Question

If \( X_1 = \begin{bmatrix} 1 \\ i \end{bmatrix} \), \( X_2 = \begin{bmatrix} i \\ 1 \end{bmatrix} \) are given vectors and \( A = \begin{bmatrix} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{bmatrix} \).
If \( P = [X_1, X_2] \), then \( P^{-1} A P \) is:

• A. \( \begin{bmatrix} e^{i\theta} & 0
  0 & e^{-i\theta} \end{bmatrix} \)
• B. \( \begin{bmatrix} e^{\theta} & 0
  0 & e^{\theta} \end{bmatrix} \)
• C. \( \begin{bmatrix} 1 & 1
  0 & 0 \end{bmatrix} \)
• D. \( \begin{bmatrix} 1 & 0
  0 & 1 \end{bmatrix} \)

Hint:

If a matrix has complex eigenvectors, you can still diagonalize it using the eigenvector matrix \( P \). For 2D rotation matrices, the eigenvalues are \( e^{i\theta} \) and \( e^{-i\theta} \), and the eigenvectors are complex-conjugate pairs.

Answer 1

The Correct Option is A

We are given: \[ A = \begin{bmatrix} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{bmatrix} \quad \text{and} \quad P = \begin{bmatrix} X_1 & X_2 \end{bmatrix} = \begin{bmatrix} 1 & i \\ i & 1 \end{bmatrix} \] We need to compute \( P^{-1} A P \), which is a similarity transformation.

Step 1: Interpret the matrix \( A \) Matrix \( A \) is a rotation matrix in 2D space. It represents a counterclockwise rotation by angle \( \theta \), and its eigenvalues are: \[ \lambda_1 = e^{i\theta}, \quad \lambda_2 = e^{-i\theta} \] So, if we can find a basis of eigenvectors for \( A \), we can diagonalize it. That is: \[ P^{-1} A P = D = \text{diag}(\lambda_1, \lambda_2) \]

Step 2: Confirm that \( X_1 \) and \( X_2 \) are eigenvectors Let us apply \( A \) to \(X_1 = \begin{bmatrix} 1  i \end{bmatrix}\): \(A X_1 = \begin{bmatrix} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{bmatrix}\)

\(\begin{bmatrix} 1  i \end{bmatrix}\) = \(\begin{bmatrix} \cos \theta + i \sin \theta \\ -\sin \theta + i \cos \theta \end{bmatrix}\) = \(e^{i\theta} \begin{bmatrix} 1 \\ i \end{bmatrix}\) = \(e^{i\theta} X_1\) So \( X_1 \) is an eigenvector with eigenvalue \( e^{i\theta} \). Similarly, apply \( A \) to \( X_2 = \begin{bmatrix} i \\ 1 \end{bmatrix} \): \[ A X_2 = \begin{bmatrix} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{bmatrix} \begin{bmatrix} i \\ 1 \end{bmatrix} = \begin{bmatrix} i \cos \theta + \sin \theta \\ -i \sin \theta + \cos \theta \end{bmatrix} = e^{-i\theta} \begin{bmatrix} i \\ 1 \end{bmatrix} = e^{-i\theta} X_2 \] Thus, \( X_2 \) is an eigenvector with eigenvalue \( e^{-i\theta} \).

Step 3: Diagonalizing the matrix The matrix \( P = [X_1 \; X_2] \) contains the eigenvectors of \( A \), so the similarity transformation: \[ P^{-1} A P = \begin{bmatrix} e^{i\theta} & 0 \\ 0 & e^{-i\theta} \end{bmatrix} \] This is the diagonal matrix with the eigenvalues of \( A \).