Finding a linear transformation from the basis of its range I'm trying to find a linear tranformation $\mathscr{L}$: $\mathbb{R}^3\mapsto\mathbb{R}^3$ such that {[1 -2 1] , [1 1 2]} is a basis for the range of $\mathscr{L}$. 
The way I thought of doing it was representing the basis as column vectors and finding a 3x3 matrix that includes this basis s.t when it reduces it will only have two pivots. However, I'm finding difficulty figuring out the third column, does anyone have any idea how to do this?

Let $\mathcal{E}$ the standard basis for $\mathbb{R}^3$, and let $\mathcal{L}$ the linear transformation sending 
$$(1,0,0)\longmapsto (1,-2,1)$$
$$(0,1,0)\longmapsto (1,1,2)$$
$$(0,0,1)\longmapsto \lambda(1,-2,1)+\mu(1,1,2)$$
where $\lambda, \mu$ are arbitrarily real numbers. For every couple of $(\lambda,\mu)$ you have an example of linear function whose range is generated by $(1,-2,1),(1,1,2)$.
