Let λ be an eigenvalue of invertible matrix A. Prove that exists and is an eigenvalue of A-¹. What does your proof also say about the corresponding eigenvector?
Let λ be an eigenvalue of invertible matrix A. Prove that exists and is an eigenvalue of A-¹. What does your proof also say about the corresponding eigenvector?
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Question
![**Question:**
Let \( \lambda \) be an eigenvalue of invertible matrix \( A \). Prove that \( \frac{1}{\lambda} \) exists and is an eigenvalue of \( A^{-1} \). What does your proof also say about the corresponding eigenvector?
**Explanation:**
This question involves concepts from linear algebra, specifically eigenvalues and eigenvectors.
1. **Definition Recall:**
- An eigenvalue \( \lambda \) of a matrix \( A \) means there exists a non-zero vector \( \mathbf{v} \) such that:
\[
A \mathbf{v} = \lambda \mathbf{v}
\]
- Matrix \( A \) is invertible, denoting that its determinant is non-zero.
2. **Proof Concept:**
- If \( A \mathbf{v} = \lambda \mathbf{v} \), then apply inverse \( A^{-1} \) to both sides:
\[
A^{-1} A \mathbf{v} = A^{-1} (\lambda \mathbf{v})
\]
- This simplifies to:
\[
\mathbf{v} = \lambda A^{-1} \mathbf{v}
\]
- Rearranging gives:
\[
A^{-1} \mathbf{v} = \frac{1}{\lambda} \mathbf{v}
\]
- Hence, \( \frac{1}{\lambda} \) is an eigenvalue of \( A^{-1} \).
3. **Implication for Eigenvector:**
- The eigenvector \( \mathbf{v} \) remains the same for both \( A \) and \( A^{-1} \).
This explanation can accompany an educational website’s text to help students understand the properties of eigenvalues and eigenvectors for invertible matrices.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F30f79184-3047-455a-b3d8-f2ad47623cdf%2Fb412e2c2-c321-41f9-93d7-0d93da83d496%2Fsnj76x2_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Question:**
Let \( \lambda \) be an eigenvalue of invertible matrix \( A \). Prove that \( \frac{1}{\lambda} \) exists and is an eigenvalue of \( A^{-1} \). What does your proof also say about the corresponding eigenvector?
**Explanation:**
This question involves concepts from linear algebra, specifically eigenvalues and eigenvectors.
1. **Definition Recall:**
- An eigenvalue \( \lambda \) of a matrix \( A \) means there exists a non-zero vector \( \mathbf{v} \) such that:
\[
A \mathbf{v} = \lambda \mathbf{v}
\]
- Matrix \( A \) is invertible, denoting that its determinant is non-zero.
2. **Proof Concept:**
- If \( A \mathbf{v} = \lambda \mathbf{v} \), then apply inverse \( A^{-1} \) to both sides:
\[
A^{-1} A \mathbf{v} = A^{-1} (\lambda \mathbf{v})
\]
- This simplifies to:
\[
\mathbf{v} = \lambda A^{-1} \mathbf{v}
\]
- Rearranging gives:
\[
A^{-1} \mathbf{v} = \frac{1}{\lambda} \mathbf{v}
\]
- Hence, \( \frac{1}{\lambda} \) is an eigenvalue of \( A^{-1} \).
3. **Implication for Eigenvector:**
- The eigenvector \( \mathbf{v} \) remains the same for both \( A \) and \( A^{-1} \).
This explanation can accompany an educational website’s text to help students understand the properties of eigenvalues and eigenvectors for invertible matrices.
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