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Let $T(z)$ be a meromorphic square matrix function, that is - a matrix whose entries are complex meromorphic function of one variable.

Recall that such a $T$ is said to have a right pole of order $r$ at direction $v$ at the point $z_0$ if $v(z)$ is a vector valued function holomorphic near $z_0$, such that $v(z_0) \ne 0$ and such that the vector valued function $T(z)v(z)$ has a pole of order $r$ at $z_0$.

It is well known that if such a $T$ is holomorphic at infinity, then it may be represented by an equation of the form

$T(z) = D+C(zI-A)^{-1}B$

Furthermore, assuming $A$ is of size $n$ and $T$ has $n$ poles (i.e, $T$ has a maximal McMillan degree - a representation with $n$ small as possible) then this representation is unique up to similarity: any other matrices $A',B',C',D'$ which represents $T$ are of the form $D'=D$, $A' = N^{-1}AN$, $B'=N^{-1}B$, $C'=CN$, where $N$ is some invertible matrix

My question: Suppose we are given two meromorphic matrix functions $T_1(z) = D+C(zI-A)^{-1}B$ and $T_2 = D'+C'(zI-A')^{-1}B'$. Suppose both of these are of maximal McMillan degree, $A$ and $A'$ are of the same size, and supopse they share the same pole data.

Does it follows that there exists an invertible matrix $N$ such that $A'=N^{-1}AN$ and $B'=N^{-1}B$?

Thanks!

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  • $\begingroup$ For future reference - the answer is yes. $\endgroup$
    – the L
    Jan 21, 2011 at 19:58
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    $\begingroup$ "the answer is yes" - could you please explain more? Are you saying you already knew the answer before you asked the question - so why did you ask it? Or, are you saying that you've now discovered the answer, after you asked the question? So you should probably answer your own question, or close it. $\endgroup$
    – Zen Harper
    Mar 3, 2011 at 6:28
  • $\begingroup$ I have discovered the answer after I asked the question. $\endgroup$
    – the L
    Mar 3, 2011 at 9:58
  • $\begingroup$ Can you please give some references regarding meromorphic matrix function theory, especially on "well-known" perpresentation $T(z) = D + C(zI-A)^{-1}B$? $\endgroup$
    – user26479
    Sep 13, 2012 at 13:19
  • $\begingroup$ See the book books.google.co.il/books/about/… $\endgroup$
    – the L
    Sep 13, 2012 at 13:30

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