Subspace of R^3

Problem 132: Determine if $H = \{(x,y,z) | x+2y + z = 0\}$ is a subspace of $\Re^3$ .

Solution: If $H$ is a subspace, then it must meets the following three conditions.

The zero vector must be in $H$ . That is, let $\vec{\mathbf{0}} = (0,0,0)$ . Thus
(1) $\begin{align*} x + 2y + z & = 0 \\ 0 + 2(0) + 0 & = 0. \end{align*}$
Since the zero vector satisfies the equation, this means that $\vec{\mathbf{0}} \in H$ .
$H$ must be closed under addition. That is, $\vec{\mathbf{u}} + \vec{\mathbf{v}}$ must be in $H$ for any two vectors $\vec{\mathbf{u}}$ and $\vec{\mathbf{v}}$ . Let $\vec{\mathbf{u}} = (x_1, y_1, z_1)$ and $\vec{\mathbf{v}} = (x_2, y_2,z_2)$ . Then
(2) $\begin{align*} x + 2y + z & = 0 \\ (x_1 + x_2) + 2(y_1 + y_2) + (z_1 + z_2) & = 0 \\ x_1 + x_2 + 2y_1 + 2y_2 + z_1 + z_2 & = 0 \\ \underbrace{x_1 + 2y_1 + z_1}_{= \ 0} + \underbrace{x_2 + 2y_2 + z_2}_{= \ 0} & = 0 \\ 0 + 0 & = 0.\end{align*}$
Hence, $\vec{\mathbf{u}} + \vec{\mathbf{v}} \in H$ .
$H$ must be closed under scalar multiplication. That is, $c \vec{\mathbf{u}}$ must be in $H$ for any scalar $c \in \Re$ and vector $\vec{\mathbf{u}}$ . That is, let $c \in \Re$ and $\vec{\mathbf{u}} = (x_1, y_1, z_1)$ so that $c \cdot \vec{\mathbf{u}} = (cx_1, cy_1, cz_1)$ . That is,
(3) $\begin{align*}cx_1 + 2(cy_1) + cz_1 & = 0 \\ cx_1 + 2cy_1 + cz_1 = 0 \\ c\underbrace{(x_1 + 2y_1 + z_1)}_{= \ 0} & = 0 \\ c \cdot 0 & = 0 \\ 0 & = 0. \end{align*}$
Therefore, $c \cdot \vec{\mathbf{u}} \in H}$ .

Indeed, $H$ is a subspace of $\Re^3$ .