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

Given:

$M$ is an $n \times n$ matrix with $M^2 = 0$
$\alpha \in \mathbb{R} \setminus {0}$

Key property: Since $M^2 = 0$, all eigenvalues of $M$ are 0.

Step 1: Finding eigenvalues of $(M + \alpha I)$:

Let $\lambda$ be an eigenvalue of $M$ with eigenvector $v$: $$Mv = \lambda v = 0 \cdot v = 0$$

Then: $$(M + \alpha I)v = Mv + \alpha v = 0 + \alpha v = \alpha v$$

Therefore, $\alpha$ is an eigenvalue of $(M + \alpha I)$.

Since all eigenvalues of $M$ are 0, and $(M + \alpha I)$ shifts all eigenvalues by $\alpha$:

All eigenvalues of $(M + \alpha I)$ are $0 + \alpha = \alpha$

Step 2: Finding eigenvalues of $(M - \alpha I)$:

Similarly: $$(M - \alpha I)v = Mv - \alpha v = 0 - \alpha v = -\alpha v$$

Therefore, $-\alpha$ is an eigenvalue of $(M - \alpha I)$.

All eigenvalues of $(M - \alpha I)$ are $0 - \alpha = -\alpha$

Step 3: Finding eigenvalues of $(\alpha I - M)$:

$$(\alpha I - M)v = \alpha v - Mv = \alpha v - 0 = \alpha v$$

Therefore, $\alpha$ is an eigenvalue of $(\alpha I - M)$.

All eigenvalues of $(\alpha I - M)$ are $\alpha$

Step 4: Evaluating options:

(A) $(M + \alpha I)$ has eigenvalue $\alpha$, but $(M - \alpha I)$ has eigenvalue $-\alpha$, not $\alpha$

(B) $(M + \alpha I)$ has eigenvalue $\alpha$, and $(\alpha I - M)$ has eigenvalue $\alpha$

(C) $(M + \alpha I)$ has eigenvalue $\alpha$, not $-\alpha$

(D) $(M + \alpha I)$ has eigenvalue $\alpha$, not $-\alpha$

Answer: (B) $\alpha$ is the only eigenvalue of $(M + \alpha I)$ and $(\alpha I - M)$