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I have skimmed this video that I found on mathoverflow: http://tube.sfu-kras.ru/video/407?playlist=397

At about 15:05 the lecturer wrote down an equality $\sum F(m_1, \ldots, m_m)z^{m_1}\ldots z^{m_m} = \frac{1}{det(zI - A)}$, where F is given in terms of coefficients of the $m$-by-$m$ matrix $A$. The definition of $F$ doesn't yet make much sense to me, but that's beyond the point.

My question is, how is this equality possible? the right hand side obviously has poles—the eigenvalues of $A$, but the left hand side is just a polinomial, it can't have poles! Is there something I'm missing?

EDIT: well, it was actually $\sum F(m_1, \ldots, m_m) z_1^{m_1}\ldots z_m^{m_m} = \frac{1}{det(z - A)}$. Here I have abused the notation slightly: the z means actually the diagonalization of the corresponding vector. However, when you take $z_1 = \ldots = z_m$, the question still stands.

The domain is real, as I understand from the context of the lecture :), so the use of the term pole isn't really correct, I just forgot the right word in English :)

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    And the formula is, as mentioned in the video, MacMahon's Master Theorem: http://en.wikipedia.org/wiki/MacMahon_Master_theorem. The summation runs over all nonnegative integer vectors, but it appears that Zeilberger made a little mistake; it should be det(I-ZA), not det(Z-A). (Otherwise it doesn't even work for Z=0.)2010-11-29

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From the comment by @HansLundmark above.


The formula is MacMahon's Master Theorem, as mentioned in the video. The summation runs over all non-negative integer vectors, but it appears that Zeilberger made a little mistake; it should be $\det(I-ZA)$, not $\det(Z-A)$. Otherwise, it doesn't even work for $Z=0$.