---
tag: 794QRK467A
kind: proof
name: Proof of the Pythagorean theorem in an inner product space
msc: [46C05]
proves: 0BVTDM872V
uses: [0VGSYBWHV2, 0KG4D8AP6G, 0TS7VZJBWD]
sources:
  - url: https://linear.axler.net/LADR4e.pdf
    locator: Theorem 6.12
---

## 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|0VGSYBWHV2]] $\ip{\cdot}{\cdot}$. Let $x, y \in V$ with $\ip{x}{y} = 0$, which means that $x$ and $y$ are [[orthogonal|0TS7VZJBWD]].
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|0KG4D8AP6G]], $\norm{x + y}^2 = \ip{x + y}{x + y}$.
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 $\norm{x + y}^2 = \ip{x}{x} + \ip{y}{y} = \norm{x}^2 + \norm{y}^2$.

## Idea

The square of the norm of a sum expands like the square of a sum of numbers. There are two cross terms, $\ip{x}{y}$ and $\ip{y}{x}$. Orthogonality makes both cross terms zero, and only the two squares remain.