Problem 28
Question
Find the Jordan canonical form \(J\) for the matrix \(A_{1}\) and determine an invertible matrix \(S\) such that \(S^{-1} A S=J\). \(A=\left[\begin{array}{rrrr}2 & -1 & 0 & 1 \\ 0 & 3 & -1 & 0 \\ 0 & 1 & 1 & 0 \\ 0 & -1 & 0 & 3\end{array}\right]\).
Step-by-Step Solution
Verified Answer
The Jordan canonical form of the given matrix A is:
\[J=\left[\begin{array}{cccc} 2 & 0 & 0 & 0 \\ 0 & 3 & 0 & 0 \\ 0 & 0 & 3 & 1 \\ 0 & 0 & 0 & 3 \end{array}\right]\]
and the invertible matrix S is:
\[S=\left[\begin{array}{cccc} 1 & 1 & 1 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 1 & 0 & 1 \\ 0 & 0 & 1 & 1 \end{array}\right]\]
1Step 1: Find Eigenvalues of Matrix A
To find the eigenvalues of a matrix A, compute the determinant of the matrix \(A-\lambda I\), where λ is a scalar and I is the identity matrix. Then find the roots of the characteristic equation \(\text{det}(A-\lambda I)=0\).
Matrix A - λI:
\[A-\lambda I=\left[\begin{array}{cccc} 2-\lambda & -1 & 0 & 1 \\ 0 & 3-\lambda & -1 & 0 \\ 0 & 1 & 1-\lambda & 0 \\ 0 & -1 & 0 & 3-\lambda\end{array}\right]\]
Computing the determinant, we get the characteristic equation:
\((2-\lambda)((3-\lambda)(1-\lambda)(3-\lambda)+1)=0\)
Solving this equation, we find the eigenvalues:
\(\lambda_1 = 2, \lambda_2 = 3\)
2Step 2: Find Eigenvectors and Generalized Eigenvectors
Now we will find the eigenvectors by solving the equation \((A-\lambda I)x = 0\) for each eigenvalue.
For \(\lambda_1=2\):
\[
(A-\lambda_1 I)x = (A-2I)x=\left[\begin{array}{cccc} 0 & -1 & 0 & 1 \\ 0 & 1 & -1 & 0 \\ 0 & 1 & -1 & 0 \\ 0 & -1 & 0 & 1\end{array}\right]\left[\begin{array}{c} x_1 \\ x_2 \\ x_3 \\ x_4 \end{array}\right] = 0
\]
Solving the system of equations, we get the eigenvector:
\(v_1 = \left[\begin{array}{c} 1 \\ 0 \\ 0 \\ 0\end{array}\right]\)
For \(\lambda_2=3\):
\[
(A-\lambda_2 I)x = (A-3I)x=\left[\begin{array}{cccc} -1 & -1 & 0 & 1 \\ 0 & 0 & -1 & 0 \\ 0 & 1 & -2 & 0 \\ 0 & -1 & 0 & 0\end{array}\right]\left[\begin{array}{c} x_1 \\ x_2 \\ x_3 \\ x_4 \end{array}\right] = 0
\]
Solving the system of equations, we get two eigenvectors and one generalized eigenvector:
\(v_2 = \left[\begin{array}{c} 1 \\ 1 \\ 1 \\ 0\end{array}\right]\),
\(v_3 = \left[\begin{array}{c} 1 \\ 0 \\ 0 \\ 1\end{array}\right]\), and
\(v_4 = \left[\begin{array}{c} 0 \\ 0 \\ 1 \\ 1\end{array}\right]\)
3Step 3: Construct Jordan Canonical Form and Invertible Matrix S
Finally, we construct the Jordan canonical form and matrix S using the eigenvalues from step 1 and eigenvectors and generalized eigenvector from step 2.
The Jordan canonical form J has diagonal elements equal to the eigenvalues of A and appropriate off-diagonal elements matching the eigenvectors and generalized eigenvectors:
\[J=\left[\begin{array}{cccc} 2 & 0 & 0 & 0 \\ 0 & 3 & 0 & 0 \\ 0 & 0 & 3 & 1 \\ 0 & 0 & 0 & 3 \end{array}\right]\]
Matrix S consists of eigenvectors and generalized eigenvector as columns:
\[S=\left[\begin{array}{cccc} 1 & 1 & 1 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 1 & 0 & 1 \\ 0 & 0 & 1 & 1 \end{array}\right]\]
Thus, we have found the Jordan canonical form J and invertible matrix S.
Key Concepts
EigenvaluesEigenvectorsInvertible Matrix
Eigenvalues
Eigenvalues are fundamental numbers associated with a square matrix that provide insightful information about its properties, especially in linear transformations.
