invariant subspace

[arthurhenry]

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

[HallsofIvy]

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.

[Monocles]

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.

[arthurhenry]

I thank you HallsofIvy,
Yes, it is clear now.