Proposition 12.6.4 (Polarisation Identity (Real)).label Let $E$ be a vector space over $\real$, $\lambda: E \times E \to \real$ be a Hermitian form, and
\[q: E \to K \quad x \mapsto \lambda(z, z)\]
be its quadratic form, then for any $x, y \in E$,
\[\lambda(x, y) = \frac{1}{4}\braks{q(x + y) - q(x - y)}\]
Proof. For each $x, y \in E$, by (H3),
\begin{align*}q(x + y)&= \dpn{x, x}{E}+ 2 \dpn{x, y}{E}+ \dpn{y, y}{E}\\ q(x - y)&= \dpn{x, x}{E}- 2 \dpn{x, y}{E}+ \dpn{y, y}{E}\end{align*}
so
\[q(x + y) - q(x - y) = 4 \dpn{x, y}{E}\]
$\square$
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