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Thread: Invariant subspaces

  1. #1
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    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.
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  2. #2
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    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 $\displaystyle \lambda\vec{v}\in E_\lambda$.
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  3. #3
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    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 $\displaystyle \lambda v$.
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  4. #4
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    Re: Invariant subspaces

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