Finding a specific basis for an endomorphism Let $E$ be $\mathbb C$-vector space of dimension $3$. Let $f$ be non zero endomorphism of $E$ such that $f^2=0$. show that there exists a basis $B=\{b_1,b_2,b_3\} $ of $E$ where the matrix of $f$ is given by :
$$\begin{array}{ccc}
0 & 1 & 0 \\ 
0 & 0 & 0 \\ 
0 & 0 & 0
\end{array}$$
My try : since $f$ is non zero, there exists $x\in E$ such that $f(x)\not =0$. Take $b_1=f(x)$ and $b_2=x$. Now $b_1,b_2$ can easily shown to be linearly independent. By the exchange lemma we can find $b_3$ in another generating system of $E$ to make $b_1,b_2,b_3$ into a basis of $E$. Now clearly $f(b_1)=f^2(x)=0$ and $f(b_2)=f(x)=b_1$. Now how to find  a $b_3$ such that $f(b_3)=0$ and $b_1,b_2,b_3$ is a basis of $E$? Thank you for your help. 
Edit: For the linear independence of $b_1,b_2$, if $\alpha x+\beta f(x)=0$ then we apply $f$ to get $\alpha f(x)=0$ and since $f(x)\not =0$ then $\alpha=0$ it remains that $\beta f(x)=0$ and again this gives that $\beta=0$.
 A: You should know that $\;x^3\;$ is the characteristic polynomial of $\;f\;$ and, by the given info, $\;x^2\;$ is its minimal polynomial, and from here that its Jordan Canonical Form has to be exactly what is written in your question...
Added by the comment below: An idea:
It can't be $\;\dim\ker f=0\,,\,3\;$ (why?), so suppose $\;\dim\ker f=1\;$ . By the dimension theorem:
$$3=\dim\ker f+\dim\,\text{Im}\,\implies \dim\,\text{Im}\,f=2$$Take now $\;u\notin\ker f\;\implies \ker f=\langle f(u)\rangle\;$ (why?) , but we also trivially have that 
$\;f(u)\in\text{Im}\,f\;$, so we can choose $\;f(v)\in\text{Im}\,f\;$ s.t. $\;\{f(u),f(v)\}\;$ basis of $\;\text{Im}\,f\;$ and thus
$\;\{u,f(u),f(v)\}\;$ is a basis of $\;E\;$, and then we have that for all $\;x\in E\;\;\exists\,a,b,c\in\Bbb C\;\;s.t.$
$$x=au+bf(u)+cf(v)\implies f(x)=f(au)+bf^2u+cf^v=f(au)\implies$$
$$x-au\in\ker f=\langle f(u)\rangle\implies x=au+tf(u)\;,\;\;t\in\Bbb C$$
and this is a contradiction (to what ?).
Thus, it must be that $\;\dim\ker f=2\;$ and rom here you can take two linearly independent $\;u,v\in \ker f\implies f(u)=f(v)=0\;$ that'll be part of a basis, as you wanted.
