$v_1=(2,2,1)\ v_2=(1,-1,0)\ v_3=(1,0,-1)$
and let $e_1=(1,0,0)\ e_2=(0,1,0)\ e_3=(0,0,1)$ be the basis for $\mathbb R^3$
so this is the part one of the question (part 2 is if the set $S={v_1v_2v_3}$ is a basis for $\mathbb R^3$)
I've shown that the set $S$ is linearly independent using gaussian elimination. I know how to show that a set is a basis but i'm not sure how to show ^ is contained in the set. I feel like i'm just confusing myself and missing something really trivial. Any hint would be great! Thank you
also this is my gaussian elimination for $v_1 v_2 v_3$ if its of any use
\begin{bmatrix}2&2&1\\1&-1&0\\1&0&-1\end{bmatrix} swap $v_2v_1,\ v_1-2v_2,\ v_3-v_2$\begin{bmatrix}1&-1&0\\0&4&1\\0&1&-1\end{bmatrix} swap $v_3v_1$ and subtract last row with 4*row2
\begin{bmatrix}1&-1&0\\0&-1&-1\\0&0&5\end{bmatrix}
sorry for the funny looking annotation on the side, im not sure how to label the vectors next to the matrix
Best Answer
If you can show that the vectors $e_1,e_2,e_3 \in \langle v_1,v_2,v_3 \rangle$ it will follows that $\mathbb{R}^3$ is spanned by $v_1,v_2,v_3$ since they generate the standard basis for $\mathbb{R}^3$.
What you have done is show that $S = \{v_1,v_2,v_3\}$ is a linearly independent set of vectors in $\mathbb{R}^3$. This is actually sufficient to claim that you have a basis for $\mathbb{R}^3$. This is due to the following
The proof of this follows from something called the replacement theorem generally, but if you haven't seen this you need to simply show that $e_1,e_2$ and $e_3$ can be expressed as linear combinations of the elements in $S$. Once that is the case, if $S$ can "create" the standard basis for $\mathbb{R}^3$ it can certainly generate all of $\mathbb{R}^3$ itself.
So consider the equation $a_1v_1+a_2v_2+a_3v_3 = e_1$. This amounts to another system of equations to find the coefficients $a_1,a_2,a_3$. Can you finish the other two?