How can we determine if every matrix of $\mathbb{R}^{2 \times 2}$ can be written as a linear combination of specific $A, B$ matrices We have these two matrices:  
$$K = \left(\begin{matrix} 2 & 1 \\ 8 & 7\end{matrix}\right), \quad L = \left(\begin{matrix} 2 & 1 \\ 2 & 7 \end{matrix} \right)$$
We have been asked if every matrix of $\mathbb{R}^{2 \times 2}$ can be written as a linear combination of $K$ and $L$ matrices. This means that the set $\{K,L\}$ is a base of $\mathbb{R}^{2 \times 2}$, right?  
I've thought of this: For $K$ and $L$ matrices to be a base of $\mathbb{R}^{2 \times 2}$ they must be linearly independent, is that correct?
$a,b$ numbers of $\mathbb{R}$
$a \cdot K + b \cdot L = 0$, where $0$ is the $\left(\begin{matrix} 0 & 0 \\ 0 & 0 \end{matrix}\right)$ matrix.  
So: 
$$\begin{array}{cccc} 
2a &+& 2b &=& 0 \\  
a  &+& b  &=& 0 \\  
8a &+& 2b &=& 0 \\  
7a &+& 7b &=& 0 \end{array}$$  
(the solution set of this system is empty set?) 
How can I think of that?
Thank you!
 A: Hint: The dimension of ${\sf M}_2({\mathbb R})$ is $4.$ So the basis has to have how many matrices?!
A: Instead of writing a $2\times 2$ matrix as
$$\begin{pmatrix} a & b \\ c & d\end{pmatrix},$$
write it unconventionally as $(a,b,c,d)$. Now do you see that the vector space of $2\times 2$ matrices, with the usual addition, is $4$-dimensional?
A: What about noting that $\begin{pmatrix} 1 & 0 \\ 0 & 0\end{pmatrix}, \begin{pmatrix} 0 & 1 \\ 0 & 0\end{pmatrix}, \begin{pmatrix} 0 & 0 \\ 1 & 0\end{pmatrix}, \begin{pmatrix} 0 & 0 \\ 0 & 1\end{pmatrix} $
forms a basis of $\mathbb{R}^{2\times 2}$ and so as all basis of a vector space have the same number of elements and as you only have 2 elements then these cannot span.
A: Your matrices $K,L$ are indeed linearly independent, but $\{K,L\}$ is not a basis in $\mathbb R^{2\times 2}$, because $\dim \mathbb R^{2\times 2} = 4$. But I think it would be an interesting exercise for you to find other two matrices 
$M,N\in \mathbb R^{2\times 2}$ such that $\{K,L,M,N\}$ is a basis. 
