Proof of the canonical isomorphism with the double dual

Proof

1. Let $V$ be a vector space over a field $F$, and let $V^*$ be its dual space. By the definition, $V^*$ has pointwise operations.
2. Fix $v \in V$. For $f, g \in V^*$ and $a, b \in F$, we have $\operatorname{ev}(v)(af + bg) = (af + bg)(v) = a f(v) + b g(v)$.
3. The right side of step 2 equals $a \operatorname{ev}(v)(f) + b \operatorname{ev}(v)(g)$. So $\operatorname{ev}(v)$ is a linear form on $V^*$, that is, $\operatorname{ev}(v) \in V^{**}$.
4. Let $v, w \in V$ and $a, b \in F$. For every $f \in V^*$, linearity of $f$ gives $\operatorname{ev}(av + bw)(f) = f(av + bw) = a f(v) + b f(w)$.
5. The right side of step 4 equals $(a \operatorname{ev}(v) + b \operatorname{ev}(w))(f)$. Since this holds for all $f$, the map $\operatorname{ev}$ is linear.
6. We show that the kernel of $\operatorname{ev}$ is $\{0\}$. Let $v \in V$ with $v \ne 0$.
7. The set $\{v\}$ is linearly independent, because $v \ne 0$. So it extends to a basis $B$ of $V$ with $v \in B$.
8. By the extension property of bases, there is a linear map $f \colon V \to F$ with $f(v) = 1$ and $f(b) = 0$ for $b \in B \setminus \{v\}$. Then $f \in V^*$.
9. Then $\operatorname{ev}(v)(f) = f(v) = 1 \ne 0$. So $\operatorname{ev}(v) \ne 0$.
10. By steps 6 to 9, the kernel of $\operatorname{ev}$ is $\{0\}$. So $\operatorname{ev}$ is injective, by the kernel criterion.
11. Now let $V$ be finite-dimensional with $n = \dim V$. By the dimension of the dual space, $V^*$ is finite-dimensional with $\dim V^* = n$.
12. Applying the same result to $V^*$, the space $V^{**}$ is finite-dimensional with $\dim V^{**} = \dim V^* = n$.
13. By the rank–nullity theorem and step 10, $\dim \im(\operatorname{ev}) = \dim V - 0 = n$.
14. Let $u_1, \dots, u_n$ be a basis of $\im(\operatorname{ev})$. These are $n$ linearly independent vectors of $V^{**}$, a space of dimension $n$.
15. By the criterion for bases, $u_1, \dots, u_n$ is a basis of $V^{**}$. So $\im(\operatorname{ev})$ contains their span $V^{**}$, and $\operatorname{ev}$ is surjective.
16. By steps 5, 10 and 15, $\operatorname{ev}$ is a bijective linear map. So it is a linear isomorphism when $V$ is finite-dimensional.