
Re: Linear independence of eigenvectors
Posted:
Dec 6, 2012 2:50 AM


On 05122012 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

