Proof of the Pythagorean theorem in an inner product space

Proof

We expand the square of the norm with the rules of the inner product.

1. Let $V$ be an inner product space over $\mathbb{K} = \R$ or $\C$, with inner product $\ip{\cdot}{\cdot}$. Let $x, y \in V$ with $\ip{x}{y} = 0$, which means that $x$ and $y$ are orthogonal.
2. For all $u, z, w \in V$, conjugate symmetry and linearity in the first argument give $\ip{u}{z + w} = \overline{\ip{z}{u} + \ip{w}{u}} = \ip{u}{z} + \ip{u}{w}$.
3. By the definition of the induced norm, $\norm{x + y}^2 = \ip{x + y}{x + y}$.
4. Linearity in the first argument and step 2 give $\ip{x + y}{x + y} = \ip{x}{x} + \ip{x}{y} + \ip{y}{x} + \ip{y}{y}$.
5. By conjugate symmetry, $\ip{y}{x} = \overline{\ip{x}{y}} = \overline{0} = 0$. Also $\ip{x}{y} = 0$ by step 1.
6. So $\norm{x + y}^2 = \ip{x}{x} + \ip{y}{y} = \norm{x}^2 + \norm{y}^2$.