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  "body": "\n## Proof\n\nWe check linearity from the uniqueness of coordinates, and then exhibit a two-sided inverse.\n\n1. Let $\\varphi \\colon V \\to F^n$ be the map $\\varphi(v) = [v]_B$. It is well defined, because each $v \\in V$ equals $a_1 b_1 + \\cdots + a_n b_n$ for exactly one $(a_1, \\dots, a_n) \\in F^n$, since $B$ is a [[basis|0KCQ9Y3QKA]].\n2. $\\varphi$ is additive. Let $[v]_B = (a_1, \\dots, a_n)$ and $[w]_B = (c_1, \\dots, c_n)$. Then $v + w = \\sum_i (a_i + c_i) b_i$.\n3. By uniqueness of [[coordinates|04CYQHN6MC]], $[v + w]_B = (a_1 + c_1, \\dots, a_n + c_n) = [v]_B + [w]_B$.\n4. $\\varphi$ is homogeneous. For $c \\in F$, $cv = \\sum_i (c a_i) b_i$, so by uniqueness $[cv]_B = c\\,[v]_B$. Hence $\\varphi$ is a [[linear map|0NRYHJPQG2]].\n5. Let $\\psi \\colon F^n \\to V$ be the map $\\psi(a_1, \\dots, a_n) = a_1 b_1 + \\cdots + a_n b_n$ on the [[coordinate space|0XM8BZD76G]].\n6. For $v \\in V$ with $[v]_B = (a_1, \\dots, a_n)$, the definition of coordinates gives $\\psi(\\varphi(v)) = \\sum_i a_i b_i = v$. So $\\psi \\circ \\varphi = \\mathrm{id}_V$.\n7. For $(a_1, \\dots, a_n) \\in F^n$, let $v = \\psi(a_1, \\dots, a_n) = \\sum_i a_i b_i$. By uniqueness of coordinates, $[v]_B = (a_1, \\dots, a_n)$. So $\\varphi \\circ \\psi = \\mathrm{id}_{F^n}$.\n8. By [[the two-sided inverse criterion|01VG2A0BHW]], $\\varphi$ is bijective, and its inverse is $\\psi$.\n9. So $\\varphi$ is a bijective linear map, that is, a [[linear isomorphism|0B8T10CE7N]]. Its inverse sends $(a_1, \\dots, a_n)$ to $a_1 b_1 + \\cdots + a_n b_n$.\n\n## Idea\n\nCoordinates exist because a basis spans, and they are unique because a basis is independent. Uniqueness makes the coordinate map linear. Rebuilding a vector from its coordinates is the inverse map.\n",
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