9.3. Find the eigenvalues and eigenvectors of the matrix [-3 -4 6] A = -4 -3 6 -4 -4 7 If A is diagonalizable, find a matrix P and a diagonal matrix D such that P-¹AP = D.
9.3. Find the eigenvalues and eigenvectors of the matrix [-3 -4 6] A = -4 -3 6 -4 -4 7 If A is diagonalizable, find a matrix P and a diagonal matrix D such that P-¹AP = D.
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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can you help me with these two problems and can you do it step by step and I have given the question and answer
![### Eigenvalues and Eigenvectors
#### Example 9.3
For a given matrix:
- The first eigenvalue \( \lambda_1 \) is 1 with a multiplicity of 2. The corresponding eigenvectors are:
\[
\begin{bmatrix}
-1 \\
1 \\
0
\end{bmatrix}
\quad \text{and} \quad
\begin{bmatrix}
3 \\
0 \\
2
\end{bmatrix}
\]
- The second eigenvalue \( \lambda_2 \) is -1 with a multiplicity of 1. The corresponding eigenvector is:
\[
\begin{bmatrix}
1 \\
1 \\
1
\end{bmatrix}
\]
Given matrices \( P \) and \( D \):
\[
P =
\begin{bmatrix}
-1 & 3 & 1 \\
1 & 0 & 1 \\
0 & 2 & 1
\end{bmatrix}
, \quad
D =
\begin{bmatrix}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & -1
\end{bmatrix}
\]
#### Example 9.4
For another given matrix:
- The first eigenvalue \( \lambda_1 \) is -1 with a multiplicity of 2. The dimension of the eigenspace \( E_{\lambda_1} \) is 1. The corresponding eigenvector is:
\[
\begin{bmatrix}
0 \\
-1 \\
1
\end{bmatrix}
\]
- The second eigenvalue \( \lambda_2 \) is -3. The corresponding eigenvector is:
\[
\begin{bmatrix}
1 \\
2 \\
-2
\end{bmatrix}
\]
However, there are only 2 linearly independent eigenvectors, thus matrix \( A \) is not diagonalizable.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fc4c2b58d-254f-40fb-b05d-8ee51be5a6b4%2F4c48a0b5-9d65-455b-8e04-a4f4c65c0a77%2Fc5dq3y8_processed.png&w=3840&q=75)
Transcribed Image Text:### Eigenvalues and Eigenvectors
#### Example 9.3
For a given matrix:
- The first eigenvalue \( \lambda_1 \) is 1 with a multiplicity of 2. The corresponding eigenvectors are:
\[
\begin{bmatrix}
-1 \\
1 \\
0
\end{bmatrix}
\quad \text{and} \quad
\begin{bmatrix}
3 \\
0 \\
2
\end{bmatrix}
\]
- The second eigenvalue \( \lambda_2 \) is -1 with a multiplicity of 1. The corresponding eigenvector is:
\[
\begin{bmatrix}
1 \\
1 \\
1
\end{bmatrix}
\]
Given matrices \( P \) and \( D \):
\[
P =
\begin{bmatrix}
-1 & 3 & 1 \\
1 & 0 & 1 \\
0 & 2 & 1
\end{bmatrix}
, \quad
D =
\begin{bmatrix}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & -1
\end{bmatrix}
\]
#### Example 9.4
For another given matrix:
- The first eigenvalue \( \lambda_1 \) is -1 with a multiplicity of 2. The dimension of the eigenspace \( E_{\lambda_1} \) is 1. The corresponding eigenvector is:
\[
\begin{bmatrix}
0 \\
-1 \\
1
\end{bmatrix}
\]
- The second eigenvalue \( \lambda_2 \) is -3. The corresponding eigenvector is:
\[
\begin{bmatrix}
1 \\
2 \\
-2
\end{bmatrix}
\]
However, there are only 2 linearly independent eigenvectors, thus matrix \( A \) is not diagonalizable.
![### Linear Algebra Exercises: Eigenvalues and Eigenvectors
#### 9.3. Find the Eigenvalues and Eigenvectors
Consider the matrix:
\[
A = \begin{bmatrix}
-3 & -4 & 6 \\
-4 & -3 & 6 \\
-4 & -4 & 7
\end{bmatrix}
\]
**Task:**
- Find the eigenvalues and eigenvectors of the matrix \( A \).
