-
The matrices in $S$ will span $M_{22}$ if for any $\begin{bmatrix} x& y\\z& w\end{bmatrix}$, there are coefficients $a, b, c, d, e$ so that
\begin{align*}
a\begin{bmatrix} 1 & 2\\ 2 & 1 \end{bmatrix} +
b\begin{bmatrix} 2 & 1\\-1 & 2 \end{bmatrix} +
c\begin{bmatrix} 0 & 1 \\ 1 & 2\end{bmatrix} +
d\begin{bmatrix} 1 & 0 \\ 1 & 1\end{bmatrix} +
e\begin{bmatrix} 1 & 4 \\ 0 & 3 \end{bmatrix}
&=
\begin{bmatrix} x & y \\ z & w \end{bmatrix}
\end{align*}
Thus, we have
\begin{align*}
\begin{bmatrix}
a + 2b + d + e & 2a + b + c + 4e\\
2a - b + c + d& a + 2b + 2c + d + 3e
\end{bmatrix}
&=
\begin{bmatrix}
x & y \\
z & w
\end{bmatrix}\\
\end{align*}
so we have the matrix equation
\begin{align*}
\begin{bmatrix}
1 & 2 & 0 & 1 & 1 \\
2 & 1 & 1 & 0 & 4\\
2 & -1 & 1 & 1 & 0\\
1 & 2 & 2 & 1 & 3
\end{bmatrix}
\colvector{a\\b\\c\\d\\e}
&=
\colvector{x\\y\\z\\w}
\end{align*}
This system will have a solution for every vector on the right side if the row-reduced coefficient matrix has a leading one in every row, since then it is never possible to have a pivot column appear in the final column of a row-reduced augmented matrix.
\begin{align*}
\begin{bmatrix}
1 & 2 & 0 & 1 & 1 \\
2 & 1 & 1 & 0 & 4\\
2 & -1 & 1 & 1 & 0\\
1 & 2 & 2 & 1 & 3
\end{bmatrix}
\rref
\begin{bmatrix}
\leading{1}& 0 & 0 & 0 & 1\\
0 & \leading{1} & 0 & 0 & 1\\
0 & 0 & \leading{1} & 0 & 1\\
0 & 0 & 0 & \leading{1} & -2
\end{bmatrix}
\end{align*}
Since there is a leading one in each row of the row-reduced coefficient matrix, there is a solution for every vector $\colvector{x\\y\\z\\w}$, which means that there is a solution to the original equation for every matrix $\begin{bmatrix} x & y\\ z & w \end{bmatrix}$. Thus, the original five matrices span $M_{22}$.
- The matrices in $S$ are linearly independent if the only solution to
\begin{align*}
a\begin{bmatrix} 1 & 2\\ 2 & 1 \end{bmatrix} +
b\begin{bmatrix} 2 & 1\\-1 & 2 \end{bmatrix} +
c\begin{bmatrix} 0 & 1 \\ 1 & 2\end{bmatrix} +
d\begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix} +
e\begin{bmatrix} 1 & 4 \\ 0 & 3 \end{bmatrix}
&=
\begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}
\end{align*}
is $a = b = c = d = e = 0$.
We have
\begin{align*}
\begin{bmatrix}
a + 2b + d + e & 2a + b + c + 4e\\
2a - b + c + d & a + 2b + 2c + d + 3e
\end{bmatrix}
&=
\begin{bmatrix}
1 & 2 & 0 & 1 & 1 \\
2 & 1 & 1 & 0 & 4\\
2 & -1 & 1 & 1 & 0\\
1 & 2 & 2 & 1 & 3 \end{bmatrix}
\colvector{a\\b\\c\\d\\e}
=
\begin{bmatrix}
0 & 0 \\
0 & 0 \end{bmatrix}
\end{align*}
so we need to find the nullspace of the matrix
\begin{align*}
\begin{bmatrix}
1 & 2 & 0 & 1 & 1 \\
2 & 1 & 1 & 0 & 4\\
2 & -1 & 1 & 1 & 0\\
1 & 2 & 2 & 1 & 3
\end{bmatrix}
\end{align*}
We row-reduced this matrix in part (a), and found a pivot column, which provides a free variable in a description of the solution set to the homogeneous system, so the nullspace is nontrivial and there are an infinite number of solutions to
\begin{align*}
a\begin{bmatrix} 1 & 2\\ 2 & 1 \end{bmatrix} +
b\begin{bmatrix} 2 & 1\\-1 & 2 \end{bmatrix} +
c \begin{bmatrix} 0 & 1 \\ 1 & 2\end{bmatrix} +
d \begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix} +
e \begin{bmatrix} 1 & 4 \\ 0 & 3 \end{bmatrix}
&=
\begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}
\end{align*}
Thus, this set of matrices is not linearly independent.