Basis of the image of the linear transformation $f:\mathbb{R^4}\rightarrow\mathbb{R^3}$? I'm asked to find a basis of the image of the linear transformation $f:\mathbb{R^4} \rightarrow \mathbb{R^3}$ defined as $f(v) = (v_1-v_3+v_4,2v_1+v_2+2v_3+v_4,3v_1-v_2+v_4)$.
I found the matrix of the linear transformation $Af =      \begin{pmatrix}
        1 & 0 & -1 & 1\\
        2 & 1 & 2 & 1\\
        3 & -1 & 0 & 1 \\
        \end{pmatrix} $ with rank equal to 3.
I tried to find a basis this way : $Im  f=\{w\in\mathbb{R^3} | \exists v\in\mathbb{R^4}s.t f(v)=w\}$ i.e 
$$
\left\{\begin{pmatrix}v_1-v_3+v_4\\2v_1+v_2+2v_3+v_4\\3v_1-v_2+v_4\end{pmatrix}\right\}=\left\{v_1\begin{pmatrix}1\\2\\3\end{pmatrix} + v_2\begin{pmatrix}-1\\2\\0\end{pmatrix} + v_3\begin{pmatrix}0\\1\\-1\end{pmatrix} + v_4\begin{pmatrix}1\\1\\1\end{pmatrix}\right\}
$$
The problem is that I have 4 vectors here, and I know the image should be of the dimension 3 since the rank of the matrix is 3.But I don't know how to compute the basis formed from 3 vectors ,I think one of them is a linear combination of another , but I don't know wich one .Any help please ? Thank you ! 

If it's not completely obvious which column in a linear combination, then just row reduce?
$\begin{pmatrix}
        1 & 0 & -1 & 1\\
        2 & 1 & 2 & 1\\
        3 & -1 & 0 & 1 \\
        \end{pmatrix} \to \begin{pmatrix}
        \color{red}{1} & 0 & -1 & 1\\
        0 & \color{red}{1} & 4 & -1\\
        0 & 0 & \color{red}{1} & \frac{-3}{7} \\
        \end{pmatrix}  $
Notice that this is row echeolen  and not reduced, since we only want to find out with columns have pivots. 
From this you see that the basis is:
$$b=\{[1,2,3],[0,1,-1],[-1,2,0]\}$$