- If \( A \) is diagonalizable, find a matrix \( P \) and a diagonal matrix \( D \) such that \( P^{-1}AP = D \).
#### 9.4. Find the Eigenvalues and Eigenvectors
Consider the matrix:
\[
A = \begin{bmatrix}
-3 & -4 & -4 \\
-4 & 3 & 4 \\
4 & -4 & -5
\end{bmatrix}
\]
**Task:**
- Find the eigenvalues and eigenvectors of the matrix \( A \).
- If \( A \) is diagonalizable, find a matrix \( P \) and a diagonal matrix \( D \) such that \( P^{-1}AP = D \).
### Instructions for Solution:
To solve these problems, follow these steps:
1. **Calculate the Eigenvalues:**
- Solve the characteristic equation \(\det(A - \lambda I) = 0\).
2. **Find the Eigenvectors:**
- For each eigenvalue \(\lambda\), solve the system \((A - \lambda I)x = 0\) to find the corresponding eigenvectors.
3. **Check Diagonalizability:**
- Verify if \( A \) is diagonalizable by ensuring that there are enough linearly independent eigenvectors to form the matrix \( P \).
4. **Construct Matrices \( P \) and \( D \):**
- If \( A \) is diagonalizable, construct \( P \) using the eigenvectors and \( D \) using the eigenvalues.
### Notes:
- Ensure to show all steps and calculations clearly.
- Eigenvalues are the roots of the characteristic polynomial.
- Eigenvectors are found by substituting the eigenvalues back into the equation \((A - \lambda I)x = 0\) and solving for \( x \).
This exercise helps to understand the concepts of eigenvalues, eigenvectors,](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fc4c2b58d-254f-40fb-b05d-8ee51be5a6b4%2F4c48a0b5-9d65-455b-8e04-a4f4c65c0a77%2F59ck5i_processed.png&w=3840&q=75)
Transcribed Image Text:### Linear Algebra Exercises: Eigenvalues and Eigenvectors
#### 9.3. Find the Eigenvalues and Eigenvectors
Consider the matrix:
\[
A = \begin{bmatrix}
-3 & -4 & 6 \\
-4 & -3 & 6 \\
-4 & -4 & 7
\end{bmatrix}
\]
**Task:**
- Find the eigenvalues and eigenvectors of the matrix \( A \).
- If \( A \) is diagonalizable, find a matrix \( P \) and a diagonal matrix \( D \) such that \( P^{-1}AP = D \).
#### 9.4. Find the Eigenvalues and Eigenvectors
Consider the matrix:
\[
A = \begin{bmatrix}
-3 & -4 & -4 \\
-4 & 3 & 4 \\
4 & -4 & -5
\end{bmatrix}
\]
**Task:**
- Find the eigenvalues and eigenvectors of the matrix \( A \).
- If \( A \) is diagonalizable, find a matrix \( P \) and a diagonal matrix \( D \) such that \( P^{-1}AP = D \).
### Instructions for Solution:
To solve these problems, follow these steps:
1. **Calculate the Eigenvalues:**
- Solve the characteristic equation \(\det(A - \lambda I) = 0\).
2. **Find the Eigenvectors:**
- For each eigenvalue \(\lambda\), solve the system \((A - \lambda I)x = 0\) to find the corresponding eigenvectors.
3. **Check Diagonalizability:**
- Verify if \( A \) is diagonalizable by ensuring that there are enough linearly independent eigenvectors to form the matrix \( P \).
4. **Construct Matrices \( P \) and \( D \):**
- If \( A \) is diagonalizable, construct \( P \) using the eigenvectors and \( D \) using the eigenvalues.
### Notes:
- Ensure to show all steps and calculations clearly.
- Eigenvalues are the roots of the characteristic polynomial.
- Eigenvectors are found by substituting the eigenvalues back into the equation \((A - \lambda I)x = 0\) and solving for \( x \).
This exercise helps to understand the concepts of eigenvalues, eigenvectors,
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