To find the eigenvalues of a matrix, say matrix \(A\), one typically solves the equation \(\text{det}(A-\lambda I)=0\), where \(\lambda\) represents an eigenvalue and \(I\) is the identity matrix.
This determinant results in a characteristic polynomial, and its roots are the eigenvalues.
The nature of these eigenvalues can give an idea about the behavior of the matrix; for instance, they can inform us about the stability of a system represented by the matrix.
These values are then used to explore other aspects of the matrix, like finding eigenvectors.
To find the eigenvalues of a matrix, say matrix \(A\), one typically solves the equation \(\text{det}(A-\lambda I)=0\), where \(\lambda\) represents an eigenvalue and \(I\) is the identity matrix.
This determinant results in a characteristic polynomial, and its roots are the eigenvalues.
The nature of these eigenvalues can give an idea about the behavior of the matrix; for instance, they can inform us about the stability of a system represented by the matrix.
- If all eigenvalues have negative real parts, the system is stable.
- Eigenvalues can be real or complex numbers.
These values are then used to explore other aspects of the matrix, like finding eigenvectors.
Eigenvectors
Eigenvectors accompany eigenvalues and describe directions in which a linear transformation represented by a matrix scales but does not change direction.
For a given eigenvalue \(\lambda\), an eigenvector \(\mathbf{v}\) satisfies the equation \((A-\lambda I)\mathbf{v} = \mathbf{0}\).
In simpler terms, applying the matrix \(A\) to the vector \(\mathbf{v}\) equals multiplying \(\mathbf{v}\) by \(\lambda\).
These vectors are crucial because they form invariant lines under the linear transformation that the matrix represents.
Notably, sometimes we also find generalized eigenvectors, expanding the framework for matrices like our example.
For a given eigenvalue \(\lambda\), an eigenvector \(\mathbf{v}\) satisfies the equation \((A-\lambda I)\mathbf{v} = \mathbf{0}\).
In simpler terms, applying the matrix \(A\) to the vector \(\mathbf{v}\) equals multiplying \(\mathbf{v}\) by \(\lambda\).
These vectors are crucial because they form invariant lines under the linear transformation that the matrix represents.
- Often more than one eigenvector exists for a given eigenvalue.
- Eigenvectors are essential in dimension reduction techniques like Principal Component Analysis.
Notably, sometimes we also find generalized eigenvectors, expanding the framework for matrices like our example.
Invertible Matrix
An invertible matrix, also known as a non-singular or non-degenerate matrix, is a square matrix that has an inverse.
A matrix is invertible if and only if its determinant is not zero.
Such matrices play a key role in linear algebra because they enable solving systems of linear equations and transforming problems into more easily solvable forms.
In the context of finding the Jordan canonical form, an invertible matrix \(S\) is used such that \(S^{-1} A S = J\), where \(J\) is the Jordan canonical form.
This matrix transforms the original matrix \(A\) into its Jordan form, which simplifies its structure while maintaining its intrinsic properties.
Thus, demonstrating the importance of understanding both the concept of invertibility and its application in canonical form transformations.
A matrix is invertible if and only if its determinant is not zero.
Such matrices play a key role in linear algebra because they enable solving systems of linear equations and transforming problems into more easily solvable forms.
In the context of finding the Jordan canonical form, an invertible matrix \(S\) is used such that \(S^{-1} A S = J\), where \(J\) is the Jordan canonical form.
- The invertibility ensures that the transformation is reversible and each step can be undone.
- Property of matrices crucial for matrix diagonalization, simplifying complex operations.
This matrix transforms the original matrix \(A\) into its Jordan form, which simplifies its structure while maintaining its intrinsic properties.
Thus, demonstrating the importance of understanding both the concept of invertibility and its application in canonical form transformations.
Other exercises in this chapter
Problem 27
Determine a basis for each eigenspace of \(A\) and sketch the eigenspaces. $$\begin{aligned} &A=\left[\begin{array}{rrr} 3 & 1 & -1 \\ 1 & 3 & -1 \\ -1 & -1 & 3
View solution Problem 28
We call a matrix \(B\) a square root of \(A\) if \(B^{2}=A\) (a) Show that if \(D=\operatorname{diag}\left(\lambda_{1}, \lambda_{2}, \ldots, \lambda_{n}\right),
View solution Problem 28
Determine all eigenvalues and corresponding eigenvectors of the given matrix. $$\left[\begin{array}{rrr}2 & -1 & 3 \\\3 & 1 & 0 \\\2 & -1 & 3\end{array}\right]$
View solution Problem 28
Determine a basis for each eigenspace of \(A\) and sketch the eigenspaces. $$A=\left[\begin{array}{rrr} -3 & 1 & 0 \\ -1 & -1 & 2 \\ 0 & 0 & -2 \end{array}\righ
View solution