Can someone help?
1. Is a generalized eigenspace an eigenspace?
2. How can we be sure that a generalized eigenspace has a full rank?
your first question is easy to answer though:
suppose that is a finite dimensional vector space, an operator on , an eigenvalue of and the generalized eigenspace corresponding to
suppose that is an eigenspace. then there exists a scalar such that . now if for some and
, then and, since we'll get thus which implies
so the only way that a generalized eigenspace becomes an eigenspace is just the trivial one.
for example, consider the linear transformation defined by then the generalized eigenspace corresponding to is the eigenspace corresponding to
Every matrix satisfies its own characteristic equation. That is, if the characteristic equation for matrix A is then (A- aI)^nv= 0 for every v. It might happen that there are n independent eigenvectors in which case the "eigenspace" has dimension n. If not, if the eigenspace has dimension m< n, for v NOT in that eigenspace, we must have (A- aI)^(n-m)v= 0 so that v is a "generalized" eigen value.