Problem 27 - Differential Equations and Linear Algebra

Question

Find the Jordan canonical form \(J\) for the matrix \(A_{1}\) and determine an invertible matrix \(S\) such that \(S^{-1} A S=J\). \(A=\left[\begin{array}{rrr}-1 & -1 & 0 \\ 0 & -1 & -2 \\ 0 & 0 & -1\end{array}\right]\).

Step-by-Step Solution

Answer

The Jordan canonical form of the given matrix A is \(J = \begin{pmatrix} -1 & 1 & 0\\ 0 & -1 & 1\\ 0 & 0 & -1 \end{pmatrix}\), and the invertible matrix S is \(S = \begin{pmatrix} 1 & 1 & 0\\ 1 & 0 & 0\\ 0 & -1 & -1 \end{pmatrix}\), such that \(S^{-1} A S=J\).

Step 1: Compute the characteristic polynomial

To find the eigenvalues, we compute the characteristic polynomial of A, given by \(\det(A - \lambda I)\), where \(I\) is the identity matrix and \(\lambda\) are the eigenvalues. In this case, let's compute the determinant: \(\det(A - \lambda I) = \det\begin{pmatrix} -1-\lambda & -1 & 0\\ 0 & -1-\lambda & -2\\ 0 & 0 & -1-\lambda \end{pmatrix}\) = \((-1-\lambda)[(-1-\lambda)^2 - 0]\) = \((-1-\lambda)^3\)

Step 2: Compute the eigenvalues

Since the characteristic polynomial is \((-1-\lambda)^3\), we have only one eigenvalue, \(\lambda = -1\). We will find the eigenvectors for this eigenvalue. #Step 2: Find eigenvectors and generalized eigenvectors#

Step 3: Find eigenvectors for \(\lambda=-1\)

To find eigenvectors corresponding to \(\lambda = -1\), we need to solve the equation, \((A - \lambda I)v = 0\), where \(v\) is the eigenvector. \(\begin{pmatrix} -1-(-1) & -1 & 0\\ 0 & -1-(-1) & -2\\ 0 & 0 & -1-(-1) \end{pmatrix} \begin{pmatrix} v_1\\ v_2\\ v_3 \end{pmatrix} = \begin{pmatrix} 0\\ 0\\ 0 \end{pmatrix}\) Solve for \(v\), we get a result \(v = \begin{pmatrix}1\\1\\0\end{pmatrix}\).

Step 4: Find a generalized eigenvector for \(\lambda=-1\)

Since we can only find one eigenvector, we need to find a generalized eigenvector by solving the equation, \((A - \lambda I)w = v\), where \(w\) is the generalized eigenvector. \(\begin{pmatrix} -1-(-1) & -1 & 0\\ 0 & -1-(-1) & -2\\ 0 & 0 & -1-(-1) \end{pmatrix} \begin{pmatrix} w_1\\ w_2\\ w_3 \end{pmatrix} = \begin{pmatrix} 1\\ 1\\ 0 \end{pmatrix}\) Solve for \(w\), we get a result \(w = \begin{pmatrix}1\\0\\-1\end{pmatrix}\). #Step 3: Find the Jordan canonical form J and the invertible matrix S#

Step 5: Form J and S using the eigenvectors and generalized eigenvectors

Using the eigenvector and generalized eigenvector, we form the Jordan canonical form J and the matrix S: \(J = \begin{pmatrix} -1 & 1 & 0\\ 0 & -1 & 1\\ 0 & 0 & -1 \end{pmatrix}\), \(S = \begin{pmatrix} 1 & 1 & 0\\ 1 & 0 & 0\\ 0 & -1 & -1 \end{pmatrix}\) Now, we have found the Jordan canonical form J and the invertible matrix S, such that \(S^{-1} A S=J\).

Key Concepts

Eigenvectors

Eigenvectors are fundamental in understanding linear transformations. When you multiply a matrix by one of its eigenvectors, the result is a scalar multiple of that eigenvector itself. This scalar is known as the eigenvalue. To find eigenvectors, we solve the equation \((A - \lambda I)v = 0\), where \(A\) is our matrix, \(\lambda\) is the eigenvalue, and \(v\) is an eigenvector.
It involves:

• Setting up the matrix equation \((A - \lambda I)\), which becomes a system of linear equations.
• Solving for the vector \(v\), where each component adds up to zero, leads us to the eigenvectors.

In our problem, with \(\lambda = -1\), solving this gives us the eigenvector \(v = \begin{pmatrix}1 \ 1 \ 0\end{pmatrix}\). This eigenvector indicates directions along which the linear transformation of matrix \(A\) scales points by the eigenvalue of \(-1\).

Generalized Eigenvectors

When a matrix has fewer independent eigenvectors than dimensions, we move to generalized eigenvectors for completing the basis of the transformation space. For our matrix \(A\), we only have one independent eigenvector. Thus, finding generalized eigenvectors is necessary.
To find generalized eigenvectors, we solve \((A - \lambda I)w = v\), where \(v\) is an existing eigenvector and \(w\) is our new goal, the generalized eigenvector.

• This involves using the earlier eigenvector equation as a right-hand side in a new linear system.
• Through this, we identify vectors that fill in the gaps left by regular eigenvectors.

Through this process, we found \(w = \begin{pmatrix} 1 \ 0 \ -1 \end{pmatrix}\) for \(\lambda = -1\). Instructively, generalized eigenvectors help form the Jordan canonical form.

Characteristic Polynomial

The characteristic polynomial is a vital algebraic tool used to find the eigenvalues of a matrix. It is calculated from \(\det(A - \lambda I)\), which must be zero for nontrivial solutions. This determinant generates a polynomial, with \(\lambda\) as a variable, whose roots give the eigenvalues.
In our exercise:

• The determinant and polynomial setup, \((A - \lambda I)\), yields \((-1 - \lambda)^3\).
• This shows our eigenvalue as \(\lambda = -1\), occurring with a multiplicity of 3.

The characteristic polynomial essentially serves as a bridge, letting us transition from matrix representation to understanding the scaling factors applicable to its systems of vectors.

Eigenvalues

Eigenvalues are specific scalars associated with a matrix that provide information on how linear transformations stretch or compress vectors. They are derived from the characteristic polynomial of a matrix. With matrix \(A\), substituting into \((A - \lambda I)\), we end with an equation like \((-1 - \lambda)^3 = 0\).
For our matrix:

• The repeated root implies a single eigenvalue, \(\lambda = -1\), with a multiplicity matching the dimension, showing a higher degree of algebraic multiplicity.
• As a further insight, each eigenvalue corresponds to a specific eigenvector or a set of generalized eigenvectors to manage multiple algebraic multiplicities.

Understanding eigenvalues allows us to comprehend the overall dynamics enacted by transformations on a vector space, such as rotations or reflections.