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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 6, 2012 1:17 PM
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On Thu, 06 Dec 2012 07:50:46 +0000, José Carlos Santos
<jcsantos@fc.up.pt> wrote:

>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.


See RC's comment.

>
>Best regards,
>
>Jose Carlos Santos





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