The Pythagorean Identity for Inner Product Spaces

## The Pythagorean Identity for Inner Product Spaces

Recall from the Orthogonal and Orthonormal Sets in Inner Product Spaces page that if $H$ is an inner product space and $S \subseteq H$ then $S$ is said to be an orthogonal set if for all $x, y \in S$ with $x \neq y$ we have that:

(1)
\begin{align} \quad \langle x, y \rangle = 0 \end{align}

Furthermore, $S$ is said to be a orthonormal set if additionally for all $x \in S$ we have that:

(2)
\begin{align} \quad \| x \| = 1 \end{align}

Where $\| \cdot \|$ is the norm induced by the inner product ($\| x \| = \langle x, x \rangle^{1/2}$). We can now prove an important identity known as the Pythagorean identity for inner product spaces.

 Theorem 1 (The Pythagorean Identity for Inner Product Spaces): Let $H$ be an inner product space and let $\{ x_1, x_2, ..., x_n \}$ be an orthonormal subset of $H$. Then for all $c_1, c_2, ..., c_n \in \mathbb{C}$ we have that $\displaystyle{\biggr \| \sum_{k=1}^{n} c_kx_k \biggr \|^2 = \sum_{k=1}^{n} |c_k|^2}$.
• Proof: We have that:
(3)
\begin{align} \quad \biggr \| \sum_{k=1}^{n} c_kx_k \biggr \|^2 &= \left \langle \sum_{k=1}^{n} c_kx_k, \sum_{k=1}^{n} c_kx_k \right \rangle \\ &= \sum_{k=1}^{n} \sum_{j=1}^{n} c_k\overline{c_j} \langle x_k, x_j \rangle \\ \end{align}
• Since $\langle x_k, x_j \rangle = 0$ for every $k \neq j$, we have that the double sum above reduces to:
(4)
\begin{align} \quad \biggr \| \sum_{k=1}^{n} c_kx_k \biggr \|^2 &= \sum_{k=1}^{n} c_k \overline{c_k} \langle x_k, x_k \rangle \\ &= \sum_{k=1}^{n} |c_k|^2 \underbrace{\| x_k \|^2}_{=1} \\ &= \sum_{k=1}^{n} |c_k|^2 \quad \blacksquare \end{align}