Prove that spans $\mathcal{V}$ spans $\text{Mat}_2(\mathbb{R})$ Let $A=\pmatrix{-1 & 0\\ 2 & 1},B=\pmatrix{2 & -1\\ 1 & 0},C=\pmatrix{1 & 2\\ 0 & -1},D=\pmatrix{0 & 1\\ -1 & 2}$
and $\mathcal{V}=(A,B,C,D)$.
My objective is to show that $\mathcal{V}$ spans $\text{Mat}_2(\mathbb{R})$ and that $\mathcal{V}$ is a basis for $\text{Mat}_2(\mathbb{R})$.
I have shown that:
$A + B −C + 3D=\pmatrix{0 & 0\\ 0 & 8}$,
$D + A − B + 3C=\pmatrix{0 & 8\\ 0 & 0}$,
$C + D − A + 3B=\pmatrix{8 & 0\\ 0 & 0}$,
$B +C − D + 3A=\pmatrix{0 & 0\\ 8 & 0}$.
From that I can see that it's possible to create linear combinations to get the following matrices:
$\pmatrix{0 & 1\\ 0 & 0},\pmatrix{0 & 0\\ 0 & 1},\pmatrix{0 & 0\\ 1 & 0},\pmatrix{1 & 0\\ 0 & 0}\in\text{span}(\mathcal{V})$.
How do I proceed?
 A: You can proceed as follows, if $$\begin{pmatrix}a&b\\c&d\end{pmatrix}\in\text{Mat}_2(\mathbb{R}),$$
then
$$\begin{pmatrix}a&b\\c&d\end{pmatrix}=a\begin{pmatrix}1&0\\0&0\end{pmatrix}+b\begin{pmatrix}0&1\\0&0\end{pmatrix}+c\begin{pmatrix}0&0\\1&0\end{pmatrix}+d\begin{pmatrix}0&0\\0&1\end{pmatrix}.$$
The L.H.S. is in $\text{Span}(\mathcal{V}),$ hence so is the $R.H.S.,$ so $\mathcal{V}$ is a spanning set. Any set of $4$ vectors spanning a $4$ dimensional vector space is automatically a basis, so $\mathcal{V}$ is a basis for $\text{Mat}_2(\mathbb{R})$.
A: You need not guess. Consider the basis
$$
\mathscr{B}=\left\{
E_1=\begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix},
E_2=\begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix},
E_3=\begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix},
E_4=\begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix}
\right\}
$$
so the coordinate vectors of $A$, $B$, $C$ and $D$ with respect to $\mathscr{B}$ are the columns of the matrix
$$
\begin{bmatrix}
-1 & 2 & 1 & 0 \\
0 & -1 & 2 & 1 \\
2 & 1 & 0 & -1 \\
1 & 0 & -1 & -2
\end{bmatrix}
$$
which is easily seen to have rank $4$. Therefore the coordinate vectors of $A,B,C,D$ form a basis of $\mathbb{R}^4$ which is equivalent to stating that $\{A,B,C,D\}$ is a basis for $\operatorname{Mat}_2(\mathbb{R})$.
You can also do in a direct way: show that $\{A,B,C,D\}$ is a linearly independent set. By the theorems on dimension of vector spaces, you can conclude that it is a basis. However, you'll discover that the linear system you have to solve has the same matrix as the $4\times4$ matrix above.
