Let $V$ be a $K$-vector space, $f: V \rightarrow V$ a linear map. Under some condition, show that $v, f(v),...,f^n(v)$ are linearly independent. 
Let $V$ be a $K$-vector space,  $f: V \rightarrow V$ a linear map. Let $v \in V$. May a number $n ≥ 0$ exist, so that: $f^n(v) \not= 0$ and $ f^{n+1}(v) = 0$. Show that $v, f(v),...,f^n(v)$ are linearly independent.

What i know:


*

*$f^0 = id, f^1 = f, f^2 = f \circ f, ...$

*$v, f(v),...,f^n(v)$ are linearly independent if:
$a_1v + a_2f(v) + \cdots+a_nf^n(v)=0 \implies a_1 = a_2 =\cdots= a_n = 0$

*$f$ is linear, hence $f(0) = 0$, $af(n) = f(an)$, $f(v_1 + v_2) = f(v_1)+f(v_2)$


My thoughts on it: the definition of linearly independent, in our case, can be rewritten as
$a_1v + f(a_2v) + f(f(a_3v)) + \cdots+f^n(a_nv)=0 \implies a_1 = a_2 =\cdots= a_n = 0$
Taking the function $f$ on both sides we can also obtain
$a_1f(v) + f(f(a_2v)) + f(f(f(a_3v))) + \cdots+f^{n+1}(a_nv)=0 \implies a_1 = a_2 =\cdots= a_n = 0$ 
...
I feel that i am on the right track, but i can't take it further from here.
So i have to show this sum only becomes $0$ when all $a_n$ are $0$. I thought about a proof by contradiction, but i wouldn't know where i can lead this to a contradiction. 
Thanks for all help
 A: Yes you are on the right way! Suppose that the $a_n$'s are such that 
$$ a_0 v +  a_1 f(v) +\dots + a_n f^n(v)=0 $$
Taking the function times on both side (because of linearity) you obtain that 
$$ a_0 f(v) +  a_1 f^2(v) +\dots + a_n f^{n+1}(v)=0 $$
and since $f^{n+1}(v)=0$ you get
$$ a_0 f(v) +  a_1 f^2(v) +\dots + a_{n-1} f^{n}(v)=0. $$
Now you can do the same with the equation above and you obtain the relation 
$$ a_0 f^2(v) +  a_1 f^3(v) +\dots + a_{n-2} f^{n}(v)=0 $$
and so on (for $n$ times). At the end you obtain the relation 
$$a_0 f^n(v)=0$$
and since $f^n(v) \not= 0$ you can conclude that $a_0=0$. Hence, going back to the equation 
$$ a_0 v +  a_1 f(v) +\dots + a_n f^n(v)=0 $$
we can conclude that
$$ a_1 f(v) +\dots + a_n f^n(v)=0. $$
Now we can restart the reasoning above using in place of $v$ the vector $f(v)$, and we arrive to the conclusion that $a_1=0$. Going forward we can conclude that all the $a_n$'s are zero and we are done. 
A: Yes, you're on the right track.
If $n=0$ the statement is obvious, because it reduces to $v\ne0$.
So suppose we have proved the statement for the pairs $(w,g)$ with $g^{n-1}(w)=0$ and $g^n(w)=0$, for $n>0$.
Let $\alpha_1v+\alpha_2f(v)+\dots+\alpha_nf^{n}(v)=0$. By applying $f$ to this equality and setting $w=f(v)$, we get
$$
\alpha_1w+\dots+\alpha_{n-1}f^{n-1}(w)=0
$$
and, by the induction hypothesis, $\alpha_1=\dots=\alpha_{n-1}=0$. Therefore also $\alpha_n=0$.
