# Commutators of unitary matrices and their unitary square roots with other matrices

Given some unitary matrix $$U$$, we can possibly find many unitary square root matrices $$S_i$$ so that $$S_i^2=U$$. Let's assume we have an additional (complex) matrix $$T$$ so that it commutes with $$U$$:

$$[U,T]=0.$$

What can we say about the commutator $$[S_i,T]$$? Do these matrices necessarily commute? If not, is there some restriction we can place on $$U$$ so that this holds for all $$S_i$$?

Note that this is an extension this question, but they are not equivalent, because it only asks for the existence of such a matrix $$S_i$$.

• They don't necessarily commute, since $\begin{pmatrix}0&1\\0&0\end{pmatrix}$ is not in the center of $M_2(k)$, you could (maybe) use the dunford decomposition to find some results? Commented Mar 19 at 17:36

As I stated before, they don't necessarily commute, since $$\begin{pmatrix}0&1\\0&0\end{pmatrix}^2 = 0$$ and $$0 \in Z(\operatorname{Mat}_2(k))$$, but $$\begin{pmatrix}0&1\\0&0\end{pmatrix} \notin Z(\operatorname{Mat}_2(k))$$.
Now suppose you have $$S^2 = U$$ and $$[U,T]=0$$ the dunford decomposition of $$S$$ (provided your field is perfect) is the unique $$S = D+N$$ with $$D$$ semi-simple and $$N$$ nilpotent such that they are both polynomials in $$S$$. You also have $$U = D' + N'$$ and these are polynomials in $$U$$ and $$S$$, so if we compute $$S^2 = D^2 + 2 DN + N^2$$, which gives you $$D^2 = D'$$ and $$2DN + N^2 = N'$$.
It is obvious that $$T$$ commutes with $$U$$ iff it commutes with both $$D$$ and $$N$$ (these are polynomials in $$A$$), so you get $$D^2$$ and $$2DN + N^2$$ commute with $$T$$.
One could get some result like the rank of the commutator has to be small, I believe at most $$n/2$$.