Prove if is an eigenvalue of an invertible matrix A with v as a corresponding 1 eigenvector, then λ = 0 and is an 2 eigenvalue of A¹, again with v as a corresponding eigenvector.
Prove if is an eigenvalue of an invertible matrix A with v as a corresponding 1 eigenvector, then λ = 0 and is an 2 eigenvalue of A¹, again with v as a 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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![**Title: Eigenvalues and Inverse Matrices**
**Content:**
**Problem Statement:**
Prove if \( \lambda \) is an eigenvalue of an invertible matrix \( A \) with \( \mathbf{v} \) as a corresponding eigenvector, then \( \lambda \neq 0 \) and \( \frac{1}{\lambda} \) is an eigenvalue of \( A^{-1} \), again with \( \mathbf{v} \) as a corresponding eigenvector.
**Explanation:**
This proof involves understanding the relationship between a square matrix \( A \), its eigenvalues \( \lambda \), its eigenvectors \( \mathbf{v} \), and how these properties relate to the inverse of the matrix \( A^{-1} \).
**Key Concepts to Consider:**
1. **Eigenvalues and Eigenvectors:**
- For a square matrix \( A \), \( \lambda \) is an eigenvalue and \( \mathbf{v} \) is a corresponding eigenvector if:
\[
A\mathbf{v} = \lambda \mathbf{v}
\]
- This equation tells us how the matrix transforms its eigenvectors.
2. **Invertibility:**
- A matrix is invertible if and only if its determinant is non-zero, which implies that none of its eigenvalues are zero (\( \lambda \neq 0 \)).
3. **Inverse Matrix and Its Eigenvalues:**
- When a matrix \( A \) is invertible, the eigenvalues of \( A^{-1} \) are the reciprocals of the eigenvalues of \( A \).
**Proof Outline:**
1. **Given:** \( A\mathbf{v} = \lambda \mathbf{v} \)
2. **Show:**
- Since \( A \) is invertible, \( \lambda \neq 0 \).
- For \( A^{-1} \), apply both sides of the initial equation by \( A^{-1} \):
\[
A^{-1}A\mathbf{v} = A^{-1}(\lambda \mathbf{v})
\]
- This simplifies to:
\[
\mathbf{v} = \lambda A^{-1}\mathbf{v}
\]
- Rearranging gives:
\[
A^{-1}\mathbf{v} = \frac](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F9496fb66-15f0-4d2e-bc7e-ca3bb9b70577%2F90f35f68-dbb8-436d-b019-6100f203f6e3%2F6l27ggr_processed.png&w=3840&q=75)
Transcribed Image Text:**Title: Eigenvalues and Inverse Matrices**
**Content:**
**Problem Statement:**
Prove if \( \lambda \) is an eigenvalue of an invertible matrix \( A \) with \( \mathbf{v} \) as a corresponding eigenvector, then \( \lambda \neq 0 \) and \( \frac{1}{\lambda} \) is an eigenvalue of \( A^{-1} \), again with \( \mathbf{v} \) as a corresponding eigenvector.
**Explanation:**
This proof involves understanding the relationship between a square matrix \( A \), its eigenvalues \( \lambda \), its eigenvectors \( \mathbf{v} \), and how these properties relate to the inverse of the matrix \( A^{-1} \).
**Key Concepts to Consider:**
1. **Eigenvalues and Eigenvectors:**
- For a square matrix \( A \), \( \lambda \) is an eigenvalue and \( \mathbf{v} \) is a corresponding eigenvector if:
\[
A\mathbf{v} = \lambda \mathbf{v}
\]
- This equation tells us how the matrix transforms its eigenvectors.
2. **Invertibility:**
- A matrix is invertible if and only if its determinant is non-zero, which implies that none of its eigenvalues are zero (\( \lambda \neq 0 \)).
3. **Inverse Matrix and Its Eigenvalues:**
- When a matrix \( A \) is invertible, the eigenvalues of \( A^{-1} \) are the reciprocals of the eigenvalues of \( A \).
**Proof Outline:**
1. **Given:** \( A\mathbf{v} = \lambda \mathbf{v} \)
2. **Show:**
- Since \( A \) is invertible, \( \lambda \neq 0 \).
- For \( A^{-1} \), apply both sides of the initial equation by \( A^{-1} \):
\[
A^{-1}A\mathbf{v} = A^{-1}(\lambda \mathbf{v})
\]
- This simplifies to:
\[
\mathbf{v} = \lambda A^{-1}\mathbf{v}
\]
- Rearranging gives:
\[
A^{-1}\mathbf{v} = \frac
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