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,



Distance between subalgebras and positive elements in matrices
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.)


