Date: Dec 6, 2012 2:50 AM
Author: Jose Carlos Santos
Subject: Re: Linear independence of eigenvectors
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