```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 commutativity.Best regards,Jose Carlos Santos
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