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,

  • $\begingroup$ Trivial typo: you presumably want $\Vert u\Vert =1$. Indeed, do you mean that $u$ should be a unitary element of $A'$? $\endgroup$
    – Yemon Choi
    Feb 4, 2019 at 21:30
  • $\begingroup$ $u$ is a unitary (almost) by definition, of course! But any contraction would do, as any element is a combination of four or them. $\endgroup$ Feb 4, 2019 at 21:46

1 Answer 1


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.)


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