True or false?$\left\{x \in\mathbb{R}^3 \mid x_1=0 \right\} \cap\left\{x\in \mathbb{R}^3 \mid x_2=0 \right\}$ is a linear subspace of $\mathbb{R}^{3}$

Question

True or false? $\left\{x \in\mathbb{R}^3 \mid x_1=0 \right\} \cap\left\{x\in \mathbb{R}^3 \mid x_2=0 \right\}$ is a linear subspace of $\mathbb{R}^{3}$

I think the statement is false because if we intersect two vectors we can get a vector of another dimension which won't be of $\mathbb{R}^{3}$?

Let's say this is of the first set $\begin{pmatrix} 0\\ 1\\ 0 \end{pmatrix}$ and this is of second set $\begin{pmatrix} 0\\ 0\\ 0 \end{pmatrix}$. As intersection we would get $\begin{pmatrix} 0\\ 0 \end{pmatrix}$ I think? And this would be of $\mathbb{R}^{2}$.

Or I understood things wrong and this is complete daft?

Answer 1

All vectors such that $x_1 = 0$ and $x_2 = 0$

Looks like $$ \begin{pmatrix} x_1 \\ 0 \\ x_3 \end{pmatrix} \text{AND} \begin{pmatrix} 0 \\ x_2 \\ x_3 \end{pmatrix} $$

or as I said up there, only $x_3$ is allowed to be non zero.

I think with a little work you can show that what is true about that space. The comments in my opinion are all correct, and you will need to use the subspace criterion to do so formally. But all vectors have dimension three.