Suppose . Then, since is a subspace of the domain of the transformation T, we can extend to a basis of . Hence, is a linear combination of elements of a basis of , so .
I have to prove that the eigenspace, E lambda is an invariant subspace.
I took some eigenvector, v in the eigenspace, then T(v) =lambda v
I know that T(v) should belong to the eigenspace , but i don't really understand why ?
Elambda is the set of all eigenvectors of T, then why would (lambda) (v) be in the eigenspace ?
I am confused.
You may need to review basic definitions! A "subspace", by definition, is "closed" under vector addition and scalar multiplication. Being "closed under scalar multiplication" means that if v is a vector in the subspace and a is a scalar (number) then av is also in the subspace. The eigenvalue, is, of course, a number, so if v is in the "eigenspace", so is .