TITLE: How to calculate $f(p)_G$ where $p$ is given by$ p(x)=4 x^2+8 x+4 ∀x∈\Bbb R.$?
QUESTION [0 upvotes]: Consider a linear map $f\in\mathcal L(\prod_2\Bbb R),\Bbb R^2)$ whose matrix $f_{G,B}$ in the basis $\mathcal B$ of $\prod_2(\Bbb R)$ and 
$G=\{(1,-1) , (2,1)\}$ of $\Bbb R^2$ is 
$$f_{G,B} = \begin{pmatrix}
-10 & -2 \\ 7 & -2\\ -5 & -9
\end{pmatrix} ^T$$
Calculate $f(p)_G$ where $p$ is given by $p(x)=4x^2+8x+4 ∀x∈\Bbb R. $
Im a bit stuck on this question especially with understanding some of the notation. Because  $f_{G,B}$ is in the basis B of a second degree polynomial then $B=\{1,x,x^2\}$. I know how to find the coordinated of $p$ in $B$ (its just $[ 4,8,4]^T$ but im assuming i need to find the coordinate of $p$ in G and then put into f. I might be very wrong though. Would really appreciate your help.
Cheers!

REPLY [1 votes]: The given $3 \times 2$ matrix cannot represent a linear map in $\mathcal L(\prod_2(\Bbb R),\Bbb R^2)$ since such a matrix would have to be $2 \times 3$.
It is possible that you should transpose your matrix:
$$ f_{G,B} = \pmatrix{
-10 & 7 & -5 \\ -2 & -2 & -9}$$
In that case we have
$$f(p)_G = f_{G,B}p_B =  \pmatrix{
-10 & 7 & -5 \\ -2 & -2 & -9}\pmatrix{4 \\ 8 \\ 4
} = \pmatrix{-4 \\ -60
}$$
Therefore:
$$f(p) = -4\pmatrix{1 \\ -1
} -60 \pmatrix{2 \\ 1
}  = \pmatrix{-124 \\ -56}$$
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