An exercise from my textbook asks me to find a basis of $\mathbb R^{4}$ containing $S = (u,v)$, where $u = (0,1,2,3), v = (2,-1,0,1)$. The method they describe involves adding vectors from the standard basis of $\mathbb R^{4}$ to $S$ until we get an independent set of dim $4$.
First they add $e_1 = (1,0,0,0)$ then $e_2 = (0,1,0,0)$ resulting in the basis $(e_1,e_2,u,v)$.
I tried adding the other vectors in the standard basis and found that $(e_1,e_3,u,v)$ is also a basis of $\mathbb R^{4}$. How can both of these vectors be a basis of $\mathbb R^{4}$ $?$ I believe this is an error in the way I was thinking about what it means to form a basis.
I was imagining that with the vectors $(u,v)$ we cannot span all of $R^{4}$, so we need to add the standard vectors in the 'directions' that cannot be formed by $u$ and $v$ alone.
So by adding $e_1$ and $e_2$ we are allowing the new set to span in the directions $(1,0,0,0)$ and $(0,1,0,0)$ which it previously did not. So Why does adding the different direction $(0,0,1,0)$ instead of $e_2$ still span $\mathbb R^{4}$. Doesn't this new set not span in the direction that $e_2$ does?
Thanks for your help.
The interesting thing is that $\operatorname{span}(e_1, e_3, u,v)$ does include the $e_2$ direction even though $e_2$ is not in either $\operatorname{span}(e_1,e_3)$ or $\operatorname{span}(u,v)$. Confirm for yourself that this equation is true:
$$e_2 = (0,1,0,0) = \frac 32e_1 -\frac 12e_3 +\frac 14u -\frac 34v$$
And in fact you can show that $\operatorname{span}(e_1, e_2, u,v) = \operatorname{span}(e_1, e_3, u,v)$.
Pretty cool, right?