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

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 Jose Carlos Santos Posts: 4,896 Registered: 12/4/04
Re: Linear independence of eigenvectors
Posted: Dec 6, 2012 2:50 AM

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

Date Subject Author
12/5/12 Jose Carlos Santos
12/5/12 David C. Ullrich
12/6/12 Jose Carlos Santos
12/6/12 David C. Ullrich
12/6/12 Robin Chapman
12/6/12 David C. Ullrich
12/5/12 Shmuel (Seymour J.) Metz
12/6/12 Jose Carlos Santos