Is T invertible - linear transformation (Linear Algebra)

Question

I'm practising some problems for my final and I'm stuck with some questions. Would be nice if I could get some help.

Let $T : P_2 (\mathbb R) \to P_2 (\mathbb R)$ be defined by $T( a + bx + cx^2) = 2c - bx +ax^2$. Is $T$ invertible? If yes, find $T^{-1}.$

So I started with the dimension of the domain ($=3$) is equal to the dimension of the codomain ($=3$). Then in my head, it seems logical that T is onto (I don't know how to explain it), which makes $T$ invertible. Now I have no idea what to use to compute the inverse. Thank you for your time.

Answer 1

Polynomials of degree at most $2$ have a natural basis, namely $\{1,x,x^2\}$. Therefore, a polynomial $a+bx+cx^2$ can be represented as the column vector $$\pmatrix{a \\ b \\ c}$$

If you write your transformation as a matrix in this basis, you have:

$$\pmatrix{0 && 0 && 2 \\ 0 && -1 && 0 \\ 1 && 0 &&0}\pmatrix{a \\ b \\ c}=\pmatrix{2c \\ -b \\ a}$$

Now notice that the determinant of your matrix is non-zero, therefore it can be inverted. Can you proceed with this idea?

Answer 2

Yes, it is invertible. It is easy to check that$$a+bx+cx^2\mapsto c-bx+\frac a2x^2$$is the inverse.

Answer 3

The matricial expression of $T$ with respect to the basis $B=\{1,x,x^2\}$ is $$\begin{bmatrix}{a'}\\{b'}\\{c'}\end{bmatrix}=\underbrace{\begin{bmatrix}{0}&{0}&{2}\\{0}&{-1}&{0}\\{1}&{0}&{0}\end{bmatrix}}_{A}\begin{bmatrix}{a}\\{b}\\{c}\end{bmatrix}$$ with $A$ invertible, so $T$ is invertible and the matricial expression of $T^{-1}$ with respect to $B$ is $$\begin{bmatrix}{a}\\{b}\\{c}\end{bmatrix}=\begin{bmatrix}{0}&{0}&{2}\\{0}&{-1}&{0}\\{1}&{0}&{0}\end{bmatrix}^{-1}\begin{bmatrix}{a'}\\{b'}\\{c'}\end{bmatrix}=\ldots$$