1) Suppose that A² = A and B is a diagonal matrix similar to A. Prove that B2 = B and that the entries on the diagonal of B are only 0 and 1.

Advanced Engineering Mathematics
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Author:Erwin Kreyszig
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Chapter2: Second-order Linear Odes
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### Problem Statement

1) **Matrix Similarity and Idempotence**  
   Suppose that \( A^2 = A \) and \( B \) is a diagonal matrix similar to \( A \). Prove that \( B^2 = B \) and that the entries on the diagonal of \( B \) are only 0 and 1.

### Explanation:

Let's break the problem into parts to understand the given conditions and the requirements:

1. \( A^2 = A \):
   - This condition means that \( A \) is an idempotent matrix, implying \( A \) multiplied by itself results in \( A \).

2. \( B \) is a diagonal matrix similar to \( A \):
   - Being similar to \( A \) means there exists an invertible matrix \( P \) such that \( B = P^{-1}AP \).
   - Since \( B \) is diagonal, it simplifies the proof as we are dealing with its eigenvalues.

#### Proof Steps:

1. **Proving \( B^2 = B \)**:
   - From the similarity transformation, \( B = P^{-1}AP \).
   - To find \( B^2 \), consider:
     \[
     B^2 = (P^{-1}AP)(P^{-1}AP) = P^{-1}A(PP^{-1})AP = P^{-1}A^2P
     \]
   - Given \( A^2 = A \), substitute \( A \) for \( A^2 \):
     \[
     B^2 = P^{-1}AP = B
     \]
   - Therefore, \( B \) is also idempotent, i.e., \( B^2 = B \).

2. **Entries on the Diagonal of \( B \)**:
   - Since \( B \) is a diagonal matrix, its entries are the eigenvalues of \( A \).
   - From \( A^2 = A \), it follows that for each eigenvalue \(\lambda\) of \( A \):
     \[
     \lambda^2 = \lambda \implies \lambda (\lambda - 1) = 0
     \]
   - Thus, \(\lambda\) must be either 0 or 1.
   - Hence, the entries on the diagonal of \( B \) must be only 0 or 1.

By following
Transcribed Image Text:### Problem Statement 1) **Matrix Similarity and Idempotence** Suppose that \( A^2 = A \) and \( B \) is a diagonal matrix similar to \( A \). Prove that \( B^2 = B \) and that the entries on the diagonal of \( B \) are only 0 and 1. ### Explanation: Let's break the problem into parts to understand the given conditions and the requirements: 1. \( A^2 = A \): - This condition means that \( A \) is an idempotent matrix, implying \( A \) multiplied by itself results in \( A \). 2. \( B \) is a diagonal matrix similar to \( A \): - Being similar to \( A \) means there exists an invertible matrix \( P \) such that \( B = P^{-1}AP \). - Since \( B \) is diagonal, it simplifies the proof as we are dealing with its eigenvalues. #### Proof Steps: 1. **Proving \( B^2 = B \)**: - From the similarity transformation, \( B = P^{-1}AP \). - To find \( B^2 \), consider: \[ B^2 = (P^{-1}AP)(P^{-1}AP) = P^{-1}A(PP^{-1})AP = P^{-1}A^2P \] - Given \( A^2 = A \), substitute \( A \) for \( A^2 \): \[ B^2 = P^{-1}AP = B \] - Therefore, \( B \) is also idempotent, i.e., \( B^2 = B \). 2. **Entries on the Diagonal of \( B \)**: - Since \( B \) is a diagonal matrix, its entries are the eigenvalues of \( A \). - From \( A^2 = A \), it follows that for each eigenvalue \(\lambda\) of \( A \): \[ \lambda^2 = \lambda \implies \lambda (\lambda - 1) = 0 \] - Thus, \(\lambda\) must be either 0 or 1. - Hence, the entries on the diagonal of \( B \) must be only 0 or 1. By following
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