Question

Suppose \( \lambda \) is an eigenvalue of matrix \( A \) and \( x \) is the corresponding eigenvector. Let \( x \) also be an eigenvector of the matrix \( B = A - 2I \), where \( I \) is the identity matrix. Then, the eigenvalue of \( B \) corresponding to the eigenvector \( x \) is equal to:

• A. \( \lambda \)
• B. \( \lambda + 2 \)
• C. \( 2\lambda \)
• D. \( \lambda - 2 \)

Hint:

When modifying a matrix by a scalar multiple of the identity matrix (such as \( A - 2I \)), the eigenvalues are adjusted by the scalar.

Answer 1

The Correct Option is D

Concept Explanation:
By definition, if \( \lambda \) is an eigenvalue of matrix \( A \) and \( x \) is the corresponding eigenvector, then the following equation holds: \[ A x = \lambda x. \] Now, let's consider the matrix \( B \) defined as: \[ B = A - 2I. \] Applying this transformation to the eigenvector \( x \): \[ B x = (A - 2I) x. \] Expanding the equation: \[ B x = A x - 2I x = \lambda x - 2x. \] Factoring out \( x \): \[ B x = (\lambda - 2) x. \] Conclusion:
From the above equation, we can see that \( x \) remains an eigenvector, and its corresponding eigenvalue for \( B \) is \( \lambda - 2 \).
Therefore, the eigenvalues of matrix \( B \) are simply the eigenvalues of \( A \), shifted by 2.
Correct Answer: Option (D) \( \lambda - 2 \). ✅
Key Takeaway:
When a scalar multiple of the identity matrix \( kI \) is added or subtracted from a matrix \( A \), its eigenvalues shift accordingly: \[ \text{Eigenvalues of } (A + kI) = \lambda + k. \] \[ \text{Eigenvalues of } (A - kI) = \lambda - k. \] However, the eigenvectors remain unchanged.
Example:
Suppose matrix \( A \) has eigenvalues \( \lambda_1 = 5 \) and \( \lambda_2 = 3 \).
If \( B = A - 2I \), then the new eigenvalues are: \[ \lambda_1 - 2 = 3, \quad \lambda_2 - 2 = 1. \]