Square of Inner Product Norm of Sum

Theorem

Let $\struct {X, \innerprod \cdot \cdot}$ be an inner product space.

Let $\norm {\, \cdot \,}$ be the inner product norm.

Let $x, y \in X$.

Then, we have:

$\norm {x + y}^2 = \norm x^2 + 2 \map \Re {\innerprod x y} + \norm y^2$

$\ds \norm {x + y}^2$

$=$

$\ds \innerprod {x + y} {x + y}$

$\ds $

$=$

$\ds \innerprod x x + \innerprod y x + \innerprod x y + \innerprod y y$

$\ds $

$=$

$\ds \norm x^2 + \overline {\innerprod x y} + \innerprod x y + \norm y^2$

$\ds $

$=$

$\ds \norm x^2 + 2 \map \Re {\innerprod x y} + \norm y^2$

$\blacksquare$