Question

Consider an \( n \times n \) matrix \( A \) and a non-zero \( n \times 1 \) vector \( p \). Their product \( Ap = \alpha^2 p \), where \( \alpha \in \mathbb{R} \) and \( \alpha \notin \{-1, 0, 1\} \). Based on the given information, the eigenvalue of \( A^2 \) is:

• A. \( \alpha \)
• B. \( \alpha^2 \)
• C. \( \sqrt{\alpha} \)
• D. \( \alpha^4 \)

Hint:

When calculating the eigenvalue of \( A^2 \) for a matrix \( A \), apply the matrix transformation twice to the eigenvector and square the eigenvalue.

Answer 1

The Correct Option is D

We are given the equation: \[ Ap = \alpha^2 p. \] This means that \( p \) is an eigenvector of the matrix \( A \) with the eigenvalue \( \alpha^2 \). To find the eigenvalue of \( A^2 \), we apply the matrix \( A \) again to \( p \): \[ A^2 p = A (Ap) = A (\alpha^2 p) = \alpha^2 A p = \alpha^2 (\alpha^2 p) = \alpha^4 p. \] Thus, \( p \) is also an eigenvector of \( A^2 \) with the eigenvalue \( \alpha^4 \). Hence, the eigenvalue of \( A^2 \) is \( \alpha^4 \), corresponding to Option (D).
Final Answer: (D) \( \alpha^4 \)