# Inverse of a linear transformation

• Dec 30th 2011, 12:00 PM
cristi92
Inverse of a linear transformation
$\displaystyle A:\mathbb{R}_{2}[x]->\mathbb{R}^{3} , A(P)=(P(-1),P(0),P(1))$
where A is a bijective linear transformation, $\displaystyle \mathbb{R}_{2}[x]$ is the set of polynomials with real coeficients and max degree=2 .
Find $\displaystyle A^{-1}(0,6,0)$

To do this I have to find the matrix of A relative to canonical basis of $\displaystyle \mathbb{R}_{2}[x]$ and $\displaystyle \mathbb{R}^{3}$ , find the inverse of that matrix and then $\displaystyle A^{-1}\begin{pmatrix}0\\ 6\\ 0\end{pmatrix}$ is the answer I'm looking for.

Please tell me if I am right or not.
• Dec 30th 2011, 12:35 PM
ILikeSerena
Re: Inverse of a linear transformation
Welcome to MHF, cristi92! :)

That looks fine to me!
• Dec 30th 2011, 12:38 PM
cristi92
Re: Inverse of a linear transformation
Thank you! :)
• Dec 30th 2011, 12:55 PM
HallsofIvy
Re: Inverse of a linear transformation
That would work but since you are NOT asked to find either A or $\displaystyle A^{-1}$ in general, it would seem easier to me to state the problem as "find numbers, a, b, c, so that $\displaystyle P(x)= ax^2+ bx+ c$, such that $\displaystyle P(-1)= a(-1)^2+ b(-1)+ c= a- b+ c= 0$, $\displaystyle P(0)= a(0^2)+ b(0)+ c= c= 6$, $\displaystyle P(1)= a(1^2)+ b(1)+ c= a+ b+ c= 0$". In other words solve the three equations a- b+ c= 0, c= 6, a+ b+ c= 0 for a, b, and c.

Of course, if you solve those equations by setting up the matrix of coeffients and row-reducing, you are doing what you describe but those are particularly easy equations to solve.
• Dec 30th 2011, 12:59 PM
cristi92
Re: Inverse of a linear transformation
Thank you!