0

I've been working through a proof for why the determinant can be calculated from any row in a square matrix. In this proof we first develop the determinant from row 1, and then (in the sub-determinants) from the arbitrary row k.

We then develop the determinant from the arbitrary row k, and then from row 1 in the sub-determinants, the idea is to prove that these expressions are equal.

I've arrived at the following expressions (please forgive me för just using copy/paste).

enter image description here

I don't quite agree with the annotation, but i is supposed to represent the number of the element in the first row, while s is supposed to represent the number of the element in the kth row. The expressions $|(A_{1i} )_{ks} |$ and $|(A_{ks} )_{1i}|$ refer to the sub-determinants of their respective expression.

In the next step, we are supposed to realize that the expressoin on the upper left and the lower right are the same. If we start with the expression on the upper left and chech which pairs of i and s are possible, we arrive at the following types of pairs:

          (2,1)
   (3.2), (3,1)

(4,3), (4,2), (4,1) (......and so forth)

If we look at the colums we will find that the only i that are possible for a certain s are those with a minimum value of s+1. The next step is a bit...hazy, but I think I'm supposed to realize that this means I can somehow switch places of the summation signs and change the indecies, so that:

$$\sum^n_{i=1}\sum_{s=1}^{i-1}=\sum_{s=1}^n\sum_{i=s+1}^n$$

...And I'm way to tired but I think that would make those expressions equal, after which the same procedure can be performed on the expressions to the upper right and lower left.

But is it really possible to swap the summation signs in this manner and in that case why? I would appreciate some kind of intuitive explanation.

Magnus
  • 703
  • 1
    Both sums are over the pairs $(s,i)$ with $1\le s<i\le n$. – Angina Seng Sep 07 '20 at 17:44
  • Are you familiar with reversing the order of integration? What you have there is the analogue for summations of $$\int_0^a\int_0^xf(x,y),dydx=\int_0^a\int_y^af(x,y),dxdy,.$$ – Brian M. Scott Sep 07 '20 at 17:56
  • @BrianM.Scott No, sorry – Magnus Sep 07 '20 at 18:03
  • @Magnus: Okay; see if the diagram in this answer and my first comment under it help at all. – Brian M. Scott Sep 07 '20 at 18:12
  • The alleged duplicate question, Proof of a double summation reduction identity, has only a vague similarity to the present question. @AnginaSeng's comment is a far better answer to this question than are any of the answers to that other question. – Calum Gilhooley Sep 08 '20 at 14:46
  • @AnginaSeng Sure, we are taking the sum of every i times every s smaller than i, as long as an s exists which is smaller. In the other term we are taking every s times every i larger than i, as long as an i exists which is larger.. Intuitively every instance of a i times a smaller s could be described as an s time a larger i. Since we are summing up all possible combinations of i with smaller s/s with larger i we could claim the expressions are equivalent. But is there a way of actually proving this? – Magnus Sep 09 '20 at 16:38

0 Answers0