$V$ is a linear space of dimension $n$ over $F$.
$T: V \to V$ is a linear transformation, s.t. $T^n$ = 0 and $T^{n-1}\ne 0$
Prove that there is $v$ from $V$ s.t. $\{v,Tv,...,T^{n-1}v\}$ is a base of $V$.
I tried to conclude things about the kernel, and to work with a base of $V$, but got stuck.
Any help appreciated.
As $T^{n-1}\ne 0$ there exists $v\in V$ such that $T^{n-1}v\ne 0$. Then, $$\lambda_1v+\lambda_2Tv+\cdots+\lambda_nT^{n-1}v=0$$ $$\Rightarrow T\left(\lambda_1v+\lambda_2Tv+\cdots+\lambda_nT^{n-1}v\right)=T(0)$$ $$\Rightarrow \lambda_1Tv+\lambda_2T^2v+\cdots+\lambda_n\underbrace{T^{n}v}_{=0}=0$$ Taking repeatedly $T$ in both sides you'll finally get $$\lambda_1=0\Rightarrow \lambda_2=0\Rightarrow \cdots \Rightarrow \lambda_n=0.$$
So, $B=\{v,Tv,...,T^{n-1}v\}$ is linearly independent and $|B|=\dim V=n,$ so $B$ is basis of $V.$
Suppose the set is linearly dependent. Then
$$a_0v+a_1Tv+a_2T^2v+...+a_{n-1}T^{n-1}v=0$$
for non-trivial set of $a_i$. First suppose $a_0 \neq 0$. Then you can rewrite the terms as
$$v=a'_1Tv+a'_2T^2v+...+a'_{n-1}T^{n-1}v$$
Now, apply $T^{n-1}$ to both sides of the equation. The right side is zero, while the left side is non-zero.
Similarly, if $a_0=0$ but $a_1 \neq 0$ then you can write
$$Tv=a'_2Tv^2+a'_3T^3v+...+a'_{n-1}T^{n-1}v$$
Apply $T^{n-2}$ to the equation, and you get same result as above. Continuing this process shows that all of the $a_i$ have to be zero, reaching a contradiction. Finally, since the number of elements in the set equals the dimension, it is a basis for $V$.