
Re: Linear independence of eigenvectors
Posted:
Dec 6, 2012 1:16 PM


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...).

