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Let \(n\) be a finite type and let \(Q \in \mathrm{O}(n)\) be an orthogonal real matrix. Then for every vector \(x \in \mathbb {R}^n\),
\[ (Qx)\cdot (Qx) = x\cdot x. \]
Proof
From \(Q^\top Q = I\) one rewrites \((Qx)\cdot (Qx) = x\cdot (Q^\top Q x) = x\cdot x\).