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

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David C. Ullrich

Posts: 21,553
Registered: 12/6/04
Re: Linear independence of eigenvectors
Posted: Dec 5, 2012 12:43 PM
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On Wed, 05 Dec 2012 16:11:21 +0000, José Carlos Santos
<jcsantos@fc.up.pt> wrote:

>Hi all,
>
>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.

???

>
>Best regards,
>
>Jose Carlos Santos





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