Why an Invariant Subspace Has an Eigenvector

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Discussion Overview

The discussion revolves around the justification for why an invariant subspace must contain an eigenvector, particularly in the context of linear transformations. The scope includes theoretical aspects of linear algebra and properties of invariant subspaces and eigenvectors.

Discussion Character

  • Technical explanation
  • Conceptual clarification

Main Points Raised

  • One participant questions the justification for the statement that an invariant subspace has an eigenvector.
  • Another participant explains that if a subspace M is invariant under a linear transformation T, and if the field is the complex numbers, then T must have at least one eigenvector in M.
  • A different perspective suggests that considering the converse, where eigenvectors span invariant subspaces, may provide additional insight. Specifically, if u is an eigenvector, then the subspace spanned by u is invariant under T.

Areas of Agreement / Disagreement

Participants present different viewpoints on the relationship between invariant subspaces and eigenvectors, with some suggesting that every invariant subspace must contain an eigenvector under certain conditions, while others explore the implications of eigenvectors spanning invariant subspaces. The discussion does not reach a consensus.

Contextual Notes

The discussion assumes the context of linear transformations over complex numbers, which may influence the existence of eigenvectors. There is also an implicit dependence on the definitions of invariant subspaces and eigenvectors that may not be fully articulated.

arthurhenry
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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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