For the following nilpotent matrix A, compute its Jordan canonical form. A: 1 3 -1 2 0 0 0 0 0 1 0 0 -0.5 -1 0.5 -1
For the following nilpotent matrix A, compute its Jordan canonical form. A: 1 3 -1 2 0 0 0 0 0 1 0 0 -0.5 -1 0.5 -1
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:**
For the following nilpotent matrix \( A \), compute its Jordan canonical form.
\[
A = \begin{pmatrix}
1 & 3 & -1 & 2 \\
0 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 \\
-0.5 & -1 & 0.5 & -1
\end{pmatrix}
\]
**Explanation and Methodology:**
To find the Jordan canonical form of a nilpotent matrix, we first need to determine its eigenvalues and corresponding Jordan blocks. For a nilpotent matrix, all eigenvalues are zero. The size and number of Jordan blocks are determined by the sizes of the chains of generalized eigenvectors.
**Steps to Find the Jordan Canonical Form:**
1. **Eigenvalues:** Since \( A \) is nilpotent, its eigenvalue is 0.
2. **Find the Kernel and Generalized Kernel:**
- \( \text{Ker}(A) \)
- \( \text{Im}(A) \)
- Higher powers of \( A \) may be used to find the chain of generalized eigenvectors.
3. **Construct Jordan Blocks:**
- Determine the sizes of Jordan blocks from the dimensions of kernels and generalized kernels of powers of \( A \).
The result will be a matrix \( J \) where the main diagonal has zeros, and each Jordan block is filled with 1's on the super-diagonal.
**Conclusion:**
After calculating, you'd find that the Jordan canonical form of matrix \( A \) is block diagonal, with each block corresponding to a chain of generalized eigenvectors.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F19556b7c-c5b6-4b4d-b523-18b6eba0d2d2%2F9449209f-e9ce-4481-9758-bdfd5c04d575%2Fgv6do6_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Problem Statement:**
For the following nilpotent matrix \( A \), compute its Jordan canonical form.
\[
A = \begin{pmatrix}
1 & 3 & -1 & 2 \\
0 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 \\
-0.5 & -1 & 0.5 & -1
\end{pmatrix}
\]
**Explanation and Methodology:**
To find the Jordan canonical form of a nilpotent matrix, we first need to determine its eigenvalues and corresponding Jordan blocks. For a nilpotent matrix, all eigenvalues are zero. The size and number of Jordan blocks are determined by the sizes of the chains of generalized eigenvectors.
**Steps to Find the Jordan Canonical Form:**
1. **Eigenvalues:** Since \( A \) is nilpotent, its eigenvalue is 0.
2. **Find the Kernel and Generalized Kernel:**
- \( \text{Ker}(A) \)
- \( \text{Im}(A) \)
- Higher powers of \( A \) may be used to find the chain of generalized eigenvectors.
3. **Construct Jordan Blocks:**
- Determine the sizes of Jordan blocks from the dimensions of kernels and generalized kernels of powers of \( A \).
The result will be a matrix \( J \) where the main diagonal has zeros, and each Jordan block is filled with 1's on the super-diagonal.
**Conclusion:**
After calculating, you'd find that the Jordan canonical form of matrix \( A \) is block diagonal, with each block corresponding to a chain of generalized eigenvectors.
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