1
$\begingroup$

The conditions are: \begin{equation} f(2,1,0)=(0,0)\\f(1,0,0)=(1,1)\\f(3,1,0)=(1,1) \end{equation} These $3$ don't completely define a linear transformation since $\{ (2,1,0),(1,0,0),(3,1,0)\}$ is not a basis of $\mathbb{R}^3$ (the first vector is the third minus the second). Then I can chose for example $(2,1,0),(1,0,0)$ and add a third linearly independent vector to form a basis. I thought simply of $(0,0,1)$. But I don't know what $f(0,0,1)$ is.

Someone can help me solve this exercise please?

$\endgroup$
  • 1
    $\begingroup$ It isn't necessary to know $f(0,0,1)$. You can make something up. $\endgroup$ – Jakob Elias Dec 11 '16 at 22:24
1
$\begingroup$

Note that if you have \begin{equation} f(1,0,0)=(1,1)\\f(3,1,0)=(1,1) \end{equation}

then

\begin{equation} f(2,1,0)=f(3,1,0)-f(1,0,0)=(1,1)-(1,1)=(0,0). \end{equation}

As you have said $(1,0,0),(3,1,0),(0,0,1)$ form a basis. So you only need to define $f(0,0,1).$ You can choose your favourite vector of $\mathbb{R}^2$.

That is, $f$ is given by

\begin{equation} f(1,0,0)=(1,1)\\f(3,1,0)=(1,1)\\f(0,0,1)=(a,b) \end{equation} where $(a,b)$ is a given vector in $\mathbb{R}^2$.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ So, practically, I leave the transformed vector of $(0,0,1)$ as an unknown and I say that it is some vector of $\mathbb{R}^2$? $\endgroup$ – M.Giacchello Dec 11 '16 at 22:26
  • $\begingroup$ Since you are asked to find a linear map you can consider $(a,b)=(1,0)$ for example. Instead of $(1,0)$ you can consider $(0,1),(0,0),(1,1),....$ $\endgroup$ – mfl Dec 11 '16 at 22:30
  • $\begingroup$ Sorry, if I bother you but it is this point that is not clear to me. I can state that by applying the map $f$ on the vector of the basis $(0,0,1)$ I can get a vector which I decide by my own volition? Why is that? $\endgroup$ – M.Giacchello Dec 11 '16 at 22:34
  • $\begingroup$ Because you have a linear map defined on $\mathbb{R}^3$ and you have two linear independet constraints (the third one can be written in terms of the other two). So you have three degrees of freedom and two constraints. That is, you have one degree of freedom. This is the reason $f(0,0,1)$ is not given by the conditions. So you can choose it as you want. $\endgroup$ – mfl Dec 11 '16 at 22:37
  • $\begingroup$ perfect, thank you very much! $\endgroup$ – M.Giacchello Dec 11 '16 at 22:38
1
$\begingroup$

Let's do a linear transformation. The matrix that represents the linear transformation will be of the structure:

$$ \left( \begin{matrix} a & b & c\\ d & e & f\\ \end{matrix} \right) $$

From your first part, you want $2a+b=0$ and $2d+e=0$. So the matrix becomes $$ \left( \begin{matrix} a & -2a & c\\ d & -2d & f\\ \end{matrix} \right) $$

From your second part, you want $a=1$ and $d=1$. So the matrix becomes $$ \left( \begin{matrix} 1 & -2 & c\\ 1 & -2 & f\\ \end{matrix} \right) $$ From this matrix, your third part happens, so any value you put for $c,f$ is valid for this transformation. With this transformation is easy to see that

$$f(0,0,1)=(c,f)$$

Where are any value you put for the matrix before.

| cite | improve this answer | |
$\endgroup$
0
$\begingroup$

In order to well define $f$ you need the image of a basis. However you don't have a basis of $\Bbb R^3$ with the three given vectors as you have said. The image by $f$ of the vector $(0,0,1)$ you have chosen can be arbitrary to finish the required answer.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.