newcommand\abs[1]{\lvert#1\rvert}\newcommand\pair[2]{\langle#1, #2\rangle}\pair A B:=\operatorname{Tr}(A^tB), \qquad \abs A^2:=\pair A A.$$

Question: is the following inequality true $$\abs{A^{\frac 12}-B^{\frac 12}}^2\leq C_N \abs{A-B}\quad ???$$ 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 $$\abs{\sqrt a-\sqrt b}^2\leq C\abs{a-b}$$, which of course is true due to $$\abs{\sqrt a-\sqrt b}^2=\abs{\sqrt a-\sqrt b}\times \abs{\sqrt a-\sqrt b}\leq \abs{\sqrt a-\sqrt b} \times \abs{\sqrt a+\sqrt b}=\abs{a-b}$$.

$$\DeclareMathOperator\diag{diag}$$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 $$\abs{A^\frac 12-B^\frac 12}^2 =\sum\limits_{i=1}^N \abs{\sqrt a_i-\sqrt b_i}^2 \leq \sum\limits_{i=1}^N \abs{a_i-b_i} \leq \sqrt N \left(\sum\limits_{i=1}^N \abs{a_i-b_i}^2\right)^\frac 12=\sqrt N \abs{A-B}.$$

Some hidden convexity seems to be involved, but in the general (non diagonal) case I am embarrassingly 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.

Suvrit's answer to Subadditivity of the square root for matrices and answer to Ratio sum comparison on operators 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 $$d(A,B)=\min\limits_U \abs{A^\frac 12-B^\frac 12U}$$ (the infimum runs over unitary matrices $$UU^t=\operatorname{Id}$$).

• Is there a reason you call your matrices $U$ unitary rather than orthogonal? (They surely are unitary, but I am used to this choice of terminology only when one does not want to insist a priori on the matrices being real, as you do.) Jan 23, 2023 at 19:13
• Not really, I came across the question while reading some papers from Quantum-based optimal transportation, and I guess in this mathematical physics context the terminology "unitary" is predominant? But maybe not, and perhaps I was only unconsciously influenced by the paper I was reading at the time. I confess I was not even aware of the subtle real/complex difference in vocabulary that you pointed out, actually I htough unitary/orthogonal was rather a question of finite/ininite dimension, matrices vs. operators. Guess I was wrong! Jan 23, 2023 at 20:24

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 $$A=\operatorname{Id},B=0$$).
• 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. May 28, 2020 at 12:50