In the space of all $2\times 2$ matrices, $M_{22}$ consider the subspace \begin{equation*} Z=\setparts{\begin{bmatrix}a&b\\c&d\end{bmatrix}}{a+3b-c-5d=0,\ -2a-6b+3c+14d=0} \end{equation*} and find a spanning set for $Z$.
We need to construct a limited number of matrices in $Z$ so that every matrix in $Z$ can be expressed as a linear combination of this limited number of matrices. Suppose that $B=\begin{bmatrix}a&b\\c&d\end{bmatrix}$ is a matrix in $Z$. Then we can form a column vector with the entries of $B$ and write \begin{equation*} \colvector{a\\b\\c\\d}\in \nsp{\begin{bmatrix}1 & 3 & -1 & -5\\-2 & -6 & 3 & 14\end{bmatrix}} \end{equation*}
Row-reducing this matrix and applying Theorem REMES we obtain the equivalent statement, \begin{equation*} \colvector{a\\b\\c\\d}\in \nsp{\begin{bmatrix}\leading{1} & 3 & 0 & -1\\0 & 0 & \leading{1} & 4\end{bmatrix}} \end{equation*}
We can then express the subspace $Z$ in the following equal forms, \begin{align*} Z&=\setparts{\begin{bmatrix}a&b\\c&d\end{bmatrix}}{a+3b-c-5d=0,\ -2a-6b+3c+14d=0}\\ &=\setparts{\begin{bmatrix}a&b\\c&d\end{bmatrix}}{a+3b-d=0,\ c+4d=0}\\ &=\setparts{\begin{bmatrix}a&b\\c&d\end{bmatrix}}{a=-3b+d,\ c=-4d}\\ &=\setparts{\begin{bmatrix}-3b+d&b\\-4d&d\end{bmatrix}}{b,\,d\in\complex{\null}}\\ &=\setparts{ \begin{bmatrix}-3b&b\\0&0\end{bmatrix}+ \begin{bmatrix}d&0\\-4d&d\end{bmatrix} }{b,\,d\in\complex{\null}}\\ &=\setparts{ b\begin{bmatrix}-3&1\\0&0\end{bmatrix}+ d\begin{bmatrix}1&0\\-4&1\end{bmatrix} }{b,\,d\in\complex{\null}}\\ &=\spn{\set{ \begin{bmatrix}-3&1\\0&0\end{bmatrix},\, \begin{bmatrix}1&0\\-4&1\end{bmatrix} }} \end{align*}
So the set \begin{equation*} Q=\set{ \begin{bmatrix}-3&1\\0&0\end{bmatrix},\, \begin{bmatrix}1&0\\-4&1\end{bmatrix} } \end{equation*} spans $Z$ by Definition SSVS.