等距变换是（向上平移的）正交线性变换

Suppose that we have an isometry in $2$-dimensional coordinate space. We would like to show that this isometry may expressed as an orthogonal linear transformation, up to translation. We would also like to generalize to higher dimensions, if possible.

Let us denote $G=T\circ F$, where $G$ is our isometry, $T$ is a final translation that we will ignore for now, and $F$ is what we care about.

Since we are disregarding the translation, the origin obviously maps to itself. Thus, $F(0,0)=(0,0)$.

Now, let us take two other points $(1,0)$ and $(0,1)$.

Both of these points must be mapped to points on the unit circle, since isometries preserve distances.

Thus, let us say that $F(1,0)=(\cos\theta,\sin\theta)$.

Since isometries preserve angles, we know that the vector from the origin to $F(1,0)$ is perpendicular to the vector from the origin to $F(0,1)$. Thus, we have one of two possibilities: $$F(0,1)=\left(\cos(\theta+\frac{\pi}{2}),\sin(\theta+\frac{\pi}{2})\right)=(-\sin\theta,\cos\theta)$$ $$F(0,1)=\left(\cos(\theta-\frac{\pi}{2}),\sin(\theta-\frac{\pi}{2})\right)=(\sin\theta,-\cos\theta)$$

Since we have $F(0,0)$, $F(1,0)$, and $F(0,1)$, the isometry is actually completely determined!

In fact, it is easy to see that the isometry must be one of the following, depending on the value of $F(1,0)$: $$F(x,y)=(x\cos\theta+y\sin\theta,y\cos\theta-x\sin\theta)$$ $$F(x,y)=(x\cos\theta+y\sin\theta,x\sin\theta-y\cos\theta)$$ These representations leave no doubt that the isometry $F$ is indeed an orthogonal linear transformation.

Thus, we have proven that in $2$-dimensional space, an isometry is an orthogonal linear transformation, up to translation.

A similar but much more complicated process may be used in any number of dimensions. First observe that the origin maps to itself, and then try to calculate the value of the function for all of the principal unit vectors.