Originally Posted by

**topspin1617** Looks good.

This subspace consists of all vectors of the form $\displaystyle \left( \begin{array}{c} \frac{-1}{2}y\\y\\z\end{array}\right)=y\left( \begin{array}{c} \frac{-1}{2}\\1\\0\end{array}\right) +z\left( \begin{array}{c} 0\\0\\1\end{array}\right)$,

so the vectors $\displaystyle \left( \begin{array}{c} \frac{-1}{2}\\1\\0\end{array}\right)$ and $\displaystyle \left( \begin{array}{c} 0\\0\\1\end{array}\right)$ span the subspace. Since they are obviously linearly independent, they form a basis for the subspace.