Show that $\mathcal{B}$ is a basis of $\mathbb{R}^4$. 
Let $v_1 = (1,-1,2,-2)$ and $v_2 = (0,0,-1,1)$.
Let $v_3 = (1,0,0,0)$ and $v_4 =(1,1,1,1)$.
Show that $\mathcal{B} = \{v_1, v_2, v_3, v_4 \}$ is a basis of $\mathbb{R}^4$.


After showing that $\mathcal{B} $ is set of independent vectors and that $\operatorname{Vect}(v_1, v_2, v_3, v_4)$ is a subspace of $\mathbb{R}^4$ and its dimension is the dimension of $\mathbb{R}^4$. I don't see how to proceed to get $\mathcal{B}$ is a basis of $\mathbb{R}^4$.
Thank you for your help.
 A: Since $\mathcal{B}$ is a set of $4$ linearly independent vectors (i.e. span $\mathbb{R}^4$) it is by definition a basis of $\mathbb{R}^4$.
Indeed note that forall $b\in\mathbb{R}^4$ the following system
$$\begin{pmatrix}1 & 0 & 1 & 1 \\ -1 & 0 & 0 & 1 \\ 2 & -1 & 0 & 1 \\ -2 & 1 & 0 & 1 \end{pmatrix}x=b$$
has an unique solution then $\mathcal{B}$ span $\mathbb{R}^4$.
A: In a four dimensional vector space every set of four linearly independent vectors constitutes a basis for the space. 
Since you have proved $$ B=\{ v_1, v_2, v_3, v_4\}$$ are linearly independent vectors,
$B$  is a basis for $\mathbb {R}^4 $  
A: To show that $\mathcal{B}$ is a set of linearly independent vectors, it's sufficient to show that :
$$\begin{pmatrix}1 & -1 & 2 & -2 \\ 0 & 0 & -1 & 1 \\ 1 & 0 & 0 & 0 \\ 1 & 1 & 1 & 1 \end{pmatrix}$$
cannot be "Gauss-Eliminated" aka it's irreducible. Since Gauss-Elimination is a starter-pack tool in Linear Algebra, you should definitely be familiar with it and by matrix operations (row swaps - subtractions), you should easily conclude it. If one is able to bring this matrix to an echelon form consisting of the $4$ unit vectors of $\mathbb R^4$, it means that not only these vectors are linearly independent (since its dimension won't be reduced) but also that they span $\mathbb R^4$. 
