A question about functionals and dual space

Question

I have a question that I'm having trouble proving

If $f_1, f_2 ,…, f_n$ are linearly independent functionals in an $n$-dimensional vector space $V$ to it's scalar field $F$ does there always exists a base $x_1, x_2,…, x_n$ of V such that $$f_i(x_j)=\delta_{ij}=\begin{cases}1 \qquad i=j \\ 0 \qquad i \ne j \end{cases}$$

I know I should put my work here but I don't know how to prove it. It's an exam problem that I have in two days and I would really appreciate some help

Answer 1

Some steps for reach the result:

• Prove that $f_1,\dots,f_n$ is a basis for $V^*$, the space of all linear functions from $V$ to $\mathbf F$.
• For each $v \in V$ define $\operatorname{ev}(v) : V^* \to \mathbf F$ by $\operatorname{ev}(v)(\phi) = \phi(v)$, and prove that $\operatorname{ev}(v) \in V^{**}$, where $V^{**}$ is the space of all linear functions from $V^*$ to $\mathbf F$.
• Prove that if $v \in V \setminus \{0_V\}$ then there exists $\phi \in V^*$ such that $\phi(v) \neq 0$. Conclude that $\operatorname{ev} : V \to V^{**}$ is injective, and then, conclude that any $\varphi \in V^{**}$ is $\operatorname{ev}(v_\varphi)$ for some $v_\varphi \in V$.
• If $\varphi_1,\dots,\varphi_n \in V^{**}$ is the dual basis for $f_1,\dots,f_n$, then for each $i$ between $1$ and $n$ let $x_i \in V$ such that $\varphi_i = \operatorname{ev}(x_i)$, and prove that $x_1,\dots,x_n$ is the desired basis for $V$.