TITLE: Nilpotent matrix and basis for $F^n$
QUESTION [5 upvotes]: Let $A \in Mat_{n \times n}(F)$ a matrix which satisfies $A^n=0$ and $A^{n-1} \ne 0$ for some positive integer $n$. Let $v \in F^n$ be such that $A^{n-1}v \ne 0$.  Prove that $\{v, Av, A^2v,...,A^{n-1}v\}$ is a basis for $F^n$.  

So, from definition of $A$, I know that $A^iv \ne 0$ for all $0\le i \le n-1$.
I know I need to prove the set is linearly independent and span $F^n$ to prove it's a basis, but I just don't know how.

REPLY [4 votes]: Layout: Suppose the set is linearly dependent.Then there exist scalars $\alpha _0, \ldots, \alpha _{n-1}$ not all null, such that $\alpha _0v+\ldots +\alpha _{n-1}A^{n-1}v=0_{n\times 1}$. 
Since not all of them are null, you can take the smallest $j\in \{0,\ldots ,n-1\}$ such that $\alpha_j \neq 0$. (See spoiler below).
This transforms the equality above in something with less terms. Multiply it by an appropriate power of $A$ and find a contradiction.
Spoiler:

Since $\alpha _0= \ldots = \alpha _{j-1}=0$ the equality can be rewritten as $\alpha _jA^jv+\ldots +\alpha _{n-1}A^{n-1}v=0_{n\times 1}$. Now multiply this by $A^{n-j-1}$, (note that $n-j-1\ge 0$). Can you conclude?