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  "body": "\n## Statement\n\nLet $V$ be a finite-dimensional inner product space over $\\R$ or $\\C$, with the inner product linear in the first argument.\nThen, for each [[linear form|07YQ1C78G0]] $f \\colon V \\to \\mathbb{K}$, there is exactly one vector $y \\in V$ with\n$$f(x) = \\langle x, y \\rangle \\qquad \\text{for all } x \\in V.$$\n\n## Remarks\n\nFor an orthonormal basis $e_1, \\dots, e_n$, the vector is $y = \\sum_i \\overline{f(e_i)}\\, e_i$.\nThe map $y \\mapsto \\langle \\cdot, y \\rangle$ from $V$ to its dual is bijective; it is linear over $\\R$ and conjugate linear over $\\C$.\nThe statement holds for continuous linear forms on every Hilbert space.\n\n## Explanations\n\n### L1 human\n\nIn an inner product space, every linear form is an inner product with a fixed vector.\nOn $\\R^3$, the form $x \\mapsto 2x_1 - x_2 + 5x_3$ is the dot product with $(2, -1, 5)$.\nSo in finite dimensions, vectors and linear forms correspond one to one.\n",
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