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

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Robin Chapman

Posts: 412
Registered: 5/29/08
Re: Linear independence of eigenvectors
Posted: Dec 6, 2012 4:09 AM
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On 05/12/2012 17:43, David C. Ullrich wrote:
> On Wed, 05 Dec 2012 16:11:21 +0000, José Carlos Santos
> <> 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. 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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