### Eigenvalues and Invertibility of MatricesConsider 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

Let A be an n×n matrix. The characteristic polynomial of A is the function f(λ) given by…

Answered: Let 1 be an eigenvalue of the n x n…

Math

Advanced Math

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

Section: Chapter Questions

Problem 1RQ

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Question

### 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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