How to check whether the sets U1, U2, U3 are subspaces of the vector space V. In the $\mathbb{R}$-vector space $V = \mathbb{R}^3$
, we consider the following subsets:
$U_1 := \{x\in\mathbb{R}^3\ |\ x_1 + x_2 − x_3 = 0\}$,
$U_2:= \{x\in\mathbb{R}^3\ |\ \exists n\in\mathbb{Z}\ :\ x_1 = n, x_2 = 2n, x_3 = 3n\}$,
$U_3:= \{x\in\mathbb{R}^3\ |\ (x_1 = 0)\vee(x_2 = 0)\vee (x_3 = 0)\}$
I need to check whether the sets $U_1,\ U_2$, and $U_3$ are subspaces of the vector space $V$, How can I properly do that?
 A: To show that $U \subset \mathbb{R}^3$ is an $\mathbb{R}$-subspace of $(\mathbb{R}^3, +, \cdot)$ with componentwise addition $+$ and componentwise scalar multiplication $\cdot$, you must show that $U$ is a vectorspace with respect to $+$ restricted to $U \times U$ and $\cdot$ restricted to $\mathbb{R} \times U$. This can be done by showing that each of the following 3 statements hold:
$$\begin{align*}
    0 \in U\\
    u, v \in U \implies u + v \in U\\
    r \in \mathbb{R}, u \in U \implies r \cdot u \in U
\end{align*}$$
Axioms like associativity and distributivity still hold in a subset $U$, no need to re-verify those.
$\textbf{1)}$ We check the axioms for $U_1$:
$0 = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}$ satisfies $0 + 0 - 0 = 0$, hence $0 \in U_1$.
If $u, v \in U_1$ then there exist $u_1, u_2, u_3, v_1, v_2, v_3 \in \mathbb{R}$ such that $u = \begin{pmatrix} u_1 \\ u_2 \\ u_3 \end{pmatrix}, v = \begin{pmatrix} v_1 \\ v_2 \\ v_3 \end{pmatrix}$ and $u_1 + u_2 - u_3 = v_1 + v_2 - v_3 = 0$. Then $u + v = \begin{pmatrix} u_1 + v_1 \\ u_2 + v_2 \\ u_3 + v_3 \end{pmatrix}$ satisfies
$$\begin{align*}
(u_1 + v_1) + (u_2 + v_2) - (u_3 + v_3) = (u_1 + u_2 - u_3) + (v_1 + v_2 - v_3) = 0 + 0 = 0.
\end{align*}$$
This means that $u + v \in U_1$ as well.
Finally, if $r \in \mathbb{R}$, then $r \cdot u = r \cdot \begin{pmatrix} u_1 \\ u_2 \\ u_3 \end{pmatrix} = \begin{pmatrix} r \cdot u_1 \\ r \cdot u_2 \\ r \cdot u_3 \end{pmatrix}$ satisfies
$$\begin{align*}
r \cdot u_1 + r \cdot u_2 - r \cdot u_3 = r \cdot (u_1 + u_2 - u_3) = r \cdot 0 = 0.
\end{align*}$$
This proves that $r \cdot u \in U$ and hence $\textbf{$U_1$ is an $\mathbb{R}$-subspace of $\mathbb{R}^3$}$ since all axioms are fulfilled.
$\textbf{2)}$ We check $U_2$: We have that $u := \begin{pmatrix} 1 \\ 2 \cdot 1 \\ 3 \cdot 1 \end{pmatrix} = \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix} \in U_2$, but $\frac{1}{2} \cdot u = \begin{pmatrix} \frac{1}{2} \\ 1 \\ \frac{3}{2} \end{pmatrix} \notin U_2$ since there is no $n \in \mathbb{Z}$ such that $\begin{pmatrix} \frac{1}{2} \\ 1 \\ \frac{3}{2} \end{pmatrix} = \begin{pmatrix} n \\ 2n \\ 3n \end{pmatrix}$. So $\textbf{$U_2$ is $\textit{not}$ an $\mathbb{R}$-subspace of $\mathbb{R}^3$}$ since $U_2$ is not closed under scalar multiplication.
$\textbf{3)}$ We check $U_3$: We have that $u := \begin{pmatrix} 0 \\ 1 \\ 1 \end{pmatrix} \in U_3$ and $v := \begin{pmatrix} 1 \\ 0 \\ 1 \end{pmatrix} \in U_3$, but $u+v = \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix} \notin U_3$. This means that $\textbf{$U_3$ is $\textit{not}$ an $\mathbb{R}$-subspace of $\mathbb{R}^3$}$ since $U_3$ is not closed under vector addition.
A: As commented above you can check whether they are subspaces. You can also check by just looking at the problem of such types. How this happens?
If $W$ be any plane in $\Bbb{R}^3$ passing through the origin, then they satisfy the subspace conditions (You can see here).
Good luck!
