Date: Dec 6, 2012 2:50 AM
Author: Jose Carlos Santos
Subject: Re: Linear independence of eigenvectors
On 05-12-2012 17:43, David C. Ullrich wrote:
>> A classical Linear Algebra exercise says: prove that _n_ eigenvectors
>> of an endomorphism of some linear space with _n_ distinct eigenvalues
>> are linear independent. Does this hold for modules over division rings?
> I believe so - I don't see where the standard proof uses
> commutattivity, or however one spells it.
> Say x_1, ... x_n are eigenvectors wrt disctinct eigenvalues
> l_j. Say
> sum c_j x_j = 0.
> Applying that endomorphism k times shows that
> sum c_j l_j^k x_j = 0
> for k = 0, 1, ... . Hence
> sum c_j P(l_j) x_j = 0
> for any polynomial P.
> Now let P be the obvious product, so that
> P(l_1) <> 0 while P(l_j) = 0 for j >= 2. Then
> c_1 P(l_1) x_1 = 0;
> since it's a division ring and x_1 <> 0 this shows that c_1 = 0.
Cute! I did not know this proof. The one I am familiar with uses
Jose Carlos Santos