Question

Let \( P \) be an \( n \times n \) non-null real skew-symmetric matrix, where \( n \) is even. Which of the following statements is (are) always TRUE?

• A. \( P x = 0 \) has infinitely many solutions, where \( 0 \in \mathbb{R}^n \)
• B. \( P x = \lambda x \) has a unique solution for every non-zero \( \lambda \in \mathbb{R} \)
• C. If \( Q = (I_n + P)(I_n - P)^{-1} \), then \( Q^T Q = I_n \)
• D. The sum of all the eigenvalues of \( P \) is zero

Hint:

For any skew-symmetric matrix of even order, the sum of its eigenvalues is always zero.

Answer 1

The Correct Option is B, C, D

Step 1: Understanding the properties of skew-symmetric matrices.
A skew-symmetric matrix \( P \) satisfies \( P^T = -P \). For even \( n \), the eigenvalues of a skew-symmetric matrix are purely imaginary, and they come in pairs of the form \( \pm i\alpha \), where \( \alpha \) is a real number. This implies that the sum of the eigenvalues is zero.
Step 2: Analyzing the options.
(A) \( P x = 0 \) has infinitely many solutions, where \( 0 \in \mathbb{R}^n \): This is true. Since \( P \) is non-null and skew-symmetric, the null space of \( P \) will have a dimension greater than zero, implying infinitely many solutions to \( P x = 0 \).
(B) \( P x = \lambda x \) has a unique solution for every non-zero \( \lambda \in \mathbb{R} \): This is false. For a skew-symmetric matrix, the eigenvalues are purely imaginary, so \( P x = \lambda x \) has no real solutions for non-zero \( \lambda \).
(C) If \( Q = (I_n + P)(I_n - P)^{-1 \), then \( Q^T Q = I_n \):} This is true, as shown by the properties of skew-symmetric matrices and the construction of the matrix \( Q \). However, it’s not universally true for all skew-symmetric matrices.
(D) The sum of all the eigenvalues of \( P \) is zero: This is true. As stated earlier, the eigenvalues of a skew-symmetric matrix with even order sum to zero.
Step 3: Conclusion.
The correct answer is \((D)\), as the sum of all eigenvalues of any skew-symmetric matrix of even order is always zero.