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  "body": "\n## Proof\n\nWe split $x$ into a combination of the $e_i$ and a part orthogonal to them, and apply the Pythagorean theorem.\n\n1. Let $V$ be an inner product space over $\\mathbb{K} = \\R$ or $\\C$, with [[inner product|0VGSYBWHV2]] $\\ip{\\cdot}{\\cdot}$ and [[induced norm|0KG4D8AP6G]] $\\norm{\\cdot}$. Let $e_1, \\dots, e_k$ be an [[orthonormal family|09VYHEF2VT]] and let $x \\in V$.\n2. For all $u, w \\in V$ and $a \\in \\mathbb{K}$, conjugate symmetry and linearity in the first argument give $\\ip{u}{a w} = \\overline{a \\ip{w}{u}} = \\overline{a} \\ip{u}{w}$. In the same way, $\\ip{u}{\\cdot}$ is additive.\n3. Set $c_i = \\ip{x}{e_i}$, $s = \\sum_{i=1}^{k} c_i e_i$ and $r = x - s$. Then $x = s + r$.\n4. For each $j$, linearity in the first argument and orthonormality give $\\ip{s}{e_j} = \\sum_i c_i \\ip{e_i}{e_j} = c_j$.\n5. So $\\ip{r}{e_j} = \\ip{x}{e_j} - \\ip{s}{e_j} = c_j - c_j = 0$ for each $j$.\n6. By step 2, $\\ip{r}{s} = \\sum_{j} \\overline{c_j} \\ip{r}{e_j} = 0$. So $r$ and $s$ are [[orthogonal|0TS7VZJBWD]].\n7. By the [[Pythagorean theorem|0BVTDM872V]], $\\norm{x}^2 = \\norm{r + s}^2 = \\norm{r}^2 + \\norm{s}^2$.\n8. By step 2 and linearity in the first argument, $\\norm{s}^2 = \\ip{s}{s} = \\sum_{i} \\sum_{j} c_i \\overline{c_j} \\ip{e_i}{e_j}$.\n9. By orthonormality, only the terms with $i = j$ remain. So $\\norm{s}^2 = \\sum_{i} c_i \\overline{c_i} = \\sum_{i=1}^{k} \\abs{c_i}^2$.\n10. Since $\\norm{r}^2 = \\ip{r}{r} \\ge 0$, steps 7 and 9 give $\\sum_{i=1}^{k} \\abs{\\ip{x}{e_i}}^2 = \\norm{s}^2 \\le \\norm{x}^2$.\n\n## Idea\n\nThe vector $s = \\sum_i \\ip{x}{e_i} e_i$ is the part of $x$ along the directions $e_i$, and $x - s$ is orthogonal to all of them. By Pythagoras, $\\norm{x}^2$ is $\\norm{s}^2$ plus a nonnegative remainder. Orthonormality makes $\\norm{s}^2$ the sum of the squared coefficients.\n",
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