Show that a set of vectors is a basis in $\mathbb{R}^4$. Can you please explain this question to me? 

Show that the vectors $\beta = \{(1,1,0,0), (0,1,1,0), (0,0,1,1), (1,0,0,1)\}$ are basis in $\mathbb{R}^4$.

Attempt: 
The set is a basis if the vectors are linearly independent. But when I solved the question, the vectors were linearly dependent. But the questions says "show that ..." which means that the vectors form a basis. I don't know where I made a mistake.
 A: The vectors are linearly dependent.  To see why, consider that if $a_{1} = (1,1,0,0)$, $a_{2} = (0,1,1,0)$ and $a_{3} = (0,0,1,1)$ and $a_{4} = (1,0,0,1)$, then $a_{4}$ can be written as $a_{1} - a_{2} + a_{3}$.
A: One of possible methods is to calculate a determinant for the matrix constructed from vectors  $(1,1,0,0),(0,1,1,0),(0,0,1,1),(1,0,0,1)$. Because it has many $0$'s it is easy to calculate it by hand or with the use of matrix calculator.
We see that the matrix has zero determinant so the vectors are linearly dependent and can't be a basis.
A: If $w, x, y, z \in \mathbb{R}^4$, all that needs to be checked is whether or not they are all linearly independent. 
Suppose that $w, x, y, z$ are linearly independent in $\mathbb{R}^4$, but somehow do not form a basis for $\mathbb{R}^4$. This contradicts the fundamental theorem of linear algebra by the fact that $dim(\mathbb{R}^n) = n$
The easiest method is to represent your vectors as column vectors in a matrix $A$, and then row reduce to see if they are linearly independent. 
