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Good evening!

I have serious doubts on how to do this exercise:

Let be $B=\{(1,2),(1,1)\}$ a base for $\mathbb{R}^2$ and I calculated $B^*=\{f_1,f_2\}$ the dual base of $B$ in $(\mathbb{R}^2)^{*}$ , Where the linear functionals of the dual base are determined by $f_1(x,y)=-x+y$ and $f_2(x,y)=2x-y$. I don't know how to find the explicit form of the elements in $B^{**}=\{h_1,h_2\}$, the dual base of $B^*$ in $(\mathbb{R}^2)^{**}$. Can someone help me with this? I'm stuck in this point. Thanks!

1 Answers 1

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You need to find two linear functionals $h_1,h_2 \colon \left( \mathbb{R}^2 \right)^{*} \rightarrow \mathbb{R}$ such that $h_i(f_j) = \delta_{ij}$. Consider the linear functionals

$$ h_1(\varphi) := \varphi(1,2),\\ h_2(\varphi) := \varphi(1,1). $$

That is, $h_i$ takes as an argument a linear functional $\varphi$ on $\mathbb{R}^2$ and evaluates it on the $i$-th basis vector. Check using the definition of the dual basis that you indeed have $h_i(f_j) = \delta_{ij}$. For example, we have

$$ h_1(f_1) = f_1(1,2) = 1, \,\,\, h_1(f_2) = f_2(1,1) = 0. $$

Remark: Note that you won't need to actually use the specific form of the basis $\mathcal{B}$ or the dual basis $f_1,f_2$. In fact, you have just proved that the dual basis of the dual basis is always given by evaluation functionals on the original basis. This is part of the duality between a vector space $V$ and its double dual $V^{**}$.

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    Excuse me but I'm new to this course, could it be a little more specific? I did not quite understand. – 2017-03-01
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    @DanielaRondón: I've added the calculation that shows that $h_1(f_1) = 1$ and $h_1(f_2) = 0$. – 2017-03-01
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    This means that $B^{**}=\{1,0\}$ I do not know?, but 0 can not be on a base. How are the functionalities sought? The function $\psi=f$? Sorry! – 2017-03-01
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    I don't understand. The dual basis is $(h_1,h_2)$. Neither $h_1$ nor $h_2$ are the zero linear functionals so all is well. – 2017-03-01
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    Then the dual basis is $\{h_1,h_2\}$ when $h_1(f)=f(1,2)$ and $h_2(f)=f(1,1)$? – 2017-03-01
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    @DanielaRondón: Yes. – 2017-03-01
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    If I consider the following linear functional in $\mathbb{R}^2$ $\psi(x,y)=x-7y$, then $h_1(\psi)=\psi(1,2)$ and $h_2(\psi)=\psi(1,1)$ form a basis of $(\mathbb{R})^{**}$? – 2017-03-01
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    The expression $h_1(\psi)$ is a number, so it doesn't make sense to ask whether $h_1(\psi),h_2(\psi)$ is a basis of $\mathbb{R}^{**}$. – 2017-03-01
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    I really regret not understanding your answer of everything, then to finalize how the base I am looking for, ie how are the functionals? – 2017-03-01
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    The basis is $(h_1,h_2)$ where $h_i$ are defined as in my answer... – 2017-03-01
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    But in your response you make a functional $\phi$ appear, my question is how is that functional determined or is arbitrary? – 2017-03-01
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    It is arbitrary because we need to define $h_i(\varphi)$ for each functional $\varphi$ in order to get a map $h_i \colon \left( \mathbb{R}^2 \right)^{*} \rightarrow \mathbb{R}$. – 2017-03-01
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    Thanks for the help! Is there a book or paper you can recommend me to learn this subject well? – 2017-03-01
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    @DanielaRondón: Not really, I don't have anything off the top of my head, sorry. – 2017-03-01