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Topic: on determinants
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Dustin Morado

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

Dustin Morado

***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!
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Message was edited by: Dustin Morado



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