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Let A be an eigenvalue of an n x n matrix A. Show that A² -2 is an eigenvalue of A² - 21₁.

Let A be an eigenvalue of an n x n matrix A. Show that A² -2 is an eigenvalue of A² - 21₁.

Advanced Engineering Mathematics
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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Linear Algebra

### Eigenvalues of Matrix Transformations **Problem Statement:** 2. Let \( \lambda \) be an eigenvalue of an \( n \times n \) matrix \( A \). Show that \( \lambda^2 - 2 \) is an eigenvalue of \( A^2 - 2I_n \). **Explanatory Solution:** Given the context of linear algebra and matrix theory: 1. **Eigenvalue Definition:** - If \( \lambda \) is an eigenvalue of matrix \( A \), there exists a non-zero vector \( \mathbf{v} \) such that \( A\mathbf{v} = \lambda\mathbf{v} \). 2. **Expression to Show:** - Given \( \lambda \) is an eigenvalue of \( A \), we want to show that \( \lambda^2 - 2 \) is an eigenvalue of \( A^2 - 2I_n \). **Steps to Solution:** 1. **Starting from the Eigenvalue Definition:** - \( A\mathbf{v} = \lambda\mathbf{v} \) 2. **Square Both Sides:** - \( A^2\mathbf{v} = A(A\mathbf{v}) = A(\lambda\mathbf{v}) = \lambda(A\mathbf{v}) = \lambda(\lambda\mathbf{v}) = \lambda^2\mathbf{v} \) This shows that \( \mathbf{v} \) is also an eigenvector of \( A^2 \) with eigenvalue \( \lambda^2 \). 3. **Transformation Involving Identity Matrix:** - Define \( B = A^2 - 2I_n \). 4. **Apply Transformation to Eigenvector:** - \( B\mathbf{v} = (A^2 - 2I_n)\mathbf{v} = A^2\mathbf{v} - 2I_n\mathbf{v} \). 5. **Substitute Known Values:** - \( B\mathbf{v} = \lambda^2\mathbf{v} - 2\mathbf{v} = (\lambda^2 - 2)\mathbf{v} \). 6. **Conclusion:** - Since \( B\mathbf{v} = (\lambda^2 - 2)\mathbf{v} \
Transcribed Image Text:### Eigenvalues of Matrix Transformations **Problem Statement:** 2. Let \( \lambda \) be an eigenvalue of an \( n \times n \) matrix \( A \). Show that \( \lambda^2 - 2 \) is an eigenvalue of \( A^2 - 2I_n \). **Explanatory Solution:** Given the context of linear algebra and matrix theory: 1. **Eigenvalue Definition:** - If \( \lambda \) is an eigenvalue of matrix \( A \), there exists a non-zero vector \( \mathbf{v} \) such that \( A\mathbf{v} = \lambda\mathbf{v} \). 2. **Expression to Show:** - Given \( \lambda \) is an eigenvalue of \( A \), we want to show that \( \lambda^2 - 2 \) is an eigenvalue of \( A^2 - 2I_n \). **Steps to Solution:** 1. **Starting from the Eigenvalue Definition:** - \( A\mathbf{v} = \lambda\mathbf{v} \) 2. **Square Both Sides:** - \( A^2\mathbf{v} = A(A\mathbf{v}) = A(\lambda\mathbf{v}) = \lambda(A\mathbf{v}) = \lambda(\lambda\mathbf{v}) = \lambda^2\mathbf{v} \) This shows that \( \mathbf{v} \) is also an eigenvector of \( A^2 \) with eigenvalue \( \lambda^2 \). 3. **Transformation Involving Identity Matrix:** - Define \( B = A^2 - 2I_n \). 4. **Apply Transformation to Eigenvector:** - \( B\mathbf{v} = (A^2 - 2I_n)\mathbf{v} = A^2\mathbf{v} - 2I_n\mathbf{v} \). 5. **Substitute Known Values:** - \( B\mathbf{v} = \lambda^2\mathbf{v} - 2\mathbf{v} = (\lambda^2 - 2)\mathbf{v} \). 6. **Conclusion:** - Since \( B\mathbf{v} = (\lambda^2 - 2)\mathbf{v} \
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