1
$\begingroup$

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.

  • 0
    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.2012-07-29
  • 0
    Oh no! Is an abbreviation for newbie :D2012-07-29

2 Answers 2

1

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 : $$f : (x,y,z) \mapsto (y,z,0)$$

  • 0
    Thanks :) I've forgotten to say that the endomorphism must be not null :(2012-07-29
  • 0
    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 $f$?2012-07-29
  • 0
    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 $\text{Im}(f) = \mathbb{R}e_3 = \text{Ker}(f)$.2012-07-29
  • 0
    @Lat_New: In general, you will need to put dollar signs around all LaTeX expressions.2012-07-29
  • 0
    @Jbeuh excuse me but $Dim(Im(f))=1$ and $Dim(Ker(f))=2$, isn'it?2012-07-29
  • 0
    Yes sorry, you have $\text{Im}(f)=\mathbb{R}e_3 \subset \text{Ker}(f) = \text{Vect}((2,-1,0),e_3)$.2012-07-29
  • 0
    excuse me again... i have $Ker(f)=<(-2, 1,0), (0, 0, 1)>$.. where is my mistake? Thanks! :)2012-07-29
  • 0
    No mistake, this gives the same Ker as my answer ($e_3=(0,0,1)$ and $(-2,1,0)=-(2,-1,0)$, this generates the same space).2012-07-29
  • 0
    Perfect! Thank you so much! :)2012-07-29
2

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 $f\neq 0$.