Classify Euclidean isometries algebraically

Try

The map F(x,y)=(−y+2,x+4) rotates a displacement by a quarter turn and translates the result. It preserves all distances. Its fixed point solves x=−y+2 and y=x+4, giving (−1,3). A transformation can have a translation term in its formula while still being a rotation about a different center.

An isometry here is a map R²→R² preserving every pairwise distance. We will prove its affine orthogonal form. Surjectivity will follow from that form rather than being silently assumed at the start.

Predict

Predict whether preserving lengths from the origin alone suffices. It does not: one could change angular positions inconsistently. Pairwise distance preservation supplies the dot products needed to force one common rigid motion.

Predict whether determinant one suffices. The shear matrix with rows (1,1) and (0,1) has determinant one but sends (0,1) to a vector of length √2. It preserves oriented area, not the metric.

Experiment

Inspect the linear parts of three maps.

Matrix	Image of (0,1)	QᵀQ	Isometry?
diag(−1,1)	(0,1)	I	Yes
Rows (0,−1),(1,0)	(−1,0)	I	Yes
Rows (1,1),(0,1)	(1,1)	Not I	No

Compute the last product explicitly. Then add an arbitrary constant vector to each map: differences of images cancel that vector, so the answer in the final column does not change. This separates the role of translation from the role of the linear part.

Understand

Let F preserve distances, b=F(0), and G(x)=F(x)−b. Then G fixes zero and preserves distances. By polarization it preserves dot products. In particular G(e₁),G(e₂) are an orthonormal basis, where e₁=(1,0), e₂=(0,1).

For every x, G(x)·G(eᵢ)=x·eᵢ=xᵢ. Coordinates in an orthonormal basis determine a vector uniquely, so G(x)=x₁G(e₁)+x₂G(e₂). Let Q have these two image basis vectors as columns. Then G(x)=Qx and QᵀQ=I. Thus F(x)=Qx+b.

Conversely, any such map preserves squared distance because ||Q(x−y)||²=(x−y)ᵀQᵀQ(x−y)=||x−y||². Since Q is invertible, the map is also bijective. This completes both directions of the classification without presupposing linearity.

Apply

An orthogonal plane matrix has determinant ±1. The positive sign preserves orientation; the negative sign reverses it. Orientation is an additional distinction, not a reason to exclude reflections from isometries.

If Q is a nonidentity rotation by θ, fixed points satisfy (I−Q)x=b. Its determinant is 2−2cos θ>0, so there is exactly one fixed point. If Q=I, the map is a translation; it has every point fixed when b=0 and no fixed point otherwise. The negative-determinant case will lead to reflections and glide reflections.

For optional depth, the proof extends to Rⁿ by using n image basis vectors. It depends on the source and target having the same finite dimension.

Continue

Use the formula as a proof tool, not just a recognition rule: derive Q, check orthogonality, then solve for fixed points. Continue to reflections and classification to factor these transformations into elementary geometric operations.