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Given a matrix valued rational function $f(\lambda)$ with poles of order $n_k$ at points $\lambda_k$, not at infinity, a book I am following says we can decompose as $$f(\lambda)=f_0+\sum_k f_k(\lambda), \quad f_k(\lambda)=\sum_{r=-n_k}^{-1}f_{k,r}(\lambda-\lambda_k)^r$$ where $f_0$ is a constant.

I seem to think that for a general rational function there would also have to be a piece that is polynomial in $\lambda$ rather than simply a constant term. Is there a way to prove this in general or does it seem like this must just be a definition rather than a general property?

Sources: Ch 3.2 of "Introduction to Classical Integrable Systems" by Babelon, Bernard, and Talon.

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  • $\begingroup$ So $\lambda \in \mathbb{C}$ ? And $f$ is just a collection of $n \times n$ rational functions $\mathbb{C} \to \mathbb{C}$. In general $f(z)$ is a rational function means $f(z)$ is a quotient of polynomials, factoring the denominator gives $\displaystyle f(z) = \frac{\sum_{m=0}^M a_m z^m}{\prod_{k=1}^K (z-b_k)^{e_k}}$. From this, partial fraction decomposition is the algorithm to "substract the poles iteratively" obtaining $\displaystyle f(z)= \sum_{m=0}^r c_m z^m + \sum_{k=1}^K \sum_{j=1}^{e_j} \frac{d_{k,j}}{(z-b_k)^j}$ for some coefficients $c_m, d_{k,j}$ $\endgroup$
    – reuns
    Nov 17, 2017 at 1:17
  • $\begingroup$ I think you meant to write $\lambda \in \mathbb{C}$, but yes that is correct. What you wrote in your second comment is how I understand partial fraction decomposition, but if we're given a general function there is no reason for the first sum to be a constant, right? $\endgroup$
    – Kenny H
    Nov 17, 2017 at 1:20
  • $\begingroup$ In what I wrote $r = 0$ iff $\lim_{z \to \infty} f(z) = c_0$ iff $M \le \sum_{k=1}^K e_k$ $\endgroup$
    – reuns
    Nov 17, 2017 at 1:21
  • $\begingroup$ Oh duh. The book authors are definitely assuming the function is finite at infinite. Thanks for helping me realize that! $\endgroup$
    – Kenny H
    Nov 17, 2017 at 1:23

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