Theorem 15, Section 3.5 of Hoffman’s Linear Algebra Definition: Let $(V,F,+_V,\cdot_V)$ and $(F,F,+_F,\cdot_F)$ be vector space over field $F$. Then $V^*=(L(V,F),F,+_L,\cdot_L)$ is called dual space of $V$.

Let $B=\{\alpha_1,…,\alpha_n\}$ be basis of $V$. By theorem 1 section 3.1, $\forall i\in J_n$, $\exists !f_i\in L(V,F)$ such that $f_i(\alpha_j)=\delta_{ij}$, $\forall j\in J_n$. We can actually write $f_i$ explicitly. Let $h:V\to F^n$ be an isomorphism map (here is proof) and $\pi_i:F^n\to F$ be a projection map. Then $\pi_i\circ h:V\to F$ satisfy our desired property, $\pi_i\circ h(\alpha_j)$ $=\pi_{i}(h(\alpha_j))$ $=\pi_{i}(e_j)$ $=\delta_{ij}$, $\forall j\in J_n$. By uniqueness, $f_i=\pi_i \circ h$. Now we’re ready to solve theorem 15 section 3.5.

Let $V$ be a finite-dimensional vector space over the field $F$, and let $B= \{\alpha_1,…, \alpha_n\}$ be a basis for $V$. Then there is a unique dual basis $B^* = \{f_1,,…, f_n\}$ for $V^*$ such that $f_i(a_j) = \delta_{ij}$. For each linear functional $f$ on $V$ we have $f=\sum_{i=1}^nf(\alpha_i)\cdot_Lf_i$ and for each vector $\alpha$ in $V$ we have $\alpha =\sum_{i=1}^nf_i(\alpha)\cdot_V \alpha_i$.

My attempt: Let $f\in L(V,F)$. let $\alpha \in V$. Since $B=\{\alpha_1,…,\alpha_n\}$ is basis of $V$, $\exists !(x_1,…,x_n)\in F^n$ such that $\alpha =\sum_{i\in J_n}x_i\cdot_V \alpha_i$. Since $f$ is a linear map, we have $f(\alpha)$ $=f(\sum_{i\in J_n}x_i\cdot_V \alpha_i)$ $= \sum_{i\in J_n}x_i\cdot_F f(\alpha_i)$. By definition of $V^*$, $\sum_{i=1}^nf(\alpha_i)\cdot_Lf_i\in L(V,F)$. So $(\sum_{i=1}^nf(\alpha_i)\cdot_Lf_i)(\alpha)$ $= \sum_{i=1}^n(f(\alpha_i)\cdot_Lf_i)(\alpha)$ $=\sum_{i=1}^nf(\alpha_i)\cdot_Ff_i(\alpha)$. Since $f_i=\pi_i\circ h$, we have $f_i(\alpha)$ $=\pi_i(h(\alpha))$ $=\pi_i((x_1,…,x_n))=x_i$. So $\sum_{i=1}^nf(\alpha_i)\cdot_Ff_i(\alpha)$ $= \sum_{i=1}^nf(\alpha_i)\cdot_F x_i$ $= \sum_{i=1}^n x_i\cdot_F f(\alpha_i)$. Thus $f(\alpha)$ $= \sum_{i=1}^n x_i\cdot_F f(\alpha_i)$ $= (\sum_{i=1}^nf(\alpha_i)\cdot_Lf_i)(\alpha)$, $\forall \alpha \in V$. Hence $f= \sum_{i=1}^nf(\alpha_i)\cdot_Lf_i$. So $\mathrm{span}(B^*)$ $= \mathrm{span}(\{f_1,…,f_n\})=L(V,F)$. By theorem 4 corollary 2 section 2.3, $\{f_1,…,f_n\}$ is basis of $L(V,F)$.
Let $\alpha\in V$. Since $B=\{\alpha_1,…,\alpha_n\}$ is basis of $V$, $\exists !(x_1,…,x_n)\in F^n$ such that $\alpha =\sum_{i\in J_n}x_i\cdot_V \alpha_i$. By definition of map $h:V\to F^n$, we have $h(\alpha)=(x_1,…x_n)$. So $f_i(\alpha)$ $=\pi_i(h(\alpha))$ $=\pi_i((x_1,…,x_n))=x_i$, $\forall i\in J_n$. Thus $f_i(\alpha)=x_i$, $\forall i\in J_n$. Hence $\alpha =\sum_{i\in J_n}f_i(\alpha)\cdot_V \alpha_i$. Is my proof correct?
 A: Let me try to illustrate something by rewriting a part of your argument.
Take $f\in V^*$ and $\alpha\in V$. Assume $\alpha = \sum x_i\alpha _i$, where $x_i\in F$ and $\alpha_i\in B$ $^{(1)}$. By linearity of $f$ we have
$$ f(\alpha) = f\left (\sum x_i\alpha _i \right ) = \sum x_i f(\alpha _i). $$
Let $h:V\to F^n$ be the canonical isomorphism (with respect to $B$) and $\pi _i :F^n\to F$ the projection of the $i$-th coordinate. Put $f_i := \pi _ih$ for all $i$ and note that
$$ 
\begin{align*}
\left (\sum f(\alpha _i)f_i\right )(\alpha) = \sum f(\alpha _i)f_i(\alpha) &= \sum f(\alpha _i) \pi _ih(\alpha) \\
&=\sum f(\alpha _i) \pi _ih\left (\sum x_j\alpha _j\right ) \\
&=\sum f(\alpha _i)\pi _i(x_1,\ldots,x_n) \\
&=\sum x_if(\alpha _i).\quad^{(2)}
\end{align*}
$$
(1) $n$ is fixed, so $J_n$ is redundant. It is clear from context that we mean the unique linear combination with respect to the given basis $B$. It is also clear that the operation $x_i\alpha _i$ refers to the action of $F$ on $V$.
(2) Stylistically it is preferable to carry out long computations in display mode. It is easier on the eyes and also easier to follow. Notice also that there is no ambiguity regarding the multiplications involved.

