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  "body": "\n## Proof\n\nWe show one inclusion, and then compare dimensions.\n\n1. Let $V$ be a finite-dimensional inner product space over $\\R$ or $\\C$, with [[inner product|0VGSYBWHV2]] $\\ip{\\cdot}{\\cdot}$. Let $W$ be a subspace of $V$ and $W^\\perp$ its [[orthogonal complement|079D68S3AW]]. Set $n = \\dim V$.\n2. Let $w \\in W$ and $x \\in W^\\perp$. By conjugate symmetry, $\\ip{w}{x} = \\overline{\\ip{x}{w}} = \\overline{0} = 0$.\n3. So $w$ is orthogonal to every vector of $W^\\perp$, that is, $w \\in (W^\\perp)^\\perp$. Hence $W \\subseteq (W^\\perp)^\\perp$.\n4. The set $W^\\perp$ contains $0$ and is closed under linear combinations, by linearity in the first argument. So it is a subspace of $V$. By the [[dimension formula|0E2NYPVSP9]], $\\dim W^\\perp = n - \\dim W$.\n5. The same formula for the subspace $W^\\perp$ gives $\\dim (W^\\perp)^\\perp = n - \\dim W^\\perp = \\dim W$.\n6. By step 3, $W$ is a subspace of the [[finite-dimensional|0QW084460E]] space $(W^\\perp)^\\perp$, and by step 5 it has the same dimension.\n7. A [[subspace of full dimension is the whole space|0WGAXEK99E]]. Hence $W = (W^\\perp)^\\perp$.\n\n## Idea\n\nEach vector of $W$ is orthogonal to each vector of $W^\\perp$, so $W$ lies inside $(W^\\perp)^\\perp$. Taking the complement twice brings the dimension back to $\\dim W$. A subspace inside another one of the same finite dimension is equal to it.\n",
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