Isometries in the plane

For the following geometric transformations in the plane, determine their matrices with respect to the standard basis and decide if they are isometries:

a) rotation in the plane around the origin by an angle $$\alpha$$ counterclockwise

b) axial symmetry along the axis passing through the origin, making the angle $$\alpha$$ with the axis $$x$$ in the first quadrant.

Observe that isometries in Euclidean spaces can be divided into two groups. Try to describe them both geometrically and algebraically.

• Solution

a) The matrix of the map is formed by the images of the vectors of the standard basis, i.e.: $$[f]_{K,K}= \begin{pmatrix} \cos\alpha & -\sin\alpha \\ \sin\alpha & \cos\alpha \end{pmatrix}$$.

Since $$[f]_{K,K}^T[f]_{K,K} = \begin{pmatrix} \cos^2\alpha+\sin^2\alpha & -\cos\alpha\sin\alpha+\sin\alpha\cos\alpha \\ -\sin\alpha\cos\alpha+\cos\alpha\sin\alpha & (-\sin\alpha)^2+\cos^2\alpha \end{pmatrix} = \mathbf I_2$$, it is an isometry.

b) The matrix of the map is formed by the images of the vectors of the standard basis, i.e.: $$[f]_{K,K}= \begin{pmatrix} \cos2\alpha & \sin2\alpha \\ \sin2\alpha & -\cos2\alpha \end{pmatrix}$$.

Since $$[f]_{K,K}^T[f]_{K,K}= = \begin{pmatrix} \cos^22\alpha+\sin^22\alpha & \cos2\alpha\sin2\alpha-\sin2\alpha\cos2\alpha \\ \sin2\alpha\cos2\alpha-\cos2\alpha\sin2\alpha & \sin^22\alpha+(-\cos2\alpha)^2 \end{pmatrix} = \mathbf I_2$$, it is also an isometry.

As $$\mathbf I=[f]_{K,K}^T[f]_{K,K}$$, we get that $$1=\det(\mathbf I)=\det([f]_{K,K}^T)\det([f]_{K,K})=\det([f]_{K,K})^2$$. Thus each isometry in an Euclidean space necessarily has a square of the determinant of its matrix equal to one.

In our case, the first matrix has the determinant $$1$$, while the second $$-1$$.

Yes, both maps are isometries.

Isometries in Euclidean spaces can be divided into those with a positive sign of the matrix determinant (preserving orientation) and those with a negative sign (reversing orientation).