The lemma

In the Leibniz determinant formula for a matrix A ∈ M[sub]n[/sub] (R) (an nxn matrix of reals), S[sub]n[/sub] yields an equal number of even and odd permutations (there are an equal number of -1s and 1s yielded by sgn(sigma))

was briefly demonstrated in my class, and I need it as part of a larger proof, but I forgot/can't figure out how to prove it

I'm going to go try now, but if anyone knows the general outline of the proof offhand and can point me in the right direction (don't do it for me please), I'd appreciate it.

A = Pi(r)^2

Am I right?

Aaaargh it's been too long since linear algebra.

Winner

Well, think about how the alternating group of degree n is made (ie, all of the even permutations in the symmetry group of degree n). The proof in my abstract algebra textbook is kind of lame, but that is the direction it goes in to prove that half of the members of S[sub]n[/sub] are even.

I had not seen this determinate formula in my linear algebra class. Kind of wish I was taught it. It would make finding determinates of large matrices easier.

http://en.wikipedia.org/wiki/Even_and_odd_permutations#Proofs_that_every_permutation_is_either_even_or_odd
That might steer you in the right direction though.