# 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 contraction with $$d(x,A)\geq\frac{1}{4}$$. 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,

Take $$A=\mathbb C\oplus\mathbb C\subset M_2(\mathbb C)$$, and $$x=\begin{bmatrix} 0&1/4\\0&0\end{bmatrix}.$$
Then $$A'=A$$. If $$u\in A'=A$$, then $$u=\begin{bmatrix} \lambda&0\\0&\mu\end{bmatrix}$$, with $$\lambda,\mu\in\mathbb T$$. Then $$uxu^*=\begin{bmatrix} 0&\lambda\overline\mu/4\\ 0&0\end{bmatrix},$$ and $$xuxu^*=uxu^*x=0$$. In particular, $$[x,uxu^*]=0$$.
• Right, make sense, thanks! To follow up though, what if $x$ is positive, or even just self-adjoint? – Alessandro Vignati Oct 16 '18 at 3:40
• If you take the same $A$ as above and $x=\begin{bmatrix} 1&1/4\\1/4&1\end{bmatrix}$, then $x\geq0$ and $xuxu^*=uxu^*x$ for all unitary $u\in A$. – Martin Argerami Oct 16 '18 at 4:31
• No..In particular in the top left corner of $uxu^*x$ you'll get $1+\frac{\lambda\overline\mu}{4}$ but in the top left corner you'll get it's adjoint, and unless $\lambda=\mu$ they're different. – Alessandro Vignati Oct 16 '18 at 12:19