Linear Isometry Preserves Inner Product

Theorem

Given an inner product space \(V\) where the inner product \(\langle \cdot, \cdot \rangle\) induces a norm \(\|\boldsymbol{x}\| = \sqrt{\langle \boldsymbol{x}, \boldsymbol{x}\rangle}\) which in turn induces a metric \(d(\boldsymbol{x}, \boldsymbol{y}) = \|\boldsymbol{x} - \boldsymbol{y}\|\), any linear isometry \(\tau : V \to V\) under this metric preserves the inner product, that is:

\[ \|\boldsymbol{x}\| = \|\tau(\boldsymbol{x})\| \quad \forall \boldsymbol{x} \in V \implies \langle \tau(\boldsymbol{x}), \tau(\boldsymbol{y}) \rangle = \langle \boldsymbol{x}, \boldsymbol{y} \rangle.\]

Proof

Note that the first line in this proof is one of the polarisation identities:

\[\begin{align*} \langle \boldsymbol{x}, \boldsymbol{y} \rangle &= \frac{1}{2}\left(\|\boldsymbol{x}\|^{2} + \|\boldsymbol{y}\|^{2} - \|\boldsymbol{x} - \boldsymbol{y}\|^{2}\right) \\ &= \frac{1}{2}\left(d(\boldsymbol{x}, \boldsymbol{0})^{2} + d(\boldsymbol{y}, \boldsymbol{0})^{2} - d(\boldsymbol{x}, \boldsymbol{y})^{2}\right) \\ &= \frac{1}{2}\left(d(\tau(\boldsymbol{x}), \boldsymbol{0})^{2} + d(\tau(\boldsymbol{y}), \boldsymbol{0})^{2} - d(\tau(\boldsymbol{x}), \tau(\boldsymbol{y}))^{2}\right) \\ &= \frac{1}{2}\left(\|\tau(\boldsymbol{x})\|^{2} + \|\tau(\boldsymbol{y})\|^{2} - \|\tau(\boldsymbol{x}) - \tau(\boldsymbol{y})\|^{2}\right) \\ &= \langle \tau(\boldsymbol{x}), \tau(\boldsymbol{y}) \rangle \end{align*}\]

noting that \(\tau(\boldsymbol{0}) = \boldsymbol{0}\) by linearity.