Let $V$ be an n-dimensional vector space over $\mathbb{R}$ and $T:V \rightarrow V$ be linear. Call a vector $v \in V$ cyclic if $V$ is spanned by $\{v, \ Tv, \ T^2v,..\}$.
Question:
Show that $v$ is cyclic for $T$ then there is some $k \geq 1$ such that
$$B=\{v, \ Tv,..T^{k-1}v \}$$ is linearly independent in $V$ and that $T^kv$ lies in the span of $B$.
Thoughts:
If we suppose that no such $k$ exists, then every subset of the form $\{v, \ Tv,..T^{l-1}v \}$ is linearly dependent, or it is linearly dependent with $T^lv$ not lying in it span.
There are in effect then two cases to consider.
Suppose every subset of the form $\{v, \ Tv,..T^{l-1}v \}$ is linearly dependent. Then $T^{l-1}$ can be expressed as a linear combination of $v,…,T^{l-2}v$. Now, $\{v, \ Tv,..T^{l-2}v \}$ is also linearly dependent by assumption and so we can express $T^{l-2}$ as a linear combination of $v,…,T^{l-3}v$ etc…
Continuing in this way, we show that $T^{l-1}v= \lambda v$ for some $\lambda \in \mathbb{R}$. Thus $$B=\{ \mu v \ | \ \mu \ \text{is an eigenvalue of} \ T^mv \ \text{for some} \ m \}$$ This is 'at best' countably infinite, and therefore could not span $V$ even if it were 1-dimensional, wince our base field is $\mathbb{R}$, which is uncountable.
Suppose then that $B$ is linearly independent for some $k$ but that $T^kv$ does not lie in its span. I can't see how to progress this argument.
Note: This question is for second year Undergraduates, and I think there must be some more simple approach; even the first part surprises me if we need to argue using the uncountability of $\mathbb{R}$, so any alternative approaches are also appreciated.
Best Answer
For the question mentioned the fact that $v$ is a cyclic vector is irrelevant, it just needs to be nonzero (and that only because you insisted that $k\geq1$). Since $V$ is finite dimensional, the infinite sequence $v,Tv,T^2v,\ldots$ cannot remain linearly independent forever; taking $k$ minimal so that $v,Tv,\ldots,T^kv$ is linearly dependent, the conclusion of you question will be satisfied. Moreover the subspace $W$ spanned by $B$ will be stable under $T$ (that is, $T(W)\subseteq W$) so that it makes sense to restrict $T$ to a map $T|_W: W\to W$, and (and this is where the notion of cyclic vector comes in) $v$ is a cyclic vector in the subspace $W$ for the linear operator$~T|_W$.