Finding the basis and dimension of a subspace of the vector space of 2 by 2 matrices I am trying to find the dimension and basis for the subspace spanned by:
$$
\begin{bmatrix}
1&-5\\
-4&2
\end{bmatrix},
\begin{bmatrix}
1&1\\
-1&5
\end{bmatrix},
\begin{bmatrix}
2&-4\\
-5&7
\end{bmatrix},
\begin{bmatrix}
1&-7\\
-5&1
\end{bmatrix}
$$
in the vector space $M_{2,2}$. I don't really care about the answer, I am just hoping for an efficient algorithm for solving problems like this for matrices.
I am not sure how to account for interdependence within the matrices. My instinct as of now is to find the maximum restriction imposed by the matrices. It is clear that the $1$ in position $a_{1,1}$ in each matrix will allow me to get any number in that position, so one vector in the basis will be:
$$
\begin{bmatrix}
1&0\\
0&0
\end{bmatrix}
$$
But depending on which of the matrices I scale, I have restrictions on the other entries. So I don't think I can include that matrix in my basis. 
It just occurred as I was writing this that I could maybe just think about these as $4$ by $1$ vectors and proceed as usual. Is there any danger in doing so?
 A: Inputs
$$
\alpha = 
\left(
\begin{array}{rr}
 1 & -5 \\ -4 & 2 
\end{array}
\right), 
\qquad
\beta = 
\left(
\begin{array}{rr}
 1 & 1 \\ -1 & 5 
\end{array}
\right), 
\qquad
\gamma = 
\left(
\begin{array}{rr}
 2 & -4 \\ -5 & 7 
\end{array}
\right), 
\qquad
\delta = 
\left(
\begin{array}{rr}
 1 & -7 \\ -5 & 1 
\end{array}
\right)
$$
Find the basis for these matrices.
Matrix of row vectors
As noted by @Bernard, compose a matrix of row vectors. Flatten the matrices in this manner
$$
\left(
\begin{array}{rr}
 1 & -5 \\ -4 & 2 
\end{array}
\right) 
\quad \Rightarrow \quad
\left(
\begin{array}{crrc}
 1 & -5 & -4 & 2 
\end{array}
\right)
$$
Compose the matrix
$$
\mathbf{A} = 
\left(
\begin{array}{crrr}
 1 & -5 & -4 & 2 \\\hline
 1 &  1 & -1 & 5 \\\hline
 2 & -4 & -5 & 7 \\\hline
 1 & -7 & -5 & 1 
\end{array}
\right)
$$
Row reduction
Column 1
$$
\left(
\begin{array}{rccc}
 1 & 0 & 0 & 0 \\
 -1 & 1 & 0 & 0 \\
 -2 & 0 & 1 & 0 \\
 -1 & 0 & 0 & 1 \\
\end{array}
\right)
%
\left(
\begin{array}{crrc}
 1 & -5 & -4 & 2 \\
 1 & 1 & -1 & 5 \\
 2 & -4 & -5 & 7 \\
 1 & -7 & -5 & 1 \\
\end{array}
\right)
%
=
%
\left(
\begin{array}{crrr}
 \boxed{1} & -5 & -4 & 2 \\
 0 & 6 & 3 & 3 \\
 0 & 6 & 3 & 3 \\
 0 & -2 & -1 & -1 \\
\end{array}
\right)
%
$$
Column 2
$$
\left(
\begin{array}{rccc}
 1 & 0 & 0 & 0 \\
 -1 & 1 & 0 & 0 \\
 -2 & 0 & 1 & 0 \\
 -1 & 0 & 0 & 1 \\
\end{array}
\right)
%
\left(
\begin{array}{crrr}
 \boxed{1} & -5 & -4 & 2 \\
 0 & 6 & 3 & 3 \\
 0 & 6 & 3 & 3 \\
 0 & -2 & -1 & -1 \\
\end{array}
\right)
%
=
%
\left(
\begin{array}{cccc}
 \boxed{1} & 0 & -\frac{3}{2} & \frac{9}{2} \\
 0 & \boxed{1} & \frac{1}{2} & \frac{1}{2} \\
 0 & 0 & 0 & 0 \\
 0 & 0 & 0 & 0 \\
\end{array}
\right)
%
$$
The fundamental rows are marked by the unit pivots.
Solution
The basis is
$$
\mathcal{B} = \left\{
\alpha, \, \beta \right\}
= \left\{
\left(
\begin{array}{rr}
 1 & -5 \\ -4 & 2 
\end{array}
\right), 
\
\left(
\begin{array}{rr}
 1 & 1 \\ -1 & 5 
\end{array}
\right)
\right\}
$$
