Determine whether $ (A_1,A_2,A_3,A_4)$ span $M_{22}$ Determine whether $\left[\begin{array}{cc}-1&1\\1&-1\end{array}\right], \left[\begin{array}{cc} 1&1\\1&1\end{array}\right], \left[\begin{array}{cc} 1&0\\0&1\end{array}\right],\left[\begin{array}{cc}0&1\\1&0\end{array}\right]$ spans $M_{22}$. If it does, construct the set of basis from the spanning set.
Pick an arbitrary matrix in $M_{22}=\left[\begin{array}{cc}x&y\\z&w\end{array}\right]$.If we find the following system to be consistent, then this set will span $M_{22}$
$a_1\left[\begin{array}{cc}-1&1\\1&-1\end{array}\right]+ a_2\left[\begin{array}{cc} 1&1\\1&1\end{array}\right]+a_3 \left[\begin{array}{cc} 1&0\\0&1\end{array}\right]+a_4\left[\begin{array}{cc}0&1\\1&0\end{array}\right]=\left[\begin{array}{cc}x&y\\z&w\end{array}\right]$ which simplifies to
$$-a_1+a_2+a_3=x$$
$$a_1+a_2+a_4=y$$
$$a_1+a_2+a_4=z$$
$$-a_1+a_2+a_3=w$$
By visually inspecting, we can see that eq #1 & 4 are the same and eq#2 & 3 are the same. So, if we perform row operations, we will end up with an inconsistent system. So, the set do no span $M_{22}$ and we cannot construct the basis.
Am  I on the right track? Appreciate your feedback.
 A: Even if the set doesn't span $M_{22}$, it span some subspace $\subseteq M_{22}$; we need to find the basis for such subspace.
To find it you can proceed as for ordinary row/colum vectors by RREF (in this way you could solve also part 1 by this) and selecting matrices corresponding to the pivot rows.
Notably, collecting the matrices by rows with the entries ordered as for $x,y,z,w$, we have to reduce the following
$$\begin{bmatrix}-1&1&1&-1\\1&1&1&1\\1&0&0&1\\0&1&1&0 \end{bmatrix}\to \begin{bmatrix}-1&1&1&-1\\1&1&1&1\\0&-1&-1&0\\0&0&0&0 \end{bmatrix}\to \begin{bmatrix}-1&1&1&-1\\0&2&2&0\\0&-1&-1&0\\0&0&0&0 \end{bmatrix}\to \begin{bmatrix}-1&1&1&-1\\0&2&2&0\\0&0&0&0\\0&0&0&0 \end{bmatrix}$$
from here we can conlude that the subspace spanned has dimension 2 and that we can take the first two matrices as a basis for it.
A: Note that the dimension of the space is 4 and if $A_1,...,A_4$ are the matrices in the question we have $A_2 = A_3+A_4$, hence the dimension of the span of the $A_k$
can be no more than 3.
Here is another approach:
Let $L$ be the linear functional that adds the top row and subtracts the bottom row. We see that $L (A_k) = 0$ for all $A_k$, but $L$ is not the zero operator, hence the $A_k$ cannot span the entire space. That is, $\ker L$ is not the entire space. 
We also observe that $A_1 = A_4-A_3$, hence the span has dimension of at most 2 and since $A_3,A_4$ do not lie on a line we see that the span has dimension 2.
So, we need to find two linearly independent matrices that do not lie in $\operatorname{sp}\{ A_3,A_4\}$. We see that $A \in \operatorname{sp}\{ A_3,A_4\}$ iff $[A]_{11} = [A]_{22}$ and $[A]_{12}=[A]_{21}$.
A straightforward selection would be $E_{11}, E_{12}$ (where $E_{ij}$ is the matrix of zeros except for a one in the ${ij}$ position).
