Extending the basis of a T-invariant subspace

1. Sep 11, 2013

Bipolarity

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. Sep 11, 2013

Erland

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.