**Title: Proof of Singularity of Matrix $ A $ with Zero Eigenvalue****Objective:**Demonstrate that if 0 is an eigenvalue of matrix $ A $, then $ A $ is singular.**Introduction:**In linear algebra, understanding the properties of matrices through their eigenvalues is fundamental. One important property to explore is the relationship between eigenvalues and the singularity of a matrix. This section focuses on showing that if a matrix has zero as an eigenvalue, then the matrix must be singular.**Statement:**Show that if 0 is an eigenvalue of $ A $, then $ A $ is singular.**Proof:**1. **Eigenvalue and Eigenvector Definition:** By definition, if $ \lambda $ is an eigenvalue of matrix $ A $, there exists a non-zero vector $ \mathbf{v} $ such that: \[ A\mathbf{v} = \lambda \mathbf{v} \] In this context, let $ \lambda = 0 $.2. **Substitute $ \lambda $ with 0:** Since $ \lambda = 0 $ is an eigenvalue, we have: \[ A\mathbf{v} = 0 \mathbf{v} = \mathbf{0} \] Here, $ \mathbf{0} $ is the zero vector.3. **Implication of Relationship:** The equation $ A\mathbf{v} = \mathbf{0} $ implies that there is a non-zero vector $ \mathbf{v} $ that, when multiplied by $ A $, results in the zero vector.4. **Definition of a Singular Matrix:** A matrix $ A $ is singular if and only if it is not invertible. One condition for a matrix to be non-invertible is the existence of a non-zero vector $ \mathbf{v} $ such that $ A\mathbf{v} = \mathbf{0} $.5. **Conclusion:** Given that $ \mathbf{v} \neq \mathbf{0} $ and $ A\mathbf{v} = \mathbf{0} $, matrix $ A $ is not invertible. Therefore, $ A $ is singular.**Conclusion:**The proof confirms that if a matrix \(

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Show that if 0 is an eigenvector of A, then A is singular.

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

**Title: Proof of Singularity of Matrix $ A $ with Zero Eigenvalue**

**Objective:**
Demonstrate that if 0 is an eigenvalue of matrix $ A $, then $ A $ is singular.

**Introduction:**
In linear algebra, understanding the properties of matrices through their eigenvalues is fundamental. One important property to explore is the relationship between eigenvalues and the singularity of a matrix. This section focuses on showing that if a matrix has zero as an eigenvalue, then the matrix must be singular.

**Statement:**
Show that if 0 is an eigenvalue of $ A $, then $ A $ is singular.

**Proof:**

1. **Eigenvalue and Eigenvector Definition:**
By definition, if $ \lambda $ is an eigenvalue of matrix $ A $, there exists a non-zero vector $ \mathbf{v} $ such that:
\[
A\mathbf{v} = \lambda \mathbf{v}
\]
In this context, let $ \lambda = 0 $.

2. **Substitute $ \lambda $ with 0:**
Since $ \lambda = 0 $ is an eigenvalue, we have:
\[
A\mathbf{v} = 0 \mathbf{v} = \mathbf{0}
\]
Here, $ \mathbf{0} $ is the zero vector.

3. **Implication of Relationship:**
The equation $ A\mathbf{v} = \mathbf{0} $ implies that there is a non-zero vector $ \mathbf{v} $ that, when multiplied by $ A $, results in the zero vector.

4. **Definition of a Singular Matrix:**
A matrix $ A $ is singular if and only if it is not invertible. One condition for a matrix to be non-invertible is the existence of a non-zero vector $ \mathbf{v} $ such that $ A\mathbf{v} = \mathbf{0} $.

5. **Conclusion:**
Given that $ \mathbf{v} \neq \mathbf{0} $ and $ A\mathbf{v} = \mathbf{0} $, matrix $ A $ is not invertible. Therefore, $ A $ is singular.

**Conclusion:**
The proof confirms that if a matrix \(

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