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Extending the basis of a T-invariant subspace

  1. Sep 11, 2013 #1
    Let ##T: V → V ## be a linear map on a finite-dimensional vector space ##V##.
    Let ##W## be a T-invariant subspace of ##V##.
    Let ##γ## be a basis for ##W##.

    Then we can extend ##γ## to ##γ \cup S##, a basis for ##V##, where ##γ \cap S = ∅ ##, so that ## W \bigoplus span(S) = V ##.

    My question:
    Is ##span(S)## a T-invariant subspace of ##V##?

    I've been trying to prove it is, but am not sure. I would like some assistance, so I know where I might steer my proof. Thanks!

    EDIT: Never mind just solved it. It's false!

    BiP
     
    Last edited: Sep 11, 2013
  2. jcsd
  3. Sep 11, 2013 #2

    Erland

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    Science Advisor

    This is not true. As a simple counterexample, let ##V=\Bbb R^2##, and let (##\bf e_1,e_2##) be the standard basis in ##\Bbb R^2##, and let ##W=span\{\bf e_1\}##, and let ##T## be defined by its action on the basis vectors: ##T(\bf e_1\rm)=\it T(\bf e_2\rm)=\bf e_1##. Then, ##W## is ##T##-invariant, and we can take ##S=\{\bf e_2\it\}##, and it satisfies your assumptions, but ##span \,S## is not ##T##-invariant.
     
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