Number of positive and negative terms in nth order Determinant

Question

I'm working through Shilovs book on linear algebra and is stuck on the following problem:

"Show that of the $n!$ terms of a determinant of order $n$, exactly half ($\frac{n!}{2}$) have a plus sign according to the definition of Sec. 1.3, while the other half have a minus sign"

The sign is defined by $(-1)^{N(\alpha_{1},....,\alpha_{n})}$ with $N(\alpha_{1},....,\alpha_{n})$ is the number of inversions in the permutation $\alpha_{1},....,\alpha_{n}$. An inversion in this case is an arrangement of two indicies such that the larger index comes before the smaller. For instance in the sequence 2,1,4,3 there are 2 inversions as 2 is before 1 and 4 is before 3.

To prove this I'll have to show that the number of even inversions in the permutations of the sequence $(\alpha_{1},....,\alpha_{n})$ is $\frac{n!}{2}$. I'm not entirely sure how to go about proving this. You can show it quite easily for n=3 by enumerating all the cases, but I can't find something to anchor a general proof in.

Any suggestions would be greatly appreciated!

Answer 1

For every even permutation consider the permutation that changes the last two elements. It must be odd.

2,1,4,3 is even and 2,1,3,4 is odd.

I hope it is clear how this implies that half of them are even.

Answer 2

I suggest that it would be proven this way:

All the n! terms of a determinant of order n are just permutations of its row indices 1,2,..., n while its column indices are ordered from 1 to n.

Any permutation can be formed from another one by swapping two elements. If two elements of a term are swapped than the new term receives an opposite sign. All the n! terms can be produced from a base one by subsequent swapping two elements.

Therefore, all the produced terms switch their signs (n!-1) times overall. Therefore, the number of terms of a determinant of order n with the positive sign is the same as the number of terms with the negative sign and equal to n!/2.

However, the hint in Shilov suggests considering a determinant where all elements are equal to 1.

It would be interesting to see the actual proof according to the hint.

Answer 3

Another proof according to the Shilov hint:

Let's consider a determinant of order n whose each element is 1.

Each term of the determinant is equal either +1 or - 1. The determinant is obviously equal to zero.

Therefore, the number of terms with opposite signs must be equal in order their sum to be zero.

Thus, the number of both positive and negative terms is equal to n!/2.