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  "body": "\n## Proof\n\nWe write both inner products with the polarisation identity, which uses only norms.\n\n1. Let $V$ and $W$ be inner product spaces over $\\R$ or $\\C$, with [[inner products|0VGSYBWHV2]] written $\\ip{\\cdot}{\\cdot}$ and [[induced norms|0KG4D8AP6G]] $\\norm{\\cdot}$. Let $T \\colon V \\to W$ be a [[linear isometry|0717XXPWM6]], so $\\norm{T(x)} = \\norm{x}$ for all $x \\in V$.\n2. Let $u, v \\in V$. First let the scalars be real. By the [[polarisation identity|05FKHRT0B4]] in $W$, $\\ip{T(u)}{T(v)} = \\frac{1}{4} \\left( \\norm{T(u) + T(v)}^2 - \\norm{T(u) - T(v)}^2 \\right)$.\n3. By linearity, $T(u) + T(v) = T(u + v)$ and $T(u) - T(v) = T(u - v)$.\n4. Since $T$ keeps norms, $\\norm{T(u + v)} = \\norm{u + v}$ and $\\norm{T(u - v)} = \\norm{u - v}$.\n5. So $\\ip{T(u)}{T(v)} = \\frac{1}{4} \\left( \\norm{u + v}^2 - \\norm{u - v}^2 \\right) = \\ip{u}{v}$, by the polarisation identity in $V$.\n6. Now let the scalars be complex, with the inner products linear in the first argument. By the polarisation identity in $W$, $\\ip{T(u)}{T(v)} = \\frac{1}{4} \\sum_{k=0}^{3} i^k \\norm{T(u) + i^k T(v)}^2$.\n7. By linearity, $T(u) + i^k T(v) = T(u + i^k v)$ for each $k$. Since $T$ keeps norms, $\\norm{T(u + i^k v)} = \\norm{u + i^k v}$.\n8. So $\\ip{T(u)}{T(v)} = \\frac{1}{4} \\sum_{k=0}^{3} i^k \\norm{u + i^k v}^2 = \\ip{u}{v}$, by the polarisation identity in $V$.\n\n## Idea\n\nThe polarisation identity expresses an inner product through norms of a few combinations of the two vectors. A linear map sends these combinations to the same combinations of the images, and an isometry keeps their norms. So it keeps the inner product.\n",
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