In Example DSM22 we showed that \begin{equation*} B= \set{ \begin{bmatrix}-2&1\\1&0\end{bmatrix},\, \begin{bmatrix}-2&0\\0&1\end{bmatrix} } \end{equation*} is a basis for the subspace $Z$ of $M_{22}$ (Example VSM) given by \begin{equation*} Z=\setparts{\begin{bmatrix}a&b\\c&d\end{bmatrix}}{2a+b+3c+4d=0,\,-a+3b-5c-d=0} \end{equation*}
This tells us that $\dimension{Z}=2$. In this example we will find another basis. We can construct two new matrices in $Z$ by forming linear combinations of the matrices in $B$. \begin{align*} 2\begin{bmatrix}-2&1\\1&0\end{bmatrix}+ (-3)\begin{bmatrix}-2&0\\0&1\end{bmatrix}&= \begin{bmatrix}2&2\\2&-3\end{bmatrix}\\ 3\begin{bmatrix}-2&1\\1&0\end{bmatrix}+ 1\begin{bmatrix}-2&0\\0&1\end{bmatrix}&= \begin{bmatrix}-8&3\\3&1\end{bmatrix} \end{align*}
Then the set \begin{equation*} C= \set{ \begin{bmatrix}2&2\\2&-3\end{bmatrix},\, \begin{bmatrix}-8&3\\3&1\end{bmatrix} } \end{equation*} has the right size to be a basis of $Z$. Let us see if it is a linearly independent set. The relation of linear dependence \begin{align*} a_1\begin{bmatrix}2&2\\2&-3\end{bmatrix}+ a_2\begin{bmatrix}-8&3\\3&1\end{bmatrix}&=\zeromatrix\\ \begin{bmatrix}2a_1-8a_2&2a_1+3a_2\\2a_1+3a_2&-3a_1+a_2\end{bmatrix}&= \begin{bmatrix}0&0\\0&0\end{bmatrix} \end{align*} leads to the homogeneous system of equations whose coefficient matrix \begin{equation*} \begin{bmatrix} 2&-8\\ 2&3\\ 2&3\\ -3&1 \end{bmatrix} \end{equation*} row-reduces to \begin{equation*} \begin{bmatrix} \leading{1}&0\\ 0&\leading{1}\\ 0&0\\ 0&0 \end{bmatrix} \end{equation*}
So with $a_1=a_2=0$ as the only solution, the set is linearly independent. Now we can apply Theorem G to see that $C$ also spans $Z$ and therefore is a second basis for $Z$.