91Ó°ÊÓ

A reflection about the x-axis in two-dimensional space is performed by changing the sign of the y-coordinate. The matrix for this transformation is \[ T = \begin{pmatrix} 1 & 0 \ 0 & -1 \end{pmatrix} \]

To determine the eigenvalues, solve the characteristic equation \[ \det(T - \lambda I) = 0 \]First, compute \(T - \lambda I\):\[ T - \lambda I = \begin{pmatrix} 1 & 0 \ 0 & -1 \end{pmatrix} - \begin{pmatrix} \lambda & 0 \ 0 & \lambda \end{pmatrix} = \begin{pmatrix} 1 - \lambda & 0 \ 0 & -1 - \lambda \end{pmatrix} \]Next, find the determinant:\[ \det(T - \lambda I) = (1 - \lambda)(-1 - \lambda) = 0 \]The characteristic equation simplifies to \[ (1 - \lambda)(-1 - \lambda) = 0 \]From this, the eigenvalues are \( \lambda_1 = 1 \) and \( \lambda_2 = -1 \).

To find the eigenvectors, solve the equation \[ (T - \lambda I) \mathbf{v} = 0 \]For \( \lambda_1 = 1 \): Solve\[ \begin{pmatrix} 1 - 1 & 0 \ 0 & -1 - 1 \end{pmatrix} \begin{pmatrix} x \ y \end{pmatrix} = \begin{pmatrix} 0 \ 0 \end{pmatrix} \]This results in\[ \begin{pmatrix} 0 & 0 \ 0 & -2 \end{pmatrix} \begin{pmatrix} x \ y \end{pmatrix} = \begin{pmatrix} 0 \ 0 \end{pmatrix} \]Here, x can be any real number while y must be 0, so the eigenvector for \( \lambda_1 = 1 \) is\[ \mathbf{v}_1 = \begin{pmatrix} x \ 0 \end{pmatrix} \]For \( \lambda_2 = -1 \): Solve\[ \begin{pmatrix} 1 + 1 & 0 \ 0 & -1 + 1 \end{pmatrix} \begin{pmatrix} x \ y \end{pmatrix} = \begin{pmatrix} 0 \ 0 \end{pmatrix} \]This results in\[ \begin{pmatrix} 2 & 0 \ 0 & 0 \end{pmatrix} \begin{pmatrix} x \ y \end{pmatrix} = \begin{pmatrix} 0 \ 0 \end{pmatrix} \]Here, x must be 0 while y can be any real number, so the eigenvector for \( \lambda_2 = -1 \) is\[ \mathbf{v}_2 = \begin{pmatrix} 0 \ y \end{pmatrix} \]