# 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.

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

1. Using the matrix: $$T^t(g_1)(1)=(af_1+cf_2)(1)=a+0=a$$
2. 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

1. Using the matrix: $$T^t(g_1)(x)=(af_1+cf_2)(x)=c$$ (because $$f_1(x)=0$$ and $$f_2(x)=1$$)
2. 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?

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, $$T^t(g_2) = g_2(1,1)f_1 + g_2(0,2)f_2 = 1f_1 + 2f_2$$.