> [!theorem] > > Let $\cx$ be an [[Inner Product|inner product space]], and $\seqf{x_k} \subset \cx$ such that $x_j \perp x_k$ for all $j \ne k$, then > $ > \norm{\sum_{k = 1}^{n}x_k}^2 = \sum_{k = 1}^{n}\norm{x_k}^2 > $ > *Proof*. > $ > \begin{align*} > \norm{\sum_{k = 1}^{n}x_k}^2 &= \angles{\sum_{k = 1}^{n}x_k, \sum_{k = 1}^{n}x_k} \\ > &= \sum_{j = 1}^{n}\angles{x_j, \sum_{k = 1}^{n}x_k} \\ > &= \sum_{j = 1}^{n}\sum_{k = 1}^{n}\angles{x_j, x_k} \\ > &= \sum_{k = 1}^{n}\angles{x_k, x_k} \\ > &= \sum_{k = 1}^{n}\norm{x_k}^2 > \end{align*} > $