Calculate a determinant related to permutation matrix

Question

Let $ M$ be a permutation $n \times n $ matrix and $[\lambda_1,\lambda_2, \ldots,\lambda_n]$ be the cycle type of the corresponding permutation, i.e. $ \lambda_i$ is the number of cycles of the lenght $i$.

How to prove that $$ \det(I \pm M \cdot z)=\prod_i (1 \pm z^i)^{\lambda_i}? $$ Or give me a link to a proof. Thanks.

Edit.

$z$ is formal variable.

Ok, $z$ is a formal variable, like as avariable in characteristic polynomial of a matrix — Leox, Dec 16 '13 at 21:50

score 2 · Accepted Answer · answered Dec 16 '13 at 21:49

2

Your permutation is a product of disjoint cycles, so $M$ is block diagonalized into corresponding matrices (which have size $i$ and have ones on the superdiagonal and on the lower left hand corner). So, all you need to show is that the characteristic polynomial of a cycle matrix has the right form. But that follows immediately from the Cayley-Hamilton theorem .

answered Dec 16 '13 at 21:49

Igor Rivin

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But characteristic polynomial has the form $\det(M-z \cdot I)$ not $\det(I \pm M \cdot z)$. What is the relation for these determinants? – Leox Dec 16 '13 at 21:56
@Leox let $z=1/u,$ and see what happens. – Igor Rivin Dec 16 '13 at 21:57
Oh, yes..I see..But it works for the case $\det(I-M\cdot z)$. For $\det(I+M\cdot z)$ I cant just put $z=-1/u$. – Leox Dec 16 '13 at 22:01
@Leox why not?? – Igor Rivin Dec 16 '13 at 22:02
since $(1-z^2)$ sends to $1-(1/u)^2$ but I need $1+(1/u)^2$. How to get reguired "+"? – Leox Dec 16 '13 at 22:03
@leox just try it and see. – Igor Rivin Dec 16 '13 at 22:15
I understood now. Thank you! – Leox Dec 16 '13 at 22:17

Calculate a determinant related to permutation matrix

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