## Isometries in the plane

### Task number: 4355

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 \( -1 \).

#### Answer

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).