Question

Let \( M \) be an \( n \times n \) matrix with real entries such that \( M^3 = I \). Suppose that \( Mv \neq v \) for any non-zero vector \( v \). Then which of the following statements is/are TRUE?

• A. \( M \) has real eigenvalues
• B. \( M + M^{-1} \) has real eigenvalues
• C. \( n \) is divisible by 2
• D. \( n \) is divisible by 3

Hint:

For matrices where \( M^3 = I \), the eigenvalues must be cube roots of unity. The order of the matrix must be divisible by 3 for consistency in the eigenvalues.

Answer 1

The Correct Option is B, C

Step 1: Eigenvalues of \( M \).
Since \( M^3 = I \), the eigenvalues of \( M \) must be cube roots of unity. These eigenvalues are \( 1, \omega, \omega^2 \), where \( \omega = e^{2\pi i / 3} \) is a primitive cube root of unity. Therefore, \( M \) does not have real eigenvalues unless the size of the matrix is divisible by 3 (as there will be a mix of real and non-real eigenvalues).
Step 2: Analyzing the matrix \( M + M^{-1} \).
The eigenvalues of \( M + M^{-1} \) are the sums of the eigenvalues of \( M \) and \( M^{-1} \). Since \( M^{-1} = M^2 \) (because \( M^3 = I \)), the eigenvalues of \( M + M^{-1} \) are \( 1 + 1 = 2 \), \( \omega + \omega^2 = -1 \), and \( \omega^2 + \omega = -1 \), all of which are real.
Step 3: Analyzing divisibility of \( n \).
For \( M \) to have eigenvalues that are cube roots of unity, the order of the matrix \( n \) must be divisible by 3, since the number of eigenvalues must match the degree of the cube roots. Therefore, \( n \) is divisible by 3.
Step 4: Conclusion.
Thus, the correct answers are (B) and (C).