Every $n$-tuple of scalars $(\alpha_1, \dots, \alpha_n)$ determines a linear functional on a finite dimensional normed space $X$.

What is the proof of this? Why does there exist a linear functional $f : X \rightarrow \Bbb R$ such that $f(e_i) = \alpha_i$ for each $n$-tuple? What are the functionals? How can linearity be preserved with every choice of $\alpha$'s?

I can see if $x = \sum_i c_i e_i$ then $f(x) = \sum_i c_i f(e_i) = \sum_i c_i \alpha_i$ where $\alpha_i = f(e_i)$, but I don't understand how this can be reversible to start with an $n$-tuple and find a functional.

  • $\begingroup$ For any $v\in X$ define $f(v)$ as the scalar product between the tuple and $v$ where $v$ is represented as $v_1e_1+...v_ne_n$. Show that is linear and then apply it to each of the $e_i$. $\endgroup$ – John Douma Dec 29 '19 at 19:09
  • $\begingroup$ @JohnDouma What is a scalar product between an $n$-tuple in $\Bbb R^n$ and a vector $\sum_i v_i e_i$ in a vector space $X$? $\endgroup$ – Oliver G Dec 29 '19 at 19:19
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    $\begingroup$ $\alpha_1v_1+...+\alpha_nv_n$ $\endgroup$ – John Douma Dec 29 '19 at 22:03

Determining the values of a linear map $T:V\rightarrow W$ on the basis elements of $V$ determines $T$, by the definition of a basis for a vector space. Your case is when $W=\mathbb{R}$.

This is well defined given that you are working with a fixed basis for all functionals, and not changing basis. For a different basis, this will give you different functionals.

  • $\begingroup$ Can you elaborate? I don't understand how this explains why any $\alpha$ has a linear map such that $T(e_j) = \alpha_j$. $\endgroup$ – Oliver G Dec 29 '19 at 18:59

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