
Invariant subspaces
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 ?
E_{lambda} is the set of all eigenvectors of T, then why would (lambda) (v) be in the eigenspace ?
I am confused.

Re: Invariant subspaces
Suppose $\displaystyle \vec{v} \in E_\lambda$. Then, since $\displaystyle E_\lambda$ is a subspace of the domain of the transformation T, we can extend $\displaystyle \{\vec{v}\}$ to a basis of $\displaystyle E_\lambda$. Hence, $\displaystyle \lambda\vec{v}$ is a linear combination of elements of a basis of $\displaystyle E_\lambda$, so $\displaystyle \lambda\vec{v}\in E_\lambda$.

Re: Invariant subspaces
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, $\displaystyle \lambda$ is, of course, a number, so if v is in the "eigenspace", so is $\displaystyle \lambda v$.

Re: Invariant subspaces
That's right, i don't know how i missed out on that one !