What happens to the number of inversions in a permutation group if we reflect the matrix across its antidiagonal?
(I suppose that Leibniz formula for determinants should be used or maybe showing by induction?)
Thank you in advance.
For instance for n=2
So the number of inversions in the two groups:
in it is 0
in it is 1
(That's why the value of its matrix is )
If we reflect it across its antidagonal:
And the number of inversions are the same. For n=3 it is a little more complicated. So I don't really know what happens with the number of inversions if n>2. I would be glad if you could help me.
I know what you meant, sort of. This is pretty much what I thought. You're taking the idea of an inversion in a permutation to be two values such that but . But could you explain this a little more? Combinatorially inversions can be seen as connecting the values of your matrix with lines and an inversion is an incidence of intersecting lines.
So, could you give a precise definition and not an example? It's clear if you interpret combinatorially (as above) that the number of inversions is invariant under such a flip.
There are some definitions:
Permutation Inversion -- from Wolfram MathWorld
Leibniz formula for determinants - Wikipedia, the free encyclopedia
Thank you that you sacrifice time for me.
Sorry, I see. The determinant of a matrix can be definied by permutations. So the inversion refers to permutation, that refers to determinant and that to the matrix.
Or the last step can be left out, we can say that we reflect the determinant across its minor diagonal.
Here are some links:
A Determinant's relation to permutations
Determinants
And thanks once again.