How do I go about proving that the solution set of
\begin{align*} w - 3 x + 4 y - z &= 0 \\ 3 w + x + 2 y + z &= 0 \end{align*}
is a subspace of $\mathbb{R}^{4}$?
I originally defined S as: \begin{equation} S = \left\{ ( w, x, y, z) \in \mathbb { R } ^ { 4 } | w - 3x + 4y -z = 0\textbf{ and }3w + x + 2y + z = 0 \right\} \end{equation}
and then determined if
$\vec{0} \in S,$
$S$ is closed under addition ($u,\>v \in S \implies u + v \in S$), and
$S$ is closed under scalar multiplication ($\alpha \in \mathbb{F} \text{ and } u \in S \implies \alpha \cdot u \in S $),
by checking if both $w - 3x + 4y -z = 0$ and $3w + x + 2y + z = 0$ satisfied the above conditions, which was rather straightforward.
Is that correct? Or am I missing something?
Another method would be to note that $S=\operatorname{ker}T$ where $T$ is represented by the following matrix (rel the standard basis) :
$\begin{pmatrix}1&-3&4&-1\\3&1&2&1\end{pmatrix}$.
The kernel of a linear transformation is always a subspace.