1
$\begingroup$

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.

$\endgroup$
2
$\begingroup$

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?

I hope this is helpful.

$\endgroup$
1
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.