T-invariant basis problem

**nqramjets** · February 16th 2010, 09:02 AM

Hello all, I am having trouble showing linear independence at the indicated step.

The setup:
Let $v\in V$ be a fixed vector in a finite-dimensional vector space $V$ . Let $W=\{v,Tv,T^2v,...\}$ .

Prove $W$ is $T$ -invariant. (Easy)

Suppose that $\dim(W)=k$ and show that $B=\{v,Tv,T^2v,\ldots,T^{k-1}v\}$ is a basis for $W$ .

It seems obvious that checking linearly independent is easier than span, but I am getting stuck trying to get all of the coefficients to be zero. Thanks to any and all help ahead of time!

**tonio** · February 16th 2010, 09:41 AM

Originally Posted by nqramjets

Hello all, I am having trouble showing linear independence at the indicated step.

The setup:
Let $v\in V$ be a fixed vector in a finite-dimensional vector space $V$ . Let $W=\{v,Tv,T^2v,...\}$ .

Prove $W$ is $T$ -invariant. (Easy)

Suppose that $\dim(W)=k$ and show that $B=\{v,Tv,T^2v,\ldots,T^{k-1}v\}$ is a basis for $W$ .

It seems obvious that checking linearly independent is easier than span, but I am getting stuck trying to get all of the coefficients to be zero. Thanks to any and all help ahead of time!

Use the lemma that says that if $\{v_1,v_2,\ldots,v_n\}$ is a lin. dep. set, then there's some vector lin. dep. on the PRECEEDING ones (i.e., the 2nd one on the 1st one, or the 3rd on the 1st and 3nd one., etc.)

So: if $\{v,Tv,\ldots\}$ is lin. dep. then there's some natural number r s.r. $T^rv$ is lin. dep. on $\{v,Tv,\ldots,T^{r-1}v\}\Longrightarrow T^rv=a_0v+a_1Tv+\ldots+a_{r-1}T^{r-1}v\,,\,\,a_i\in\mathbb{F}$

As $\{v,Tv,\ldots\}$ is a generating set for $W\,\,\,and\,\,\,\dim W=k$ , it's clear from the above that $\{v,Tv,\ldots,T^nv\}$ cannot be lin. dep. for any $n<k-1$ (why?), so it must be that $\{v,Tv,\ldots,T^{k-1}v\}$ is lin. independent...and now end the argument.

Tonio

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