---
tag: A03D8FXB92
kind: proof
name: Proof of the canonical isomorphism with the double dual
msc: [15A04]
proves: 0ZJM9D7NPS
uses: [07JBSZE0V4, 0M856Q1AMZ, 07YQ1C78G0, 0NRYHJPQG2, 0CRMGPAH83, 0M4SXE5WW3, 0R8X9XA83A, 0BJBYCGMWE, 00889YD15K, 0W4FSW1TVB, 096V7H13H9, 0B8T10CE7N]
sources:
  - url: https://en.wikipedia.org/wiki/Dual_space
    locator: Injection into the double-dual
---

## Proof

1. Let $V$ be a vector space over a field $F$, and let $V^*$ be its [[dual space|07JBSZE0V4]]. By the [[definition|0M856Q1AMZ]], $V^*$ 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|07YQ1C78G0]] 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|0NRYHJPQG2]].
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|0CRMGPAH83]], because $v \ne 0$. So it [[extends to a basis|0M4SXE5WW3]] $B$ of $V$ with $v \in B$.
8. By the [[extension property of bases|0R8X9XA83A]], 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|0BJBYCGMWE]].
11. Now let $V$ be finite-dimensional with $n = \dim V$. By the [[dimension of the dual space|00889YD15K]], $V^*$ 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|0W4FSW1TVB]] 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|096V7H13H9]], $u_1, \dots, u_n$ 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|0B8T10CE7N]] when $V$ is finite-dimensional.

## Idea

To see that $v \ne 0$ is not sent to zero, we need one linear form that does not vanish at $v$. Extending $v$ to a basis gives such a form. In finite dimensions, $V$ and $V^{**}$ have the same dimension, so an injective linear map between them is onto.