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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 1:16 PM
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On Thu, 06 Dec 2012 09:09:38 +0000, Robin Chapman
<R.J.Chapman@ex.ac.uk> wrote:

>On 05/12/2012 17:43, David C. Ullrich wrote:
>> 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.

>
>I think you are unconsciously assuming commutativity here.
>If your polynomials have coefficients over the division
>ring you can't push them through the c_j.


Ah, right, thanks. I suspected I might be using commutativity
somewhere...


> It works for
>polynomials with coefficients in the centre of your division
>ring, but such polynomials won't split up the eigenvectors.
>Over H one cannot find a real polynomial with f(i) = 0
>and f(j) nonzero. This gives a clue to finding a counterexample.
>Take your favourite real matrix with characteristic
>polynomial X^2 + 1. Over H its eigenvalues include i, j and
>k (and a lot more...).





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