Consider the following Laurent polynomial matrix-valued function in the variable $x\in\mathbb{C}$
$$
A(x) = \begin{bmatrix} 0 & x \\ x^{-1} & 0\end{bmatrix}.
$$


I'm interested in finding a factorization of $A(x)$ of the form
$$\tag{$\star$} \label{fact}
A(x) = C^\top(x^{-1})\begin{bmatrix} 0 & 1 \\ 1 & 0\end{bmatrix}C(x),
$$
where $C(x)$ is a suitable $2\times 2$ rational matrix-valued function and $\bullet^\top$ denotes transposition. An example of such a factorization is given, for instance, by
$$
C(x)=\begin{bmatrix} 1 & 0 \\ 0 & x\end{bmatrix}.
$$



> 
> **My question.** Does there exist a factorization of $A(x)$ as in \eqref{fact} such that the factor $C(x)$ possesses the two additional properties below?
> 
> 
> 1. The entries of $C(x)$ have no singularities at $x\in\mathbb{C}$, $|x|\le 1$, and
> 2. $C(x)$ has full rank for every $x\in\mathbb{C}$ such that $|x|\le 1$.
> 
> 
> 


I can find factors $C(x)$ that satisfy either property 1 or property 2, but not both. Thus, any comment/suggestion is really appreciated. 



Finding a particular matrix factor
This is impossible. Let 
\begin{equation\*}
C(x) = \left(
\begin{array}{cc}
a(x) & b(x) \\
c(x) & d(x) 
\end{array} \right)
\end{equation\*}


and $\theta(x)=\det C(x)$. From 
$$a(x^{-1})d(x)+b(x)c(x^{-1})=x,\,a(x)c(x^{-1})+c(x)a(x^{-1})=0$$
we have
$$a(x^{-1})=\frac{xa(x)}{\theta(x)}.$$
Assuming (1) and (2), the function $a(x)$ would be analytic both for $|x|\le 1$ and for $|x|\ge 1$, which means it is a constant. Obviously, $a=0$ then (take $x=0$). 


Then we have 
$$c(x^{-1})=-\frac{xc(x)}{\theta(x)},$$ 
and $c=0$ by the same argument.


