# Invariant subspaces

• Oct 6th 2013, 05:58 PM
mrmaaza123
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 ?

Elambda is the set of all eigenvectors of T, then why would (lambda) (v) be in the eigenspace ?

I am confused.
• Oct 6th 2013, 06:34 PM
SlipEternal
Re: Invariant subspaces
Suppose $\vec{v} \in E_\lambda$. Then, since $E_\lambda$ is a subspace of the domain of the transformation T, we can extend $\{\vec{v}\}$ to a basis of $E_\lambda$. Hence, $\lambda\vec{v}$ is a linear combination of elements of a basis of $E_\lambda$, so $\lambda\vec{v}\in E_\lambda$.
• Oct 6th 2013, 07:01 PM
HallsofIvy
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, $\lambda$ is, of course, a number, so if v is in the "eigenspace", so is $\lambda v$.
• Oct 6th 2013, 07:43 PM
mrmaaza123
Re: Invariant subspaces
That's right, i don't know how i missed out on that one !