Let 1 be an eigenvalue of the n x n matrix A. Let B = A- AI. Show that B is not an invertible matrix.
Let 1 be an eigenvalue of the n x n matrix A. Let B = A- AI. Show that B is not an invertible matrix.
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
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![### Eigenvalues and Invertibility of Matrices
Consider the \( n \times n \) matrix \( A \).
**Problem Statement:**
Let \( \lambda \) be an eigenvalue of the matrix \( A \).
Let \( B \) be defined as \( B = A - \lambda I \), where \( I \) is the identity matrix.
**Objective:**
Show that \( B \) is not an invertible matrix.
**Explanation:**
- **Eigenvalue Definition:**
An eigenvalue \( \lambda \) of a square matrix \( A \) is a scalar such that there exists a non-zero vector \( \mathbf{v} \) (called an eigenvector) that satisfies:
\[
A\mathbf{v} = \lambda\mathbf{v}.
\]
- **Invertibility Condition:**
A matrix \( B \) is invertible if there exists another matrix \( C \) such that:
\[
BC = CB = I.
\]
A matrix is not invertible (or singular) if its determinant is zero.
- **Matrix \( B \):**
Given \( B = A - \lambda I \), where \( I \) is the identity matrix.
- **Showing Non-Invertibility:**
To show that \( B \) is not invertible, we need to demonstrate that \( \det(B) = 0 \).
\[
\begin{align*}
\text{If } \lambda \text{ is an eigenvalue of } A, & \text{ then } A\mathbf{v} = \lambda\mathbf{v}. \\
\text{With } B = A - \lambda I, & \text{ we get } B\mathbf{v} = (A - \lambda I)\mathbf{v} = A\mathbf{v} - \lambda I\mathbf{v} = \lambda\mathbf{v} - \lambda\mathbf{v} = \mathbf{0}.
\end{align*}
\]
Thus, \( \mathbf{v} \) is in the null space of \( B \), indicating that \( B \) has a non-trivial null space and therefore is not invertible. This is because an invertible matrix must have a trivial null space (i.e., the only solution to \( B\mathbf{v} = \mathbf{0](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F15c5ec76-8bab-480b-9121-367c8408675d%2F23d0a3df-3848-4a72-9b3d-da9ce1993956%2F8anu14_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Eigenvalues and Invertibility of Matrices
Consider the \( n \times n \) matrix \( A \).
**Problem Statement:**
Let \( \lambda \) be an eigenvalue of the matrix \( A \).
Let \( B \) be defined as \( B = A - \lambda I \), where \( I \) is the identity matrix.
**Objective:**
Show that \( B \) is not an invertible matrix.
**Explanation:**
- **Eigenvalue Definition:**
An eigenvalue \( \lambda \) of a square matrix \( A \) is a scalar such that there exists a non-zero vector \( \mathbf{v} \) (called an eigenvector) that satisfies:
\[
A\mathbf{v} = \lambda\mathbf{v}.
\]
- **Invertibility Condition:**
A matrix \( B \) is invertible if there exists another matrix \( C \) such that:
\[
BC = CB = I.
\]
A matrix is not invertible (or singular) if its determinant is zero.
- **Matrix \( B \):**
Given \( B = A - \lambda I \), where \( I \) is the identity matrix.
- **Showing Non-Invertibility:**
To show that \( B \) is not invertible, we need to demonstrate that \( \det(B) = 0 \).
\[
\begin{align*}
\text{If } \lambda \text{ is an eigenvalue of } A, & \text{ then } A\mathbf{v} = \lambda\mathbf{v}. \\
\text{With } B = A - \lambda I, & \text{ we get } B\mathbf{v} = (A - \lambda I)\mathbf{v} = A\mathbf{v} - \lambda I\mathbf{v} = \lambda\mathbf{v} - \lambda\mathbf{v} = \mathbf{0}.
\end{align*}
\]
Thus, \( \mathbf{v} \) is in the null space of \( B \), indicating that \( B \) has a non-trivial null space and therefore is not invertible. This is because an invertible matrix must have a trivial null space (i.e., the only solution to \( B\mathbf{v} = \mathbf{0
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