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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.

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  • $\begingroup$ Have you tried to first solve the functional equation $f(x)f(x^{-1}) = 1$ for the determinant $f(x) := \det(C(x))$ of $C(x)$? $\endgroup$
    – Jochen Glueck
    Commented Jun 25, 2018 at 4:45
  • $\begingroup$ @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. $\endgroup$
    – Ludwig
    Commented Jun 25, 2018 at 14:41
  • $\begingroup$ 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. $\endgroup$
    – Jochen Glueck
    Commented Jun 25, 2018 at 16:43
  • $\begingroup$ 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. $\endgroup$
    – Achim Krause
    Commented Jun 26, 2018 at 8:51

1 Answer 1

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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.

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  • $\begingroup$ 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? $\endgroup$
    – Ludwig
    Commented Jun 26, 2018 at 14:56
  • $\begingroup$ Without (2) the argument doesn't work because the denominator could turn zero. $\endgroup$
    – Alex Gavrilov
    Commented Jun 26, 2018 at 15:43
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    $\begingroup$ 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. $\endgroup$
    – Alex Gavrilov
    Commented Jun 26, 2018 at 15:47

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