Question

e\(^A\) denotes the exponential of a square matrix A. Suppose \( \lambda \) is an eigenvalue and \( \mathbf{v} \) is the corresponding eigen-vector of matrix A. Consider the following two statements: Statement 1: e\(^\lambda\) is an eigenvalue of e\(^A\). Statement 2: \( \mathbf{v} \) is an eigen-vector of e\(^A\). Which one of the following options is correct?

• A. Statement 1 is true and statement 2 is false.
• B. Statement 1 is false and statement 2 is true.
• C. Both the statements are correct.
• D. Both the statements are false.

Hint:

For a matrix exponential \( e^A \), if \( \lambda \) is an eigenvalue of \( A \), then \( e^\lambda \) is an eigenvalue of \( e^A \), and the eigenvectors remain the same.

Answer 1

The Correct Option is C

Step 1: Understanding the eigenvalue and eigenvector properties for matrix exponentials.
For a matrix \( A \) with eigenvalue \( \lambda \) and corresponding eigenvector \( \mathbf{v} \), the following properties hold:
- Statement 1: If \( A \mathbf{v} = \lambda \mathbf{v} \), then it can be shown that \( e^A \mathbf{v} = e^\lambda \mathbf{v} \). Thus, \( e^\lambda \) is indeed an eigenvalue of \( e^A \). Therefore, Statement 1 is true.
- Statement 2: Since \( \mathbf{v} \) is an eigenvector of \( A \), it is also an eigenvector of \( e^A \) with eigenvalue \( e^\lambda \), as shown in Statement 1. Therefore, Statement 2 is also true.
Step 2: Conclusion.
Both statements are correct, so the correct answer is (C).