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




Re: Linear independence of eigenvectors
Posted:
Dec 6, 2012 4:09 AM


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. 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...).