Thus $f(\alpha)$ $= \sum_{i=1}^n x_i\cdot_F f(\alpha_i)$ $= (\sum_{i=1}^nf(\alpha_i)\cdot_Lf_i)(\alpha)$, $\forall \alpha \in V$.

Correct.

Hence $f= \sum_{i=1}^nf(\alpha_i)\cdot_Lf_i$. So $\mathrm{span}(B^*)$ $= \mathrm{span}(\{f_1,…,f_n\})=L(V,F)$.

Stay consistent with your style. You quantified the previous statement. Do so here as well. It holds that $f = \sum f(\alpha _i)f_i$ for every $f\in V^*$, thus the $f_i$ span $V^*$.

By Corollary 2 section 2.3, $\{f_1,…,f_n\}$ is basis of $L(V,F)$

I don't see how Corollary 2 supports this statement. Corollary 2 states the following. Let $V$ be an $n$-dimensional vector space. Then

*

*any collection of more than $n$ vectors is linearly dependent.

*no collection of less than $n$ vectors spans $V$.

It is true that for finite dimensional vector spaces $V\cong V^*$, but how do you know a priori that $\dim V^* = n$? You only know that the $f_i$ span $V^*$. You should also check the $f_i$ are linearly independent.

Let $\alpha\in V$. Since $B=\{\alpha_1,…,\alpha_n\}$ is basis of $V$, $\exists !(x_1,…,x_n)\in F^n$ such that $\alpha =\sum_{i\in J_n}x_i\cdot_V \alpha_i$. By definition of map $h:V\to F^n$, we have $h(\alpha)=(x_1,…x_n)$. So $f_i(\alpha)$ $=\pi_i(h(\alpha))$ $=\pi_i((x_1,…,x_n))=x_i$, $\forall i\in J_n$. Thus $f_i(\alpha)=x_i$, $\forall i\in J_n$. Hence $\alpha =\sum_{i\in J_n}f_i(\alpha)\cdot_V \alpha_i$.

This answers $f_i(\alpha _j) = \delta _{ij}$ (although you should make this computation explicit) and $\alpha = \sum f_i(\alpha) \alpha_i$.

Then there is a unique dual basis $B^* = \{f_1,,…, f_n\}$ for $V^*$ such that $f_i(a_j) = \delta_{ij}$

What about this part? Again, I don't see how uniqueness (immediately) follows from 3.1 Theorem 1.
A: Alternative proof: Let $f\in L(V,F)$. Let $j\in J_n$. Then $(\sum_{i\in J_n}f(\alpha_i)\cdot_L f_i)(\alpha_j)$ $=\sum_{i\in J_n}(f(\alpha_i)\cdot_Lf_i)(\alpha_j)$ $=\sum_{i\in J_n}f(\alpha_i)\cdot_F f_i(\alpha_j)$ $= \sum_{i\in J_n}f(\alpha_i)\cdot_F \delta_{ij}$ $=f(\alpha_j)$. Since $j$ was arbitrary, we have $(\sum_{i\in J_n}f(\alpha_i)\cdot_L f_i)(\alpha_j)$ $=f(\alpha_j)$, $\forall j\in J_n$. By theorem 1 section 3.1, $f=\sum_{i\in J_n}f(\alpha_i)\cdot_L f_i$.
This proof is much more efficient than showing $f(\alpha)=(\sum_{i\in J_n}f(\alpha_i)\cdot_L f_i)(\alpha)$, $\forall \alpha \in V$.
