# 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 =

__+____\\ \quad \quad \| u + v \|^2 + \| u – v \| ^2 =____+____+__ + + __+__

__+ + \\ \quad \quad \| u + v \|^2 + \| u – v \| ^2 = 2____+ 2 (-1)(\overline{-1} +____+__ + \overline{-1} __–__ \\ \quad \quad \| u + v \|^2 + \| u – v \| ^2 = 2

__+ 2__ +

__+__ –

__–__ \\ \quad \quad \| u + v \|^2 + \| u – v \| ^2 = 2 \| u \|^2 + 2 \| v \|^2 \quad \blacksquare \end{align}