Question

Let \( M \) be an \( n \times n \) ( \( n \ge 2 \) ) non-zero real matrix with \( M^2 = 0 \) and let \( \alpha \in \mathbb{R} \setminus \{0\}. \) Then

• A. \( \alpha \) is the only eigenvalue of \( (M + \alpha I) \) and \( (M - \alpha I) \)
• B. \( \alpha \) is the only eigenvalue of \( (M + \alpha I) \) and \( (\alpha I - M) \)
• C. \(-\alpha\) is the only eigenvalue of \( (M + \alpha I) \) and \( (M - \alpha I) \)
• D. \(-\alpha\) is the only eigenvalue of \( (M + \alpha I) \) and \( (I - \alpha M) \)

Hint:

Nilpotent matrices have all eigenvalues zero; adding or subtracting a scalar multiple of the identity shifts all eigenvalues by that scalar.

Answer 1

The Correct Option is B

Step 1: Given \( M^2 = 0 \).
This implies that all eigenvalues of \( M \) are zero.

Step 2: Find eigenvalues of \( (M + \alpha I). \)
If \( \lambda \) is an eigenvalue of \( M \), then eigenvalue of \( (M + \alpha I) \) is \( \lambda + \alpha. \) Since \( \lambda = 0 \), eigenvalue is \( \alpha. \)

Step 3: Similarly, for \( (M - \alpha I). \)
Eigenvalue \( = \lambda - \alpha = -\alpha. \) However, since \( M \neq 0 \) but nilpotent, its only eigenvalue is 0; so both \( M + \alpha I \) and \( M - \alpha I \) have single eigenvalues \( \alpha \) and \(-\alpha\) respectively. Hence, option (A) correctly captures \( \alpha \) as the only eigenvalue of both, considering consistent shift.

Final Answer: (A).