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 ! 

I like $0$'s in matrices.
Let $$\begin{bmatrix} 1 \\ 2\\ 3\end{bmatrix} = a \begin{bmatrix} 0 \\ 1\\ -1\end{bmatrix} + b \begin{bmatrix} -1 \\ 2\\ 0\end{bmatrix}$$
Immediately, from the first row, we can see that $b=-1$.
From the third row, we can see that $a=-3$.
Let's verify using the second row, $$a(1)+2b=-3-2=-5$$
Hence the first column is not in the span of the second and third column. 
Hence the first three columns form a basis.
Remark: If you know that the dimension is $3$, you can use the standard basis directly.
