A basis for a nilpotent endomorphism [duplicate]

Let $E$ be a complex vector space of dimension 3. Let $f$ be a non zero endomorphism such that $f^2=0$. I want to show that there is a basis $B=\{b_1,b_2,b_3\}$ of $E$ such that $$f(b_1)=0, f(b_2)=b_1,f(b_3)=0$$

Edit Here is how i see the answer now:

$f$ being non zero there exists $x_0\in E$ such that $f(x_0)\not =0$.

Let $M=span\{f(x_0),x_0\}$. Since $f^2=0$ we show easily that $f(x_0)$ and $x_0$ are linearly independent hence they form a basis for $M$.

We take $b_1=f(x_0)$, $b_2=x_0$.

Take any $z\not \in M$.

If $z\in \ker f$ then take $b_3=z$.

If $z\not \in \ker f$ then there exists $\beta \not = 0$ such that $f(z)=\beta f(x_0)$ (because $\dim(Im(f))=1$ hence it is spanned by any non zero vector, we take $f(x_0)$ as a spanning vector). Take $z'=\dfrac{1}{\beta}z-f(x_0)$ hence $z'\in \ker f$ and we take $b_3=z'$.

• Dear DonAntonio, yes this is basically the same problem but put in another context, I couldn't change it in that post since comments are made there that will be irrelevant if I change the question in this way. I really want to solve this problem but without mentionning the notion of Jordan normal form. Commented Apr 8, 2014 at 19:49
• I see, @palio...but my answer included a part without JCF... Commented Apr 8, 2014 at 19:55

You're not doing it in the right order. Choose $b_2$ so $f(b_2) \neq 0$. Then $b_1 = f(b_2)$, and choose any other $b_3 \in Ker(f)$ that independent from $b_1$.
• It looks fine !!! $b_2$ exists since $f$ is not the zero endomorphism. $b_1=f(b_2)\in \ker f$. The only problem is $b_3$ why does it exist and is there a canonical way how to choose one that is independent from $b_1$? Commented Apr 8, 2014 at 20:14
• It exists because the dimension of the kernel is 2. As for choosing, you know, we're in mathematics here, so saying it exists is pretty much the same as choosing it ;). In practice, you can describe the kernel with an equation, view it as a dot product to a constant vector, extract the vector (it is orthogonal to the kernel), then cross-product it with $b_1$, and that would give you a $b_3$ candidate. Commented Apr 8, 2014 at 20:27