5
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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 Jun 25 '18 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 Jun 25 '18 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 Jun 25 '18 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 Jun 26 '18 at 8:51
1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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 Jun 26 '18 at 14:56
  • $\begingroup$ Without (2) the argument doesn't work because the denominator could turn zero. $\endgroup$ – Alex Gavrilov Jun 26 '18 at 15:43
  • 1
    $\begingroup$ 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 Jun 26 '18 at 15:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.