Question

The eigenvalues of the matrix: \[ \begin{pmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{pmatrix} \] are ________

• A. \( \cos\theta \) and \( \sin\theta \)
• B. \( e^{i\theta} \) and \( e^{-i\theta} \), where \( i = \sqrt{-1} \)
• C. \( \cos\theta \) and \( -\sin\theta \)
• D. \( \cos^2\theta \) and \( \sin^2\theta \)

Hint:

The eigenvalues of a rotation matrix are always complex conjugates, representing a rotation by an angle \( \theta \).

Answer 1

The Correct Option is B

The given matrix is a rotation matrix, which represents a rotation by an angle \( \theta \). The general form of the 2x2 rotation matrix is: \[ \begin{pmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{pmatrix} \] Step 1: Eigenvalue equation.
To find the eigenvalues of this matrix, we solve the characteristic equation: \[ \det\left( \begin{pmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{pmatrix} - \lambda I \right) = 0 \] where \( I \) is the identity matrix and \( \lambda \) represents the eigenvalue. This leads to the equation: \[ \det\begin{pmatrix} \cos\theta - \lambda & -\sin\theta \\ \sin\theta & \cos\theta - \lambda \end{pmatrix} = 0 \] Step 2: Solving the determinant.
Expanding the determinant: \[ (\cos\theta - \lambda)^2 + \sin^2\theta = 0 \] Simplifying: \[ \lambda^2 - 2\lambda\cos\theta + 1 = 0 \] Step 3: Finding the eigenvalues.
Using the quadratic formula: \[ \lambda = \frac{2\cos\theta \pm \sqrt{(2\cos\theta)^2 - 4(1)(1)}}{2} = \cos\theta \pm i\sin\theta \] Thus, the eigenvalues are: \[ \lambda = e^{i\theta} \text{ and } e^{-i\theta} \] Conclusion: The correct answer is (B).