Proof that a linear isometry preserves the inner product by polarisation

Proof

We write both inner products with the polarisation identity, which uses only norms.

1. Let $V$ and $W$ be inner product spaces over $\R$ or $\C$, with inner products written $\ip{\cdot}{\cdot}$ and induced norms $\norm{\cdot}$. Let $T \colon V \to W$ be a linear isometry, so $\norm{T(x)} = \norm{x}$ for all $x \in V$.
2. Let $u, v \in V$. First let the scalars be real. By the polarisation identity in $W$, $\ip{T(u)}{T(v)} = \frac{1}{4} \left( \norm{T(u) + T(v)}^2 - \norm{T(u) - T(v)}^2 \right)$.
3. By linearity, $T(u) + T(v) = T(u + v)$ and $T(u) - T(v) = T(u - v)$.
4. Since $T$ keeps norms, $\norm{T(u + v)} = \norm{u + v}$ and $\norm{T(u - v)} = \norm{u - v}$.
5. 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 $V$.
6. Now let the scalars be complex, with the inner products linear in the first argument. By the polarisation identity in $W$, $\ip{T(u)}{T(v)} = \frac{1}{4} \sum_{k=0}^{3} i^k \norm{T(u) + i^k T(v)}^2$.
7. By linearity, $T(u) + i^k T(v) = T(u + i^k v)$ for each $k$. Since $T$ keeps norms, $\norm{T(u + i^k v)} = \norm{u + i^k v}$.
8. 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 $V$.