Exercise 1894Topology › Topology of ℝⁿM1Added 3 Nov 2026

Isometries of $\mathbb{R}^n$

#isometry#Euclidean#rigid motion

Statement

Show that every isometry $f:\mathbb{R}^n\to\mathbb{R}^n$ is of the form $f(x)=Qx+b$ with $Q$ orthogonal.

Hint

Translate to assume $f(0)=0$ ; then $Q=f$ preserves the norm; use the polarization identity to show $Q$ is linear.

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Solution

Norm preservation. Since $f$ is an isometry and $f(0)=0$ , $\|f(x)\|=\|x\|$ .

Linear. Polarization identity: $\langle f(x),f(y)\rangle=\langle x,y\rangle$ . This linearises $f$ : for $\lambda\in\mathbb{R}$ , $f(\lambda x)=\lambda f(x)$ and $f(x+y)=f(x)+f(y)$ .

Orthogonal. $Q=f$ satisfies $Q^TQ=I$ .

Conclusion. Any isometry of $\mathbb{R}^n$ is an orthogonal transformation plus a translation.

Isometries of $\mathbb{R}^n$

Statement

Solution

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