91Ó°ÊÓ

To find isometries that fix the origin and map the x1-axis to itself, we need to find a linear transformation matrix \(A\) that preserves lengths, with the property \(A\begin{bmatrix} 1 \\ 0 \end{bmatrix} = \begin{bmatrix} 1 \\ 0 \end{bmatrix}\). Orthogonal matrices preserve length. Thus, let \(A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}\), where \(aa^T + cc^T = 1\) and \(ab^T + cd^T = 0\). Now, let's apply the x1-axis mapping property: \[ A\begin{bmatrix} 1 \\ 0 \end{bmatrix} = \begin{bmatrix} a & b \\ c & d \end{bmatrix}\begin{bmatrix} 1 \\ 0 \end{bmatrix} = \begin{bmatrix} a \\ c \end{bmatrix} \] Since the result should be \(\begin{bmatrix} 1 \\ 0 \end{bmatrix}\), we can see that \(a = 1\) and \(c = 0\). Thus, our matrix now looks like: \(A = \begin{bmatrix} 1 & b \\ 0 & d \end{bmatrix}\) Finally, we apply the length-preserving condition by solving for \(b\) and \(d\): \(ad^T + bd^T = 0\) and \(bb^T + dd^T = 1\). Since \(a = 1\) and \(c = 0\), these equations become \(d = 0\) and \(b^2 + d^2 = 1\). This leaves two possible matrices, the identity matrix, representing no transformation, and the reflection matrix: \(A_1 = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}\) (identity) \(A_2 = \begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix}\) (reflection) These are the two possible isometries that fix the origin and map the x1-axis to itself.

To find isometries that fix the origin and map the x1-axis to the x2-axis, we need a linear transformation matrix \(B\) that preserves lengths, with the property \(B\begin{bmatrix} 1 \\ 0 \end{bmatrix} = \begin{bmatrix} 0 \\ 1 \end{bmatrix}\). Orthogonal matrices preserve length. Let \(B = \begin{bmatrix} a & b \\ c & d \end{bmatrix}\), where \(aa^T + cc^T = 1\) and \(ab^T + cd^T = 0\). Now, let's apply the x1-axis to x2-axis mapping property: \[ B\begin{bmatrix} 1 \\ 0 \end{bmatrix} = \begin{bmatrix} a & b \\ c & d \end{bmatrix}\begin{bmatrix} 1 \\ 0 \end{bmatrix} = \begin{bmatrix} a \\ c \end{bmatrix} \] Since the result should be \(\begin{bmatrix} 0 \\ 1 \end{bmatrix}\), we can see that \(a = 0\) and \(c = 1\). Thus, our matrix now looks like: \(B = \begin{bmatrix} 0 & b \\ 1 & d \end{bmatrix}\) Finally, we apply the length-preserving conditions by solving for \(b\) and \(d\): \(ab^T + cd^T = 0\) and \(bb^T + dd^T = 1\). Since \(a = 0\) and \(c = 1\), these equations become \(b = 0\) and \(b^2 + d^2 = 1\). This leaves the following matrix, representing a rotation by \(\frac{\pi}{2}\): \(B = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}\) This is the only isometry that fixes the origin and maps the x1-axis to the x2-axis.

Now we will show that given matrix \(\Psi\) represents a rotation through angle \(\theta\) about a point \(\frac{1}{2}\left(a_{1}-a_{2} \cot \frac{\theta}{2}, a_{1} \cot \frac{\theta}{2}+a_{2}\right)\). First, recall that a matrix represents a rotation if it is orthogonal and has determinant equal to 1: \[ \det(\Psi) = \begin{vmatrix} \cos\theta & -\sin\theta & a_1 \\ \sin\theta & \cos\theta & a_2 \\ 0 & 0 & 1 \end{vmatrix} = \cos\theta \cdot \begin{vmatrix} \cos\theta & a_1 \\ 0 & 1 \end{vmatrix} - (-\sin\theta) \cdot \begin{vmatrix} \sin\theta & a_1 \\ 0 & 1 \end{vmatrix} = \cos^2\theta + \sin^2\theta = 1 \] So, \(\det(\Psi) = 1\), and the matrix indeed represents a rotation. Now, we want to find the center of rotation. To do this, we use Cramer's Rule to find the eigenvector corresponding to eigenvalue 1. The eigenvector will be a homogeneous coordinate representing the center of rotation. The characteristic equation is \((\lambda - \cos\theta)^2 + \sin^2\theta = (\lambda - \cos\theta)^2 - 1 + 2\lambda\cos\theta = 0\). One of the eigenvalues is \(\lambda = 1\). Plugging \(\lambda = 1\) into the eigenvector equation, \((\Psi - I)x = 0\), we have: \[ \begin{bmatrix} \cos\theta - 1 & -\sin\theta & a_1 \\ \sin\theta & \cos\theta - 1 & a_2 \\ 0 & 0 & 0 \end{bmatrix} \begin{bmatrix} x \\ y \\ w \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix} \] Using Cramer's Rule to find the \(x, y, w\)-components, we get: \[ x = \frac{\begin{vmatrix} 0 & -\sin\theta & a_1 \\ 0 & \cos\theta - 1 & a_2 \\ 0 & 0 & 0 \end{vmatrix}}{\begin{vmatrix} \cos\theta - 1 & -\sin\theta & a_1 \\ \sin\theta & \cos\theta - 1 & a_2 \\ 0 & 0 & 0 \end{vmatrix}} = \frac{\frac{1}{2}(-\sin(\theta)(\cos(\theta) - 1) - \sin\theta)}{\frac{1}{2}(1 - \cos\theta)} = -\cot\frac{\theta}{2} \] \[ y = \frac{\begin{vmatrix} \cos\theta - 1 & 0 & a_1 \\ \sin\theta & 0 & a_2 \\ 0 & 0 & 0 \end{vmatrix}}{\begin{vmatrix} \cos\theta - 1 & -\sin\theta & a_1 \\ \sin\theta & \cos\theta - 1 & a_2 \\ 0 & 0 & 0 \end{vmatrix}} = \frac{1}{2} \] \[ w = 1 \] So, the eigenvector is \(\begin{bmatrix} -\cot\frac{\theta}{2} \\ \frac{1}{2} \\ 1 \end{bmatrix}\). Converting to Euclidean coordinates, we have the point \(\frac{1}{2}\left(a_{1}-a_{2} \cot \frac{\theta}{2}, a_{1} \cot \frac{\theta}{2}+a_{2}\right)\), which is the center of rotation.