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  "body": "\n## Proof\n\nWe expand both vectors in the basis and use orthonormality.\n\n1. Let $V$ be a finite-dimensional inner product space over $\\mathbb{K} = \\R$ or $\\C$, with [[inner product|0VGSYBWHV2]] $\\ip{\\cdot}{\\cdot}$ and [[induced norm|0KG4D8AP6G]] $\\norm{\\cdot}$. Let $(e_1, \\dots, e_n)$ be an [[orthonormal basis|0FWJ0VW561]] of $V$.\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 $x, y \\in V$, and set $a_i = \\ip{x}{e_i}$ and $b_j = \\ip{y}{e_j}$. By the [[formula for coordinates|09DZ8CD1AK]], $x = \\sum_{i=1}^{n} a_i e_i$ and $y = \\sum_{j=1}^{n} b_j e_j$.\n4. Linearity in the first argument and step 2 give $\\ip{x}{y} = \\sum_{i=1}^{n} \\sum_{j=1}^{n} a_i \\overline{b_j} \\ip{e_i}{e_j}$.\n5. Since $\\ip{e_i}{e_j} = \\delta_{ij}$, only the terms with $i = j$ remain. So $\\ip{x}{y} = \\sum_{i=1}^{n} a_i \\overline{b_i} = \\sum_{i=1}^{n} \\ip{x}{e_i} \\overline{\\ip{y}{e_i}}$.\n6. This is the first identity. Now take $y = x$. Then $\\norm{x}^2 = \\ip{x}{x} = \\sum_{i=1}^{n} \\ip{x}{e_i} \\overline{\\ip{x}{e_i}}$.\n7. For each $z \\in \\C$, the product of $z$ and its [[complex conjugate|0BJV24C4SY]] is $z \\overline{z} = \\abs{z}^2$. Hence $\\norm{x}^2 = \\sum_{i=1}^{n} \\abs{\\ip{x}{e_i}}^2$, which is the second identity.\n\n## Idea\n\nIn an orthonormal basis, the coordinates of a vector are its inner products with the basis vectors. Expanding $\\ip{x}{y}$ in these coordinates gives a double sum. Orthonormality kills every mixed term, so only the diagonal terms remain.\n",
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