We know that eigenspaces of distinct eigenvalues are disjoint. That is, no vector is an eigenvector of two different eigenvalues. This extends to generalized eigenvectors as well. If we have a nonzero 
then we know for sure that 
for 
.
To see this, let’s say that we have a 
that’s in two generalized eigenspaces, and start hitting it with 
over and over again. Eventually, it disappears, but before it does, it’s a nonzero eigenvector with eigenvalue 
. I say that this (not generalized) eigenvector is also in the generalized eigenspace of 
. Indeed, each time we hit with we write it as the sum 
. Since 
, the same is true of the scalar multiple 
. And since 
is a polynomial in 
, its kernel is invariant. Thus we have 
as well. And thus 
is invariant under the transformation , and we can assume (without loss of generality) that is an actual eigenvector with eigenvalue , as well as a generalized eigenvector with eigenvalue .
But now let’s start hitting with 
. Applying it once we find 
. So no matter how many times we apply the transformation, we just keep multiplying by 
, and we’ve assumed that this is nonzero. So cannot possibly be in the generalized eigenspace of . And so generalized eigenspaces corresponding to distinct eigenvalues can only intersect trivially.
February 25, 2009
Posted by John Armstrong |
Algebra, Linear Algebra |
3 Comments
