Question

Let \( \mathcal{O} = \{ P : P \text{ is a } 3 \times 3 \text{ real matrix satisfying } P^T P = I_3 \text{ and } \det(P) = 1 \}, \)
where \( I_3 \) denotes the identity matrix of order 3. Then which of the following options is/are correct?

• A. There exists a \( P \in \mathcal{O} \) with \( \lambda = \frac{1}{2} \) as an eigenvalue.
• B. There exists a \( P \in \mathcal{O} \) with \( \lambda = 2 \) as an eigenvalue.
• C. If \( \lambda \) is the only real eigenvalue of \( P \in \mathcal{O} \), then \( \lambda = 1 \).
• D. There exists a \( P \in \mathcal{O} \) with \( \lambda = -1 \) as an eigenvalue.

Hint:

When dealing with orthogonal matrices, remember that their eigenvalues must lie on the unit circle in the complex plane, and their determinant is the product of their eigenvalues.

Answer 1

The Correct Option is C, D

Step 1: Properties of orthogonal matrices. 
A matrix \( P \) is orthogonal if \( P^T P = I_3 \), meaning that \( P \) preserves the length of vectors. This implies that all eigenvalues of \( P \) lie on the unit circle in the complex plane, i.e., they have an absolute value of 1. Since \( \det(P) = 1 \), the product of the eigenvalues is 1. 
Step 2: Analyzing the options.
Option (A): A matrix \( P \in \mathcal{O} \) with \( \lambda = \frac{1}{2} \) as an eigenvalue is impossible because the eigenvalues must have an absolute value of 1.
Option (B): A matrix \( P \in \mathcal{O} \) with \( \lambda = 2 \) as an eigenvalue is also impossible for the same reason.
Option (C): If \( \lambda \) is the only real eigenvalue of \( P \in \mathcal{O} \), the other two eigenvalues must be complex conjugates, and the product of all three eigenvalues must be 1. Hence, \( \lambda = 1 \).
Option (D): Since \( \det(P) = 1 \) and \( P \) is orthogonal, it is possible for \( \lambda = -1 \) to be an eigenvalue (this would correspond to a reflection matrix). Thus, the correct answer is \( \boxed{(C)} \) and \( \boxed{(D)} \).