Computing the dual change of coordinate matrix $[T^t]^{\beta *}_{\gamma *}$ I am trying to understand the computation of $[T^t]^{\beta *}_{\gamma *}$ from Friedberg linear algebra. 
$T:P_1(R)→ R^2$ and $T(p(x))=(p(0),p(2))$
$\beta$ and $\gamma$ are the standard ordered bases for $P_1(R)$ and $R^2$ respectively and the asterisk denotes their dual bases, and $T^t$ denotes the transpose of the linear transformation $T$.
We are to compute $[T^t]^{\beta *}_{\gamma *}$ directly without using the fact that $[T^t]^{\beta *}_{\gamma *}=([T]^{\gamma}_{\beta})^t$.

First we let $[T^t]^{\beta ^*}_{\gamma ^*}=\left( \begin{matrix}
    a & b \\
    c & d \\
   \end{matrix} \right )$
Then we see that $T^t(g_1)=af_1+cf_2$ where $\beta ^*=\{f_1,f_2\}$ and $\gamma ^*=\{g_1,g_2\}$


Friedberg goes on to show that $T^t(g_1)(1)=(af_1+cf_2)(1)=af_1(1)+cf_2(1)=a(1)+c(0)=a$
My question is why did he choose $1$? Is it so that $a$ could be isolated from this equation and how would one know that? Also, why is $f_1(1)=1$ and $f_2(1)=0$. I know $f_{i}(x_j)=\delta _{ij}$ but I cannot put these two facts together since I cannot see how $1$ is $x_j$ where $x_j$ is the jth vector of a basis.
Then Friedberg goes on to show  $(T^t(g_1))(1)=g(1)(T(1))=g_1(1,1)=1$ and that proves that $a=1$.
I don't understand why $g_1(1,1)=1$, and I'm not really sure what $g_1(1,1)$ even means.
Then using similar computations not provided in the book, $b,c,$ and $d$ are found. How is this done? Is it by performing the above with $g_2$ and $1$ and $0$?Any guidance or explanation of the proof is greatly appreciated.
 A: I believe that your main issue is that you are used to think of bases in an abstract fashion. That is, if $\beta:=\{x_1, \ldots, x_n\}$ is a basis for a vector space $X$ then the dual basis $\beta^*=\{f_1, \ldots, f_n\}$ are linear functionals such that $f_{i}(x_j)=\delta_{i,j}$. However, for this question you have some concrete vector spaces and some well known bases for each of them.
First of all since $\beta$ is the standard ordered bases for $P_1(\Bbb{R})$ we actually have $\beta=\{1, x\}$. Thus, the dual basis is $\beta^*=\{f_1, f_2\}$, where $f_1, f_2 : P_1(\Bbb{R}) \to \Bbb{R}$ are such $f_1(1)=1$, $f_1(x)=0$, $f_2(1)=0$ and $f_2(x)=1$ (think of $1$ as $x_1$ and $x$ as $x_2$ in the abstract fashion above). Hopefully this answers one of your questions.
Similarly, $\gamma=\{(1,0), (0,1)\}$ is the standard basis for $\Bbb{R}^r$ and therefore the dual basis is $\gamma^*:=\{ g_1 ,g_2\}$ where $g_1, g_2: \Bbb{R}^2 \to \Bbb{R}$ are such that $g_1(1,0)=1$, $g_1(0,1)=0$, $g_2(1,0)=0$ and $g_2(0,1)=1$ (think of $(1,0)$ as $x_1$ and $(0,1)$ as $x_2$ in the abstract fashion above). Therefore, since $g_1$ is linear
$$
g_1(1,1)=g_1( (1,0)+(0,1) ) = g_1(1,0)+g_1(0,1)=1+0=1
$$
This should answer what $g(1,1)$ is and why is it equal to $1$.
Finally your main goal is to find the entries $a,b,c$ and $d$ for the matrix of the linear transformation $T^t$ with respect to the bases $\gamma^*$ and $\beta
^*$. To do this you have to use that there are two ways to compute $T^t(g_1)(1)$, namely

*

*Using the matrix: $T^t(g_1)(1)=(af_1+cf_2)(1)=a+0=a$

*By definition of $T^t$: $T^t(g_1)(1)=g_1(T(1))= g_1(1,1) = 1$
This gives you the value of $a$. Analogously there are two ways to compute $T^t(g_1)(x)$, namely

*

*Using the matrix: $T^t(g_1)(x)=(af_1+cf_2)(x)=c$ (because $f_1(x)=0$ and $f_2(x)=1$)

*By definition of $T^t$: $T^t(g_1)(x)=g_1(T(x))= g_1(0,2) = 2g_1(0,1)= 0$
This now gives the value of $c$. Similarly, when computing both $T^t(g_2)(1)$ and $T^t(g_2)(x)$ using matrix way and the definition way you should be able to find the values for $b$ and $d$.
Do you think you can take it from here now?
I hope this is helpful.
A: Above they already showed you the reasoning, but I want to show you another way of doing it. I will first observe one thing in "abstract".

If $V$ is a finite-dimensional vector space, $\alpha = \{v_1,\dots,v_n\}$ is a basis for $V$, and $\alpha^* = \{\phi_1,\dots,\phi_n\}$ the corresponding dual basis, then any $f \in V^*$ can be written as $f = f(v_1)\phi_1 + \cdots + f(v_n) \phi_n$.

This is easy to see, for if $v \in V$, then $v = \phi_1(v)v_1 + \cdots + \phi_n(v)v_n$, and then
$$f(v) = \phi_1(v)f(v_1) + \cdots + \phi_n(v)f(v_n) = \big( f(v_1)\phi_1 + \cdots + f(v_n) \phi_n \big)(v).$$
So, in this concrete example we have to write the linear functionals $T^t(g_1)$ and $T^t(g_2)$ as a linear combination of $f_1$ and $f_2$, and because $\{f_1,f_2\}$ is the dual basis of $\{1,x\}$ we have:
\begin{align}
T^t(g_1) &= T^t(g_1)(1)f_1 + T^t(g_1)(x)f_2 \\
&= g_1(T(1))f_1 + g_1(T(x))f_2 \\
&= g_1(1,1)f_1 + g_1(0,2)f_2 \\
&= 1f_1 + 0f_2
\end{align}
and similarly, $T^t(g_2) = g_2(1,1)f_1 + g_2(0,2)f_2 = 1f_1 + 2f_2$.
