# Distance between subalgebras and positive elements in matrices

I repost here from stackexchange, as I was not given an answer there. (https://math.stackexchange.com/questions/2956530/distance-between-subalgebras-and-commutants-in-matrix-algebras)

This is a question on how to relate two different distances in the matrix setting. Everywhere below, $$M_n$$ denotes the square matrices $$n\times n$$ whose entries are in $$\mathbb C$$. We consider the operator norm $$\|\cdot\|$$ on $$M_n$$.

By a subalgebra we mean a $$C^*$$-subalgebra of $$M_n$$. If $$A\subseteq M_n$$ is a subalgebra and $$x\in M_n$$, define the distance from $$x$$ to $$A$$ as the real number $$d(x,A)=\inf_{a \in A}\|x-a\|$$.

The question is the following: suppose that $$A$$ is a unital $$C^*$$-subalgebra of $$M_n$$ and that $$x\in M_n$$ is a positive contraction with $$d(x,A)\geq\frac{1}{2}$$. Does there exist $$u\in A'$$ (the commutant of $$A$$) such that $$\|[x,uxu^*]\|\geq\frac{1}{16}$$ (or, for what matter, $$\frac{1}{64}$$, the only important thing is that this number doesn't depend on the choice of $$n$$, $$A$$, or $$x$$)?

Beware, I am not asking whether I can find a $$u$$ with $$\|[u,x]\|\geq\frac{1}{16}$$. This is clearly possible since $$y=\int_{\mathcal U(A')}uxu^*d\mu(u)$$, when I am integrating over the Haar measure on the unitary group of $$A'$$, is the conditional expectation onto $$A''=A$$. Since $$y\in A$$, we have $$\|x-y\|\geq\frac{1}{4}$$, and thefore there must be a unitary $$u\in \mathcal U(A')$$ such that $$\|x-uxu^*\|\geq\frac{1}{16}$$, or in other words, such that $$\|[x,u]\|\geq\frac{1}{16}$$.

Thanks and best,

• Trivial typo: you presumably want $\Vert u\Vert =1$. Indeed, do you mean that $u$ should be a unitary element of $A'$? – Yemon Choi Feb 4 at 21:30
• $u$ is a unitary (almost) by definition, of course! But any contraction would do, as any element is a combination of four or them. – Alessandro Vignati Feb 4 at 21:46

YES. Let $$x=x^*$$ and put $$C:=d(x,A)$$. As you observed, there is a unitary element $$v\in A'$$ such that $$\| [x,v] \|\geq C$$. By decomposing $$v$$ into the real and imaginary part, one finds a self-adjoint contraction $$h\in A'$$ such that $$\|[x,h]\|\geq C/2$$. Since $$h$$ is a convex combination of $$p - p^\perp$$, $$p$$ spectral projections of $$h$$, there is a projection $$p\in A'$$ such that $$\|[x,p]\|\geq C/4$$. (This means that the hyperreflexive constant of a finite-dimensional C*-algebra is at most $$4$$---I think the optimal constant is $$2$$, but didn't find a reference). Note that $$\|[x,p]\|=\|p^\perp x p\|$$. Put $$u := p+\sqrt{-1}p^\perp \in U(A')$$. Then, $$p (x uxu^* - uxu^* x) p = pxuxp - pxu^*xp = 2\sqrt{-1} pxp^\perp xp$$ and so $$\| [x,uxu^*]\| \geq 2\|p^\perp x p\|^2\geq C^2/8.$$ (This proof is probably not optimal, but I like working on operator matrices. The displayed formula has a better looking in the operator matrix form.)