How to find the inverse of a linear transformation?

Question

I have a linear transformation which is defined as

$$H:\mathbb{R_{≤ 2}[x]} \to \mathbb{R^{2,2}} \ ; \ ax^2+bx+c \to \left[ \begin{array}{cc} b-a& b+c \\ a-b & b \\ \end{array}\right]$$

and how am I supposed to find the inverse of $H$? I can calculate an inverse of an $n,m$ matrix and I'm guessing that I'm supposed to make an augmented matrix from the matrix but I'm not sure.

Answer 1

The domain is of dimension $3$, the codomain of dimension $4$, therefore there is no inverse strictly speaking.

However, you map is injective, which means that it has a left inverse:$$\exists\ g\ |\ g\circ f=\operatorname{Id}_{\Bbb R_{\leq 2}[X]}$$ which means that given $f(P)$ you are able to find out what $P$ was.

Hint: given $f(P)$ it is quite easy to get what $b$ was, and therefrom what were the other two coefficients: $$g\left(\left[\matrix{A&B\\C&D}\right]\right)=\ldots x^2+Dx+\ldots$$

Answer 2

Note that the range is of the form $$\pmatrix {A&B\\-A&C}$$ so that it is $3-$dimensional. Also, we have $$ b=C\\ c=B-C\\ a=-A+C $$