Finding the inverse of a function $F:\mathbb{P}_2 \rightarrow \mathbb{R}^2$

Question

Here is the problem:

Define a function, $F: \mathbb{P} _2 \rightarrow \mathbb{R}^3$ via the rule, $$\text{if}\ p(x)=a_0+a_1x+a_2x^2,\text{then }\ F(p)=(a_0,a_0+a_2,a_0+3a_1+4a_2).$$ The function $F$ given above is bijective, which you can assume without proof. Find (and give) the assignment rule for $F^{-1}$ and specify its domain and codomain. Confirm that for a generic $p\in\mathbb P_2$ with $p(x)=a_0+a_1x+a_2x^2$, that you have $[F^{-1}(F(p))](x)=a_0+a_1x+a_2x^2$.

I am having a lot of trouble getting started with forming an inverse function that takes us from a point in 3 dimensions to a quadratic equation. It almost seems impossible to create an assignment rule $\mathbb{R}^3 \rightarrow \mathbb{P}^2$ for such a situation. Am I missing something? How should I approach this?

Thanks.

Answer 1

Write everything coordinatewise

$$F\begin{pmatrix}a_0\\a_1\\a_2\end{pmatrix}=F(a_0+a_1x+a_2x^2):=\begin{pmatrix}a_0\\a_0+a_2\\a_0+3a_1+4a_2\end{pmatrix}=\begin{pmatrix}1&0&0\\1&0&1\\1&3&4\end{pmatrix}\begin{pmatrix}a_0\\a_1\\a_2\end{pmatrix}$$

and thus $\;F^{-1}\;$ will be expressed by the inverse matrix of the above one. Try to take it from here.