# Finding a particular matrix factor

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.

• Have you tried to first solve the functional equation $f(x)f(x^{-1}) = 1$ for the determinant $f(x) := \det(C(x))$ of $C(x)$? – Jochen Glueck Jun 25 '18 at 4:45
• @JochenGlueck: If you solve your determinant equation then a solution satisfying the desired requirements is given by any $C(x)$ such that $\det(C(x))=1$. But perhaps I didn't understand you question. – Ludwig Jun 25 '18 at 14:41
• Yes. Now, if we could prove that every holomorphic solution of the functional equation which has no poles and no zeros in the unit disk is, say, constant (i.e. identically $1$ or $-1$) this would impose a severe restriction on the possible choices of $C(x)$. But unfortunately I don't see at the moment whether each such solution of the functional equation is constant. – Jochen Glueck Jun 25 '18 at 16:43
• Doesn't it follow immediately from Liouville? $f(x) = 1/f(x^{-1})$ together with the fact that $f$ has no zero at $0$ shows that it is bounded. – Achim Krause Jun 26 '18 at 8:51

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.
• Thanks for the answer! Could you please elaborate a little more on your sentence "assuming (1) and (2), the function $a(x)$ would be analytic both for $|x|\le 1$ and for $|x|\ge 1$"? In particular, how is requirement (2) applied here? – Ludwig Jun 26 '18 at 14:56
• For $|x|\le 1$ the function $a(x)$ is analytic by the assumption (1), and $a(x^{-1})$ is analytic because numerator and denominator are analytic and the latter is not zero. – Alex Gavrilov Jun 26 '18 at 15:47