Confusion about matrix polynomials

Question

In this answer the point is made that since a unitary and symmetric matrix $U$ commutes with all $U^j$ (I assume those are the matrix powers $U^2$, $U^3$ etc?) it must also commute with $S=\sqrt{U}$, where $\sqrt{\cdot}$ denotes the matrix squre root. I do not understand this argument. The matrix square root is not necessarily unique. I can see that if $S$ commutes with some matrix, so must $U$, but not necessarily the other way around.

In other words can there (or why can there not) be a matrix $\mathcal{S}$, which does not commute with $U$, but which nevertheless fullfills $\mathcal{S}^2=U$?

Answer 1

I believe a stronger assumption needed on $U$. Since $U$ is unitary , so by spectral theorem, all its eigenvalue must be normed $1$.

But if there exists some symmetric matrix $S$ which is the "square root" of $U$, this means for any $\textbf{v}\in\textbf{k}^n$, we must have $$\textbf{v}^TU\textbf{v}=\textbf{v}^TSS\textbf{v}=\textbf{v}^TS^TS\textbf{v}=(S\textbf{v})\cdot (S\textbf{v})\ge0$$ which shows $U$ must be at least positive semidefinite (definite). This cannot be guaranteed.

Answer 2

In fact the claim is not true, even for unitary, symmetric matrices $U$ and $S$. For example, $$U= \begin{pmatrix} -1 & 0 \\ 0 & -1 \end{pmatrix}$$ $$S= \begin{pmatrix} i & 0 \\ 0 & -i \end{pmatrix}$$ are unitary and symmetric, and $S^2=U$, but $U$ commutes with

$$M= \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}$$

while $S$ does not.