Search All of the Math Forum:

Views expressed in these public forums are not endorsed by NCTM or The Math Forum.

Notice: We are no longer accepting new posts, but the forums will continue to be readable.

Topic: on determinants
Replies: 0

 Dustin Morado Posts: 1 From: Georgia Tech Registered: 5/8/09
on determinants
Posted: May 8, 2009 11:03 PM

I am trying to teach myself linear algebra using Mirsky's an introduction to linear algebra and unfortunately i am stuck at a very early theorem. I am stuck due to wording, and so i am asking for a clarification. The theorem states:

Let (v1,...,vn) vary over all arrangements of (1,...,n) and let (k1,...,kn) be a fixed arrangement of (1,...,n). Then (v(subscript)k1,...v(subscript)kn) varies over all arrangements of (1,...,n)

In the proof to this theorem he says take two different choices of (v1,...vn); (a1,...an) and (b1,...,bn) then (v1,...,vn) is the the same arrangement [??? where i'm getting lost] and therefore a(sub)k1=b(sub)k1 ..., a(sub)kn=b(sub)k1 and therefore a1=b1 ..., an=bn which is contradictory and therefore (v1,...vn) must vary over the n! arrangements of (1,...,n)

I know this is kind of dense and maybe a pain, but i would really appreciate any help. If there is any confusion in my notation (I tried to be as clear as possible) please let me know and I'll clarify

Thanks!

***Post Script 05/11/09***

Hey I figured this out after some more reading and the reasoning is like this... (v1,...,vn) varies so choose two variations (a1,...,an) and (b1,...,bn) and say they lead to a fixed value of (v(k1),...,v(kn)) the only way this occurs is if a1=b1,...,an=bn. Since this is not the case it follows that (v(k1),...,v(kn)) must vary over all arrangements of (1,...,n).

Thanks!
*****************************

Message was edited by: Dustin Morado