Consider the subset of $\complex{5}$ defined as \begin{equation*} W =\setparts{\colvector{x_1\\x_2\\x_3\\x_4\\x_5}}{ \begin{array}{l} 3x_1+x_2-5x_3+7x_4+x_5=0,\\ 4x_1+6x_2+3x_3-6x_4-5x_5=0,\\ -2x_1+4x_2+7x_4+x_5=0 \end{array} } \end{equation*}
It is possible to show that $W$ is a subspace of $\complex{5}$ by checking the three conditions of Theorem TSS directly, but it will get tedious rather quickly. Instead, give $W$ a fresh look and notice that it is a set of solutions to a homogeneous system of equations. Define the matrix \begin{equation*} A=\begin{bmatrix} 3&1&-5&7&1\\ 4&6&3&-6&-5\\ -2&4&0&7&1 \end{bmatrix} \end{equation*} and then recognize that $W=\nsp{A}$. By Theorem NSMS we can immediately see that $W$ is a subspace. Boom!