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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 $S_i$.

sqrt6
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  • 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? – julio_es_sui_glace Mar 19 '24 at 17:36

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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 $n/2$.

julio_es_sui_glace
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