Let $T:P_1(\mathbb{R})\rightarrow \mathbb{R}^{2}$ where $P_1$ are the polynomial with the most degree 1.
And $T(P(x))=(P(0),P(2))$ with the standard basis $b=\left \{ 1,x \right \},g=\left \{ e_1,e_2 \right \}$
I want to find $\left [ T \right ]^{g}_{b}$
so I take from the first element of the $b$ basis $p(1)=1+1=2$ $(p(x)=1+x)$ so $T(P(1))=(2,2)$ then $T(P(x))=(P(0),P(2))=(1+0,1+2)=(1,3)$ so $\left [ T \right ]^{g}_{b}=\begin{pmatrix} 2 & 1\\ 2 & 3 \end{pmatrix}$ the answer I should get is $\begin{pmatrix} 1 & 0\\ 1 & 2 \end{pmatrix}$ I am assuming I just do row operations to take that matrix or am i doing something wrong?
Let $p_1(x)=1$ and $p_2(x)=x.$ Then
$$T(p_1)=(1,1)=1 \cdot e_1+1 \cdot e_2.$$
Hence the first column of $\left [ T \right ]^{g}_{b}$ is given by $(1,1)^T.$
We have
$$T(p_2)=(0,2)=0 \cdot e_1+2 \cdot e_2.$$
Hence the second column of $\left [ T \right ]^{g}_{b}$ is given by $(0,2)^T.$