Can a set of vectors in $\mathbb{R}^3$ be a base for $\mathbb{R}^2$?

Question

For example, could the vectors $(0, 1, 2)$ and $(1, 0, 2)$ potentially span $\mathbb{R}^2$ and be linearly independent?

I've seen a bunch of articles that say two vectors cannot span $\mathbb{R}^3$ which makes sense to me, but it seems to me that if the last coordinates of the vectors were $0$ then they could span a plane.

I know that normally a candidate basis is disqualified if it has more vectors than needed, but in this case the vector would have more entries than needed so I am not sure if that still holds.

I am asking for my homework as I'm pretty confused.

Answer 1

The vectors cannot span $\mathbb R^{2}$ because they are not vectors in $\mathbb R^{2}$!. However they span a two dimensional subspace of $\mathbb R^{3}$.

Answer 2

In fact, any two linearly independent vectors will span a plane in $\mathbb{R}^3$, and any such plane is isomorphic$^\dagger$ as a vector space to $\mathbb{R}^2$. But do note that $\{(x, y, 0) \ | \ x, y \in \mathbb{R} \} \neq \{(x, y) \ | \ x, y \in \mathbb{R} \} = \mathbb{R}^2$. This is to say, vectors in $\mathbb{R}^n$ must have $n$ components by definition.

In this same vein, one should not be tempted to think of $\mathbb{R}$ as a subset of $\mathbb{R}^2$ (a common error) even though the horizontal axis of the Cartesian plane is isomorphic to $\mathbb{R}$. Or, as a topologist would say, $\mathbb{R}$ is embedded in $\mathbb{R}^2$, just as $\mathbb{R}^2$ is inside $\mathbb{R}^3$.

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$^\dagger$Explicitly, if the plane has basis vectors $\mathbf{v}$ and $\mathbf{w}$, then the isomorphism is determined by $\mathbf{v} \mapsto \langle 1, 0 \rangle$ and $\mathbf{w} \mapsto \langle 0, 1 \rangle$.