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  "body": "\n## Proof\n\nWe build $y$ from an orthonormal basis, and then prove uniqueness with $x = y - y'$.\n\n1. Let $V$ be a finite-dimensional [[inner product space|01EK923SKP]] over $\\mathbb{K} = \\R$ or $\\C$, with [[inner product|0VGSYBWHV2]] $\\ip{\\cdot}{\\cdot}$ linear in the first argument. Let $f \\colon V \\to \\mathbb{K}$ be a [[linear form|0RT96J4D3H]].\n2. For all $u, w \\in V$ and $a \\in \\mathbb{K}$, conjugate symmetry and linearity in the first argument give $\\ip{u}{a w} = \\overline{a} \\ip{u}{w}$. In the same way, $\\ip{u}{\\cdot}$ is additive.\n3. Let $e_1, \\dots, e_n$ be an [[orthonormal basis|0Y85GYEGSF]] of $V$. Set $y = \\sum_{i=1}^{n} \\overline{f(e_i)}\\, e_i$.\n4. Let $x \\in V$. By the [[formula for coordinates|09DZ8CD1AK]], $x = \\sum_{i=1}^{n} \\ip{x}{e_i} e_i$.\n5. Since $f$ is linear, step 4 gives $f(x) = \\sum_{i=1}^{n} \\ip{x}{e_i} f(e_i)$.\n6. By step 2, $\\ip{x}{y} = \\sum_{i=1}^{n} \\overline{\\overline{f(e_i)}} \\ip{x}{e_i} = \\sum_{i=1}^{n} f(e_i) \\ip{x}{e_i}$.\n7. Steps 5 and 6 give $f(x) = \\ip{x}{y}$ for all $x \\in V$. This proves existence.\n8. For uniqueness, let $y, y' \\in V$ with $\\ip{x}{y} = f(x) = \\ip{x}{y'}$ for all $x \\in V$.\n9. By step 2, $\\ip{x}{y - y'} = \\ip{x}{y} - \\ip{x}{y'} = 0$ for all $x \\in V$.\n10. Take $x = y - y'$. Then $\\ip{y - y'}{y - y'} = 0$, and positivity of the inner product gives $y - y' = 0$. So $y = y'$.\n\n## Idea\n\nA linear form is fixed by its values on a basis. For an orthonormal basis, the form $x \\mapsto \\ip{x}{y}$ takes the value $\\overline{c_i}$ at $e_i$ when $y = \\sum_i c_i e_i$. So we choose $c_i = \\overline{f(e_i)}$. Uniqueness holds because only $0$ is orthogonal to every vector.\n",
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