Proving a set of 2x2 matrices are linearly independent. I'm trying to prove the following matrices are linearly independent.
$ \beta = \left\{\begin{bmatrix} 1 & -1 \\ 0 & 2 \end{bmatrix},\begin{bmatrix} 0 & 1 \\ 3 & 0 \end{bmatrix}, \begin{bmatrix} 1 & 0 \\ 1 & 0 \end{bmatrix}, \begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix} \right\} $. 
My lecture notes say that because the matrix below is row equivalent to the identity matrix, the vectors in $\beta$ must be linearly independent. 
$ \begin{bmatrix} 1 & -1 & 0 & 2 \\ 0 & 1 & 3 & 0 \\ 1 & 0 & 1 & 0 \\ 1 & 1 & 1 & 1 \end{bmatrix}$
Could somebody please explain why this is?
Thanks,
Jack
 A: Linear independence of matrices is essentially their linear independence as vectors. So you are trying to show that the vectors $(1,-1,0,2), (0,1,3,0),(1,0,1,0)$ and $(1,1,1,1)$ are linearly independent. 
These are precisely the rows of the matrix that you have given. If this matrix is indeed row equivalent to the identity matrix (a fact which I'm assuming) then the vector space the above four vectors will generate will have dimension four (recall that, row or column operations don't change the rank of a matrix). This shows that they are linearly independent.
A: Hint
The basis of $\mathbb{M}_{2\times 2}(\mathbb{R})$ is as follow
$$B=\left\{e_1=\begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix},e_2=\begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}, e_3=\begin{bmatrix} 0 & 0 \\ 1 & 0 \end{bmatrix}, e_4=\begin{bmatrix} 0 & 0 \\0 & 1 \end{bmatrix}\right\}$$ 
and 
$$\begin{bmatrix} 1 & -1 \\ 0 & 2 \end{bmatrix}=e_1-e_2+0\times e_3+2e_4\\
\qquad\begin{bmatrix} 0 & 1 \\ 3 & 0 \end{bmatrix}=0\times e_1+e_2+3e_3+0\times e_4\\
 \qquad\begin{bmatrix} 1 & 0 \\ 1 & 0 \end{bmatrix}=e_1+0\times e_2+e_3+0\times e_4\\
\begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix}=e_1+e_2+e_3+e_4$$
Edit
let 
$$\alpha\begin{bmatrix} 1 & -1 \\ 0 & 2 \end{bmatrix}+\beta\begin{bmatrix} 0 & 1 \\ 3 & 0 \end{bmatrix}+\gamma\begin{bmatrix} 1 & 0 \\ 1 & 0 \end{bmatrix}+\theta\begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix}=0_{2\times 2}$$
therefore
$$(\alpha+\gamma+\theta)e_1+(-\alpha+\beta+\theta)e_2+(3\beta+\gamma+\theta)e_3+(2\alpha+\theta)e_4=0_{2\times 2}$$
in other words
$$ \begin{bmatrix} 1 & 0 & 1 & 1 \\ -1 & 1 & 0 & 1 \\ 0 & 3& 1 & 1 \\ 2 & 0 & 0 & 1 \end{bmatrix}\begin{bmatrix}\alpha\\\beta\\\gamma\\\theta\end{bmatrix}=0$$
