# 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 unambigusouly via the spectral theorem. Also, I use the conventional Frobenius scalar product and norm $$:=Tr(A^tB), \qquad |A|^2:=$$

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 $$UU^t=Id$$)

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