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Suppose I have the Euclidean space $\mathbb{R}^3$, then let $\mathbb{S}$ be a subspace for $\mathbb{R}^3$ and let $B$ be a basis for $\mathbb{S}$

This may seem like a stupid question,

But $\mathbb{S}$ has the vector equation form of $\overrightarrow{x} = [a_1, a_2, a_3]$ in other words, 3 rows as well? Same with the basis?

What I am trying to ask is, can a subspace be in another "dimension" eg a 2d vector for a subspace in $\mathbb{R}^3$?

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    Every member of $S$ has 3 co-ordinates because it is a member of $\mathbb R^3.$ If $B$ has 3 members then $S=\mathbb R^3. $ And $B$ has 2 members iff $S$ is a plane in $\mathbb R^3$ that contains the point $(0,0,0).$ And $B$ has 1 member iff $S$ is a line in $\mathbb R^3 $ that contains the point ($0,0,0).$ And you should say linear subspace, as in other contexts there are other kinds of subspaces of $\mathbb R^3.$ – 2017-01-27
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    Since $S \subset \mathbb{R}^3$, then every element of $S$ must have three coordinates. For example, the $x$ axis is the collection of elements $S= \{(t,0,0) | t \in \mathbb{R} \}$. In the latter case, we can identify $S$ with the real line in the obvious way, but it is not the real line. – 2017-01-27

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