---
tag: 6FE43HC13F
kind: proof
name: Proof of the parallelogram law
msc: [46C05]
proves: 026HPGYGS4
uses: [0QVNCJXRY3, 0NB0AWGJBW, 0MFJJV97RJ]
sources:
  - url: https://en.wikipedia.org/wiki/Parallelogram_law
    locator: The parallelogram law in inner product spaces
---

## Proof

Let $\mathbb{K}$ be the scalar field, $\R$ or $\C$. Write $\bar a$ for the conjugate of $a \in \mathbb{K}$, so $\bar a = a$ over $\R$.

1. By axioms 1 and 2 of the [[inner product|0NB0AWGJBW]], for $u, v, w \in V$ we get $\langle w, u + v \rangle = \overline{\langle u, w \rangle + \langle v, w \rangle}$. This equals $\langle w, u \rangle + \langle w, v \rangle$.
2. Likewise $\langle w, -v \rangle = \overline{-\langle v, w \rangle} = -\langle w, v \rangle$.
3. By the [[induced norm|0MFJJV97RJ]], axiom 1 and step 1, $\norm{x + y}^2 = \langle x + y, x + y \rangle$. This equals $\langle x, x \rangle + \langle x, y \rangle + \langle y, x \rangle + \langle y, y \rangle$.
4. By axiom 1 and steps 1 and 2, $\norm{x - y}^2 = \langle x, x \rangle - \langle x, y \rangle - \langle y, x \rangle + \langle y, y \rangle$.
5. Adding steps 3 and 4, the mixed terms cancel. So $\norm{x + y}^2 + \norm{x - y}^2 = 2 \langle x, x \rangle + 2 \langle y, y \rangle = 2 \norm{x}^2 + 2 \norm{y}^2$.

## Idea

Expand both squared norms as inner products. The mixed terms $\langle x, y \rangle + \langle y, x \rangle$ appear with a plus sign in one and a minus sign in the other. So they cancel in the sum.