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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 $b_3=z'$.

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  • $\begingroup$ 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. $\endgroup$
    – palio
    Commented Apr 8, 2014 at 19:49
  • $\begingroup$ I see, @palio...but my answer included a part without JCF... $\endgroup$
    – DonAntonio
    Commented Apr 8, 2014 at 19:55

2 Answers 2

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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 $b_1$.

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    $\begingroup$ Although I do encourage you to look at the Jordan Normal Form :) $\endgroup$ Commented Apr 8, 2014 at 19:59
  • $\begingroup$ 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 $b_1$? $\endgroup$
    – palio
    Commented Apr 8, 2014 at 20:14
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    $\begingroup$ 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. $\endgroup$ Commented Apr 8, 2014 at 20:27
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You can use Jordan-Normal form to solve this easily. In fact the proof would go along the lines of proving the general Jordan-Normal form theorem directly, so you may want to look at the proof for that if you don't want to use the theorem directly.

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  • $\begingroup$ Thank you !! Is there any good reference where I can find this proof to apply it to my problem? $\endgroup$
    – palio
    Commented Apr 8, 2014 at 19:52
  • $\begingroup$ Any sensible linear algebra textbook has a proof. $\endgroup$ Commented Apr 8, 2014 at 20:01

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