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  "body": "\n## Proof\n\nWe use an orthonormal basis of $W$ to split each vector, and then show that $W \\cap W^\\perp = \\set{0}$.\n\n1. Let $V$ be an inner product space over $\\mathbb{K} = \\R$ or $\\C$, with [[inner product|0VGSYBWHV2]] $\\ip{\\cdot}{\\cdot}$. Let $W$ be a finite-dimensional [[subspace|0EPTJA5QV4]] of $V$, and let $W^\\perp$ be its [[orthogonal complement|079D68S3AW]].\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. $W^\\perp$ is a subspace. Indeed, $\\ip{0}{w} = 0$ for all $w$. If $x, y \\in W^\\perp$ and $a, b \\in \\mathbb{K}$, then $\\ip{a x + b y}{w} = a \\ip{x}{w} + b \\ip{y}{w} = 0$ for all $w \\in W$.\n4. The restriction of $\\ip{\\cdot}{\\cdot}$ to $W$ is an inner product on $W$. So $W$ is a finite-dimensional inner product space, and it has an [[orthonormal basis|0Y85GYEGSF]] $e_1, \\dots, e_k$.\n5. Let $v \\in V$. Set $w = \\sum_{i=1}^{k} \\ip{v}{e_i} e_i$, which lies in $W$, and $w' = v - w$. Then $v = w + w'$.\n6. For each $j$, linearity in the first argument and orthonormality give $\\ip{w}{e_j} = \\ip{v}{e_j}$. So $\\ip{w'}{e_j} = 0$.\n7. Let $u \\in W$. By the [[formula for coordinates|09DZ8CD1AK]] in $W$, $u = \\sum_j a_j e_j$ with $a_j \\in \\mathbb{K}$. By step 2, $\\ip{w'}{u} = \\sum_j \\overline{a_j} \\ip{w'}{e_j} = 0$.\n8. Hence $w' \\in W^\\perp$. So $V = W + W^\\perp$.\n9. Let $x \\in W \\cap W^\\perp$. Since $x \\in W^\\perp$ and $x \\in W$, we get $\\ip{x}{x} = 0$. By positivity of the inner product, $x = 0$.\n10. Let $u \\in W$ and $u' \\in W^\\perp$ with $u + u' = 0$. Then $u = -u'$ lies in $W$ and in the subspace $W^\\perp$. By step 9, $u = 0$, and so $u' = 0$.\n11. By steps 8 and 10 and the equivalent form of the definition of the [[direct sum|041JTZCK9Q]], $V = W \\oplus W^\\perp$.\n\n## Idea\n\nFor an orthonormal basis $e_1, \\dots, e_k$ of $W$, the vector $\\sum_i \\ip{v}{e_i} e_i$ is the part of $v$ in $W$. The remainder has zero inner product with each $e_i$, so it lies in $W^\\perp$. Uniqueness holds because a vector orthogonal to itself is zero.\n",
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