Let A € M4(R) be such that A is not invertible. Which of the following cannot be the characteristic polynomial of A? Explain your answer. p(t) = 1¹-1², g(t)=t¹-1, r(t) = t4
Let A € M4(R) be such that A is not invertible. Which of the following cannot be the characteristic polynomial of A? Explain your answer. p(t) = 1¹-1², g(t)=t¹-1, r(t) = t4
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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![### Problem Statement
**Let \( A \in M_4(\mathbb{R}) \) be such that \( A \) is not invertible. Which of the following cannot be the characteristic polynomial of \( A \)? Explain your answer.**
\[ p(t) = t^4 - t^2 \]
\[ q(t) = t^4 - 1 \]
\[ r(t) = t^4 \]
### Explanation
There are three given polynomials:
1. \( p(t) = t^4 - t^2 \)
2. \( q(t) = t^4 - 1 \)
3. \( r(t) = t^4 \)
To determine which of these polynomials cannot be the characteristic polynomial of the matrix \( A \) when \( A \) is not invertible, we need to consider the properties of characteristic polynomials of non-invertible (singular) matrices.
For a matrix \( A \in M_4(\mathbb{R}) \) to be non-invertible, its determinant must be zero. This implies that 0 is an eigenvalue of \( A \). Hence, the characteristic polynomial of \( A \) must have \( t = 0 \) as a root.
- **For \( p(t) = t^4 - t^2 \):**
\[ p(t) = t^2(t^2 - 1) = t^2(t - 1)(t + 1) \]
This polynomial has roots \( t = 0, t = 1, t = -1 \). Therefore, 0 is a root, and \( p(t) \) can be a characteristic polynomial of \( A \).
- **For \( q(t) = t^4 - 1 \):**
\[ q(t) = (t^2 - 1)(t^2 + 1) = (t - 1)(t + 1)(t^2 + 1) \]
The roots are \( t = 1, t = -1, t = i, t = -i \). Since 0 is not a root, \( q(t) \) cannot be the characteristic polynomial of \( A \).
- **For \( r(t) = t^4 \):**
\[ r(t) = t^4 \]
This polynomial has a root at](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fe5f558a7-14fc-4024-84d6-4debb1adc6f6%2Fe9d1bcb6-1eb4-4796-9e8a-cb7a38d304bc%2F8rpfaho_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Problem Statement
**Let \( A \in M_4(\mathbb{R}) \) be such that \( A \) is not invertible. Which of the following cannot be the characteristic polynomial of \( A \)? Explain your answer.**
\[ p(t) = t^4 - t^2 \]
\[ q(t) = t^4 - 1 \]
\[ r(t) = t^4 \]
### Explanation
There are three given polynomials:
1. \( p(t) = t^4 - t^2 \)
2. \( q(t) = t^4 - 1 \)
3. \( r(t) = t^4 \)
To determine which of these polynomials cannot be the characteristic polynomial of the matrix \( A \) when \( A \) is not invertible, we need to consider the properties of characteristic polynomials of non-invertible (singular) matrices.
For a matrix \( A \in M_4(\mathbb{R}) \) to be non-invertible, its determinant must be zero. This implies that 0 is an eigenvalue of \( A \). Hence, the characteristic polynomial of \( A \) must have \( t = 0 \) as a root.
- **For \( p(t) = t^4 - t^2 \):**
\[ p(t) = t^2(t^2 - 1) = t^2(t - 1)(t + 1) \]
This polynomial has roots \( t = 0, t = 1, t = -1 \). Therefore, 0 is a root, and \( p(t) \) can be a characteristic polynomial of \( A \).
- **For \( q(t) = t^4 - 1 \):**
\[ q(t) = (t^2 - 1)(t^2 + 1) = (t - 1)(t + 1)(t^2 + 1) \]
The roots are \( t = 1, t = -1, t = i, t = -i \). Since 0 is not a root, \( q(t) \) cannot be the characteristic polynomial of \( A \).
- **For \( r(t) = t^4 \):**
\[ r(t) = t^4 \]
This polynomial has a root at
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