Find a linear transformation $f:\mathbb{R}^3\rightarrow\mathbb{R}^2$ such that $(2,1,0)\in\ker(f), f(1,0,0)=(1,1)$ and $f(3,1,0)=(1,1)$. 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?
 A: 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 $\mathbb{R}^2$.
A: 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.
A: 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.
