Why an Invariant Subspace Has an Eigenvector

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An invariant subspace M of a vector space V for a linear transformation T ensures that if a vector u is in M, then Tu is also in M. This property allows for the restriction of T to M, where, under complex numbers, every linear transformation must have at least one eigenvector. Consequently, T will have an eigenvector that lies within the invariant subspace M. Additionally, eigenvectors themselves can span invariant subspaces, as they are collinear with their transformations. The discussion clarifies that invariant subspaces inherently possess eigenvectors due to these properties.
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I am following a proof in the text "Algebras of Linear Transformations" and having problem justifying this line: ... M is an invariant subspace so it has an eigenvector. Why should an invariant subspace have an eigenvector? Thank you

I have a feeling this is a very simple result, if so I am sorry
 
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A subspace, M, of vector space, V, is an "invariant subspace" for linear transformation T if and only if whenever u is in M, Tu is also in u. That means we can restrict T to M- think of T as a linear transformation on M alone. Now, if we are working over the complex numbers, every linear transformation has at least one eigenvector so T has at least one eigenvector in M.
 
Considering the converse scenario may help as well, i.e., that eigenvectors span invariant subspaces. Consider that if u is an eigenvector of T, Tu = cu for some constant c. Thus, u and cu are collinear. Therefore, the subspace spanned by u is an invariant subspace of T.
 
I thank you HallsofIvy,
Yes, it is clear now.
 

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