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  "body": "\n## Proof\n\n1. 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.\n2. Fix $v \\in V$. For $f, g \\in V^*$ and $a, b \\in F$, we have $\\operatorname{ev}(v)(af + bg) = (af + bg)(v) = a f(v) + b g(v)$.\n3. 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, $\\operatorname{ev}(v) \\in V^{**}$.\n4. Let $v, w \\in V$ and $a, b \\in F$. For every $f \\in V^*$, linearity of $f$ gives $\\operatorname{ev}(av + bw)(f) = f(av + bw) = a f(v) + b f(w)$.\n5. 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]].\n6. We show that the kernel of $\\operatorname{ev}$ is $\\{0\\}$. Let $v \\in V$ with $v \\ne 0$.\n7. The set $\\{v\\}$ is [[linearly independent|0CRMGPAH83]], because $v \\ne 0$. So it [[extends to a basis|0M4SXE5WW3]] $B$ of $V$ with $v \\in B$.\n8. 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 $f \\in V^*$.\n9. Then $\\operatorname{ev}(v)(f) = f(v) = 1 \\ne 0$. So $\\operatorname{ev}(v) \\ne 0$.\n10. By steps 6 to 9, the kernel of $\\operatorname{ev}$ is $\\{0\\}$. So $\\operatorname{ev}$ is injective, by the [[kernel criterion|0BJBYCGMWE]].\n11. Now let $V$ be finite-dimensional with $n = \\dim V$. By the [[dimension of the dual space|00889YD15K]], $V^*$ is finite-dimensional with $\\dim V^* = n$.\n12. Applying the same result to $V^*$, the space $V^{**}$ is finite-dimensional with $\\dim V^{**} = \\dim V^* = n$.\n13. By the [[rank–nullity theorem|0W4FSW1TVB]] and step 10, $\\dim \\im(\\operatorname{ev}) = \\dim V - 0 = n$.\n14. 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 $n$.\n15. 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.\n16. 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.\n\n## Idea\n\nTo 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.\n",
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