Does there exist a basis for the set of $2\times 2$ matrices such that all basis elements are invertible? As the title says, I'm wondering whether there exists a basis for the set of $2\times 2$ matrices (with entries from the real numbers) such that all basis elements are invertible.
I have a gut feeling that it is false, but don't know how to prove it. I know that for a matrix to be invertible, it must be row equivalent to the identity matrix and I think I may be able to use this in the proof, but I don't know how.
Thanks in advance for any help,
Jack 

Consider the counter-example basis: 
$$\beta:=\left\{\begin{pmatrix} 1&0\\0&1\end{pmatrix},\begin{pmatrix} 0&1\\1&0\end{pmatrix},\begin{pmatrix} 1&0\\1&1\end{pmatrix},\begin{pmatrix} 0&1\\1&1\end{pmatrix}\right\}$$
Let $A=\begin{pmatrix} x_1&x_2\\x_3&x_4\end{pmatrix}$, let the following equation:
$$A=a\begin{pmatrix} 1&0\\0&1\end{pmatrix}+b\begin{pmatrix} 0&1\\1&0\end{pmatrix}+c\begin{pmatrix} 1&0\\1&1\end{pmatrix}+d\begin{pmatrix} 
0&1\\1&1\end{pmatrix}$$
Then
$$\begin{pmatrix} 
1&0&1&0\\0&1&0&1\\0&1&1&1\\1&0&1&1\end{pmatrix}\begin{pmatrix} 
a\\b\\c\\d\end{pmatrix}=\begin{pmatrix} 
x_1\\x_2\\x_3\\x_4\end{pmatrix}$$
Check $\det\begin{pmatrix} 
1&0&1&0\\0&1&0&1\\0&1&1&1\\1&0&1&1\end{pmatrix}=1$ . So there is an unique solution for $A$, hence $\beta$ is a basis of $2$ by $2$ matrix, where all basis elements are invertible.
