Question

Let \( P \) be a \( 2 \times 2 \) real matrix such that every non-zero vector in \( \mathbb{R}^2 \) is an eigenvector of \( P \). Suppose that \( \lambda_1 \) and \( \lambda_2 \) denote the eigenvalues of \( P \) and \( P \left[ \begin{array}{c} \sqrt{2} \\ \sqrt{3} \end{array} \right] = \left[ \begin{array}{c} 2 \\ t \end{array} \right] \) for some \( t \in \mathbb{R} \). Which of the following statements is (are) TRUE?

• A. \( \lambda_1 \neq \lambda_2 \)
• B. \( \lambda_1 \lambda_2 = 2 \)
• C. \( \sqrt{2} \) is an eigenvalue of \( P \)
• D. \( \sqrt{3} \) is an eigenvalue of \( P \)

Hint:

If every non-zero vector in \( \mathbb{R}^2 \) is an eigenvector of a matrix, the matrix must be a scalar multiple of the identity matrix.

Answer 1

The Correct Option is B, C

Step 1: Understanding the problem. 
The matrix \( P \) has the property that every non-zero vector is an eigenvector. This implies that \( P \) must be a scalar multiple of the identity matrix, i.e., \( P = \lambda I \) for some scalar \( \lambda \). This is because the only way for every vector to be an eigenvector of \( P \) is for \( P \) to act uniformly on all vectors, which happens when \( P \) is a scalar multiple of the identity matrix. 
Step 2: Analyzing the options. 
(A) \( \lambda_1 \neq \lambda_2 \): This is not necessarily true. In fact, both eigenvalues must be equal since \( P \) is a scalar multiple of the identity matrix. 
(B) \( \lambda_1 \lambda_2 = 2 \): This is true. Since \( P \) is a scalar matrix, the eigenvalues \( \lambda_1 \) and \( \lambda_2 \) are both equal to \( \lambda \), and hence \( \lambda_1 \lambda_2 = \lambda^2 \). The given equation suggests that \( \lambda^2 = 2 \), so \( \lambda_1 \lambda_2 = 2 \). 
(C) \( \sqrt{2} \) is an eigenvalue of \( P \): This is not true because \( P \) only has one eigenvalue, which is \( \lambda = \sqrt{2} \). 
(D) \( \sqrt{3} \) is an eigenvalue of \( P \): This is also false, as the eigenvalues are equal and must be \( \sqrt{2} \). 
Step 3: Conclusion. 
The correct answer is B, as the eigenvalues are equal and satisfy \( \lambda_1 \lambda_2 = 2 \).