The Parallelogram Identity for Inner Product Spaces

We will now look at an important theorem. If $V$ is an inner product space and $u, v \in V$ then the sum of squares of the norms of the vectors $u + v$ and $u - v$ equals twice the sum of the squares of the norms of the vectors $u$ and $v$, that is:

(1)

\begin{align} \quad \| u + v \|^2 + \| u - v \|^2 = 2 \| u \|^2 + 2 \| v \|^2 \end{align}

This important identity is known as the Parallelogram Identity, and has a nice geometric interpretation is we're working on the vector space $\mathbb{R}^2$:

We will now prove this theorem.

Theorem 1 (The Parallelogram Identity): Let $V$ be an inner product space. If $u, v \in V$ then $\| u + v \|^2 + \| u - v \|^2 = 2\| u \|^2 + 2 \| v \|^2$.

• Proof: Let $V$ be an inner product space and let $u, v \in V$. Noting that $\overline{-1} = -1$ and we have that:

(2)

\begin{align} \quad \quad \| u + v \|^2 + \| u - v \| ^2 = <u + v, u + v> + <u - v, u - v> \\ \quad \quad \| u + v \|^2 + \| u - v \| ^2 = <u, u> + <u, v> + <v , u> + <v, v> + <u , u> + <u, -v> + <-v, u> + <-v, -v> \\ \quad \quad \| u + v \|^2 + \| u - v \| ^2 = 2 <u, u> + 2 (-1)(\overline{-1}<-v, -v> + <u, v> + <v, u> + \overline{-1}<u, v> -<v, u> \\ \quad \quad \| u + v \|^2 + \| u - v \| ^2 = 2 <u, u> + 2<v, v> + <u, v> + <v, u> -<u, v> -<v, u> \\ \quad \quad \| u + v \|^2 + \| u - v \| ^2 = 2 \| u \|^2 + 2 \| v \|^2 \quad \blacksquare \end{align}