Can $2$ vectors in $\mathbb{R}^4$ generate a $3$ dimensional space?

Question

Say I have $2$ vectors: $(1,2,0,1)$ and $(1,3,1,1)$, generating a subspace in R4.

Doing some calculations I get $\{(x,y,z,w) \in \mathbb{R}^4: w=x \land z=y-2x\}$ (I don't know what this is called in english, sorry. Bear with me).

So that's: $(x,y,z,w) \land (x,y,y-2x,w)$ which is equivalent to: $x(1,0,0,1) + y(0,1,0,0) + z(0,0,1,0) \land x(1,0,-2,0) + y(0,1,1,0) + w(0,0,0,1)$

So am I wrong in inferring that this returns $3$ dimensions or did I otherwise do something wrong?

Thanks

Answer 1

\begin{align}\{(x,y,z,w) \in \mathbb{R}^4: w=x , z=y-2x\}&=\{(x,y,y-2x,x) \in \mathbb{R}^4: x, y \in \mathbb{R}\} \\ &=\{x(1,0,-2,1)+y(0,1,1,0) : x,y \in \mathbb{R} \}\end{align}

The subspace that you described is $2$ dimensional.

Answer 2

No, a basis for $\mathbb{R}^3$ requires 3 linear independent vectors.