How to generate a vector space by repeated powers of a square matrix? Let $K$ be a field, like $\mathbb R$ or $\mathbb C$, and let $V$ be an $n$-dimensional $K$-vector space. Let us fix a nonzero vector $v\in V$ and a square matrix $A$ of order $n$.

Under what conditions on $A$ do we have that the set of vectors
  $\{A\cdot v\,|\,a\geq 0\}$ spans $V$?

I was trying to prove that if $A$ has maximal rank then 
$$\{A^i\cdot v\,|\,0\leq i\leq n-1\}\subset V$$ is a basis, but maybe this is false.
Any hint would be very much appreciated!
J.
 A: Yes, if $A$ has maximal rank, or equivalently, $\ker A=\{0\}$, or,  $\det A\ne 0$.
Try to prove e.g. that if a linear map $\phi:V\to W$ is injective (i.e. $\ker\phi=\{0\}$) then it maps linearly independent sets to linearly independent sets.

Wait, wait.. I might have misunderstood the question. 
You mean $\{A^i\cdot v\ \mid\ i\ge 0\}$ for a fixed $v$.
Well, for that, the condition that $A$ is invertible is not even necessary , and neither is sufficient. 
E.g. with a basis $v_1,..,v_n$, consider the linear map that maps $v_i\to v_{i+1}$ and $v_n\to 0$. This has nontrivial kernel, but - together with $v_1=A^0v_1$ - it satisfies the condition.
For the other direction, take $A$ the unit matrix, then $\{A^iv\mid i\ge0\}=\{v\}$ can be a basis only in dimension $1$.

Hint: For the given vector $v$, define $v_n:=A^nv$, and denote $r$ the first number such that $v_0,\dots,v_r$ is linearly dependent. 
Show that $v_0,\dots,v_{r-1}$ generates all $v_n$'s, hence $\dim\langle v_n\rangle_{n\ge0}=r$, so we must have $r=n$ by hypothesis. What is the matrix when transforming $A$ to basis $(v_0,\dots,v_{r-1})$?
