Polarization Identity/Real Vector Space

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Theorem

Let $\struct {V, \innerprod \cdot \cdot}$ be an inner product space over $\R$.

Let $\norm \cdot$ be the inner product norm for $V$.


Then we have:

$4 \innerprod x y = \norm {x + y}^2 - \norm {x - y}^2$

for all $x, y \in V$.


Proof

We have:

\(\ds \norm {x + y}^2 - \norm {x - y}^2\) \(=\) \(\ds \innerprod {x + y} {x + y} - \innerprod {x - y} {x - y}\) Definition of Inner Product Norm
\(\ds \) \(=\) \(\ds \paren {\innerprod x {x + y} + \innerprod y {x + y} } - \paren {\innerprod x {x - y} - \innerprod y {x - y} }\) since an inner product is linear in the first argument
\(\ds \) \(=\) \(\ds \paren {\innerprod {x + y} x + \innerprod {x + y} y} - \paren {\innerprod {x - y} x - \innerprod {x - y} y}\) since a real inner product is symmetric
\(\ds \) \(=\) \(\ds \paren {\innerprod x x + \innerprod y x + \innerprod x y + \innerprod y y} - \paren {\innerprod x x - \innerprod y x - \innerprod x y + \innerprod y y}\) using linearity in the first argument
\(\ds \) \(=\) \(\ds 2 \innerprod x y + 2 \innerprod y x\)
\(\ds \) \(=\) \(\ds 4 \innerprod x y\) since a real inner product is symmetric

$\blacksquare$


Sources