Base for Vector Space I have problem with one assignment and I am pretty sure that it's not that hard. So, I have 4 vectors :
$$
v_{1}=\begin{pmatrix}
    1 \\ 1 \\ 1
     \end{pmatrix},
  v_{2}=\begin{pmatrix}
    0 \\ 3 \\ 1
     \end{pmatrix},
  v_{3}=\begin{pmatrix}
    1 \\ -2 \\ 0
     \end{pmatrix},
  v_{4} =\begin{pmatrix}
    -2 \\ 1 \\ -1      
      \end{pmatrix} 
$$
and I have to find one base for the $span\{v_{1}, v_{2}, v_{3}, v_{4} \}$. Problem is that every combination of 3 vectors is linearly dependent (also all 4 are linearly dependent).
Have I misunderstood something or am I doing something wrong? 
 A: Notice first that $v_1$ and $v_2$ are linearly independent since they are not collinear i.e. there's not $\lambda\in\Bbb R$ such that $v_1=\lambda v_2$. Now since $v_3=v_1-v_2$ and $v_4=-2v_1+v_2$ so we can't extend the family $(v_1,v_2)$ on a basis for $\Bbb R^3$.
A: Here is one general method that involves matrix reduction (I assume you've covered that?)
Set up the matrix: $$\begin{bmatrix} 1 & 0 & 1 & -2 \\ 1 & 3 & -2 & 1 \\ 1 & 1 & 0 & -1 \end{bmatrix}$$
Then use column reduction:
$$\begin{bmatrix} 1 & 0 & 1 & -2 \\ 1 & 3 & -2 & 1 \\ 1 & 1 & 0 & -1 \end{bmatrix} \sim \begin{bmatrix} 1 & 0 & 0 & 0 \\ 1 & 3 & -3 & 3 \\ 1 & 1 & -1 & 1 \end{bmatrix} \sim \begin{bmatrix} 1 & 0 & 0 & 0 \\ 1 & 3 & 0 & 0 \\ 1 & 1 & 0 & 0 \end{bmatrix}$$
The nonzero columns will then span your space.
That is $\text{span}(v_1, v_2, v_3, v_4) = \text{span}\left(\begin{bmatrix} 1 \\ 1\\ 1\end{bmatrix}, \begin{bmatrix} 0 \\ 3\\ 1\end{bmatrix}\right)$.  So one of the infinite number of bases of your space is $\left\{\begin{bmatrix} 1 \\ 1\\ 1\end{bmatrix}, \begin{bmatrix} 0 \\ 3\\ 1\end{bmatrix}\right\}$.
