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  "body": "\n## Proof\n\nLet $\\mathbb{K}$ be the scalar field, $\\R$ or $\\C$. Write $\\bar a$ for the conjugate of $a \\in \\mathbb{K}$, so $\\bar a = a$ over $\\R$.\n\n1. By axioms 1 and 2 of the [[inner product|0NB0AWGJBW]], for $u, v, w \\in V$ we get $\\langle w, u + v \\rangle = \\overline{\\langle u, w \\rangle + \\langle v, w \\rangle}$. This equals $\\langle w, u \\rangle + \\langle w, v \\rangle$.\n2. Likewise $\\langle w, -v \\rangle = \\overline{-\\langle v, w \\rangle} = -\\langle w, v \\rangle$.\n3. By the [[induced norm|0MFJJV97RJ]], axiom 1 and step 1, $\\norm{x + y}^2 = \\langle x + y, x + y \\rangle$. This equals $\\langle x, x \\rangle + \\langle x, y \\rangle + \\langle y, x \\rangle + \\langle y, y \\rangle$.\n4. By axiom 1 and steps 1 and 2, $\\norm{x - y}^2 = \\langle x, x \\rangle - \\langle x, y \\rangle - \\langle y, x \\rangle + \\langle y, y \\rangle$.\n5. Adding steps 3 and 4, the mixed terms cancel. So $\\norm{x + y}^2 + \\norm{x - y}^2 = 2 \\langle x, x \\rangle + 2 \\langle y, y \\rangle = 2 \\norm{x}^2 + 2 \\norm{y}^2$.\n\n## Idea\n\nExpand both squared norms as inner products. The mixed terms $\\langle x, y \\rangle + \\langle y, x \\rangle$ appear with a plus sign in one and a minus sign in the other. So they cancel in the sum.\n",
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        "locator": "The parallelogram law in inner product spaces",
        "url": "https://en.wikipedia.org/wiki/Parallelogram_law"
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