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  "body": "\n## Proof\n\nWe take the inner product of a vanishing combination with each vector of the family.\n\n1. Let $V$ be an inner product space over $\\mathbb{K} = \\R$ or $\\C$, with [[inner product|0VGSYBWHV2]] $\\ip{\\cdot}{\\cdot}$. Let $(e_i)_{i \\in I}$ be an [[orthonormal family|09VYHEF2VT]] in $V$, so $\\ip{e_i}{e_j} = \\delta_{ij}$ for all $i, j \\in I$.\n2. By the definition of [[linear independence|0EVHJFTG9Z]] of a family, it is enough to show that each finite subfamily is linearly independent.\n3. Let $i_1, \\dots, i_k$ be distinct indices in $I$. Let $a_1, \\dots, a_k \\in \\mathbb{K}$ with $a_1 e_{i_1} + \\cdots + a_k e_{i_k} = 0$.\n4. Fix $m$ with $1 \\le m \\le k$. The inner product of $0$ with $e_{i_m}$ is $\\ip{0}{e_{i_m}} = \\ip{0 \\cdot 0}{e_{i_m}} = 0 \\cdot \\ip{0}{e_{i_m}} = 0$, by linearity in the first argument.\n5. Linearity in the first argument also gives $\\ip{\\sum_{l=1}^{k} a_l e_{i_l}}{e_{i_m}} = \\sum_{l=1}^{k} a_l \\ip{e_{i_l}}{e_{i_m}}$.\n6. Since the indices are distinct, $\\ip{e_{i_l}}{e_{i_m}} = \\delta_{lm}$. So the sum in step 5 equals $a_m$.\n7. By steps 3, 4 and 5, this sum is also $0$. Hence $a_m = 0$.\n8. This holds for each $m$, so the only vanishing combination is the trivial one. Thus $(e_i)_{i \\in I}$ is linearly independent.\n\n## Idea\n\nThe inner product with $e_m$ picks out the coefficient of $e_m$ in a combination, because all the other vectors are orthogonal to $e_m$ and $e_m$ has length one. If the combination is zero, each picked coefficient is zero.\n",
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        "locator": "Corollary 6.25",
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