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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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