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  "body": "\n## Proof\n\nWe compute the inner product of $v$ with each basis vector.\n\n1. Let $V$ be a finite-dimensional inner product space over $\\mathbb{K} = \\R$ or $\\C$, with [[inner product|0VGSYBWHV2]] $\\ip{\\cdot}{\\cdot}$. Let $(e_1, \\dots, e_n)$ be an [[orthonormal basis|0FWJ0VW561]] of $V$, so $\\ip{e_j}{e_i} = \\delta_{ji}$.\n2. Let $v \\in V$. Since $(e_1, \\dots, e_n)$ is a [[basis|0VTQTPQGAV]], there are scalars $a_1, \\dots, a_n \\in \\mathbb{K}$ with $v = \\sum_{j=1}^{n} a_j e_j$.\n3. Fix $i$ with $1 \\le i \\le n$. Linearity in the first argument gives $\\ip{v}{e_i} = \\sum_{j=1}^{n} a_j \\ip{e_j}{e_i}$.\n4. By [[orthonormality|09VYHEF2VT]], every term with $j \\ne i$ is $0$, and the term with $j = i$ is $a_i$. So $\\ip{v}{e_i} = a_i$.\n5. Substituting step 4 into step 2 gives $v = \\sum_{i=1}^{n} \\ip{v}{e_i} e_i$.\n6. The coordinates of $v$ in a basis are unique, so they are the numbers $a_i = \\ip{v}{e_i}$.\n\n## Idea\n\nWrite $v$ in the basis with unknown coefficients. The inner product with $e_i$ kills every term except the one with $e_i$, because the basis vectors are orthogonal unit vectors. So this inner product is exactly the coefficient $a_i$.\n",
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        "locator": "Theorem 6.30",
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