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Topic:
Linear independence of eigenvectors
Replies:
7
Last Post:
Dec 6, 2012 1:17 PM




Re: Linear independence of eigenvectors
Posted:
Dec 6, 2012 1:17 PM


On Thu, 06 Dec 2012 07:50:46 +0000, José Carlos Santos <jcsantos@fc.up.pt> wrote:
>On 05122012 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.
See RC's comment.
> >Best regards, > >Jose Carlos Santos



