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I have a problem:

Let's find an endomorphism of $\mathbb{R}^3$ such that $\operatorname{Im}(f) \subset \operatorname{Ker}(f)$.

How would you do it?

The endomorphism must be not null.

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By the way, the "New" in your name is perhaps an indication of the existence of a previous account on this site? If so please comment with a link to the previous account (if you have such link) and the moderators could merge these accounts. –  Asaf Karagila Jul 29 '12 at 15:10
    
Oh no! Is an abbreviation for newbie :D –  Lat_New Jul 29 '12 at 15:21
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2 Answers

up vote 1 down vote accepted

Well, you could always take $f$ to be the null function... Not the only solution, but certainly the simplest :)

If you want $f$ to be non-null, then you just need to make sure that $f^2=0$. Either try to find a $3 \times 3$ matrix for which this holds (look up nilpotent matrices) or look at this endomorphism and see how it can be adapted : $$f : (x,y,z) \mapsto (y,z,0)$$

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Thanks :) I've forgotten to say that the endomorphism must be not null :( –  Lat_New Jul 29 '12 at 15:05
    
Thanks again :) could you confirm me that $$\left( \begin{array}{ccc} 0 & 0 & 0 \\ 0 & 0 & 0\\ 1 & 2 & 0 \end{array} \right)$$ is a valid matrix for my $f$? –  Lat_New Jul 29 '12 at 15:27
    
Yes, it works : if $(e_1,e_2,e_3)$ is the canonical basis of $\mathbb{R}^3$ then for the $f$ you define you have $\text{Im}(f) = \mathbb{R}e_3 = \text{Ker}(f)$. –  Jbeuh Jul 29 '12 at 15:31
    
@Lat_New: In general, you will need to put dollar signs around all LaTeX expressions. –  Zev Chonoles Jul 29 '12 at 15:36
    
@Jbeuh excuse me but $Dim(Im(f))=1$ and $Dim(Ker(f))=2$, isn'it? –  Lat_New Jul 29 '12 at 15:37
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Observe that such operator has the property that $f(f(\vec x))=0$, since $f(\vec x)\in\ker f$. Any matrix $M$ in $M_3(\mathbb R)$ such that $M^2=0$ will give you such operator.

For example, then, $f(x_1,x_2,x_3)=(x_3,0,0)$, for the canonical basis the matrix looks like this:$$\begin{pmatrix} 0&0&1\\0&0&0\\0&0&0\end{pmatrix}$$

See now that $f^2=0$, but $f\neq 0$.

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Thanks, I haven't thought about this! ;) –  Lat_New Jul 29 '12 at 15:22
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