Checking if this transformation $L: R^3 \to P_4$ exists? I've been given three vectors: $\begin{bmatrix}1&2&3\end{bmatrix}$, $\begin{bmatrix}0&1&1\end{bmatrix}$, and $\begin{bmatrix}1&1&1\end{bmatrix}$ and asked if it is possible for there to be a linear transformation $L:\mathbb{R}^3\rightarrow P_4$ such that $\begin{bmatrix}1&2&3\end{bmatrix}\mapsto x^2$, $\begin{bmatrix}0&1&1\end{bmatrix}\mapsto x^2-1$, and $\begin{bmatrix}1&1&1\end{bmatrix} \mapsto 1$.
My professor goes about saying that this must be true since the vectors are linearly independent, but I'm having trouble figuring out why linear independence means this transformation must be possible. Can anyone help clarify why this is the case?

Let $x_1=[1,2,3]^T,$ $x_2=[0,1,1]^T,$ $x_3=[1,1,1]^T.$ $1,x,x^2$ is linear independent in $P_4.$ So let $y_1=x^2=[0,0,1]^T,$ $y_2=x^2-1=[-1,0,1]^T,$ $y_3=1=[1,0,0]^T.$
You want to find a linear transformation $A,$ such that $Ax_i=y_i,$ $i=1,2,3.$ Write it into $$A[x_1,x_2,x_3]=[y_1,y_2,y_3]\Rightarrow AX=Y$$
Here $X=[x_1,x_2,x_3]$ is invertible since $\{x_1,x_2,x_3\}$ is linear independent. So $A=YX^{-1}$ is a unique solution.
