# Every $\alpha \in \Bbb R^n$ determines a linear functional in a finite dimensional vector space

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.

• 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 $e_i$. Dec 29, 2019 at 19:09
• @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 $X$? Dec 29, 2019 at 19:19
• $\alpha_1v_1+...+\alpha_nv_n$ Dec 29, 2019 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 $$W=\mathbb{R}$$.
• Can you elaborate? I don't understand how this explains why any $\alpha$ has a linear map such that $T(e_j) = \alpha_j$. Dec 29, 2019 at 18:59