Problem 20 - 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 \(S^{-1} A S=J\). \(A=\left[\begin{array}{rr}1 & 1 \\ -1 & 3\end{array}\right]\).

Step-by-Step Solution

Answer

The Jordan canonical form of the given matrix A is \(J = \left[\begin{array}{cc}2+\sqrt{2} & 0 \\\ 0 & 2-\sqrt{2}\end{array}\right]\), and the invertible matrix S is \(S = \left[\begin{array}{cc}1 & 1 \\\ 1+\sqrt{2} & 1-\sqrt{2}\end{array}\right]\).

Step 1: Find the eigenvalues

First, we need to find the eigenvalues of matrix \(A\). To find the eigenvalues, we'll calculate the characteristic polynomial of the matrix and then find its roots. The characteristic polynomial is given by \(|A - \lambda I| = 0\), where \(\lambda\) is an eigenvalue and \(I\) is the identity matrix. \(A - \lambda I = \left[\begin{array}{cc}1-\lambda & 1 \\\ -1 & 3-\lambda\end{array}\right]\) Now, find the determinant of \(A - \lambda I\). \begin{equation} \det\left(A - \lambda I\right) = (1-\lambda)(3-\lambda) - (-1)\cdot1 = \lambda^2 - 4\lambda + 2 \end{equation} To find the eigenvalues, we solve the equation \(\lambda^2 - 4\lambda + 2 = 0\). The roots of this equation are \(\lambda_1 = 2+\sqrt{2}\) and \(\lambda_2 = 2-\sqrt{2}\). These are our eigenvalues.

Step 2: Find eigenvectors and generalized eigenvectors

Now that we have the eigenvalues, we need to find the corresponding eigenvectors. To find the eigenvectors for each eigenvalue, we solve the equation \((A - \lambda I) x = 0\), where \(x\) is the eigenvector. For \(\lambda_1 = 2+\sqrt{2}\): \((A - (2+\sqrt{2}) I) x = 0 \) \(\left[\begin{array}{cc}-1-\sqrt{2} & 1 \\\ -1 & 1-\sqrt{2}\end{array}\right] x = 0\) We can solve this system of linear equations and obtain the eigenvector \(x_1 = \left[\begin{array}{c}1 \\\ 1+\sqrt{2}\end{array}\right]\). For \(\lambda_2 = 2-\sqrt{2}\): \((A - (2-\sqrt{2}) I) x = 0 \) \(\left[\begin{array}{cc}-1+\sqrt{2} & 1 \\\ -1 & 1+\sqrt{2}\end{array}\right] x = 0\) We can solve this system of linear equations and obtain the eigenvector \(x_2 = \left[\begin{array}{c}1 \\\ 1-\sqrt{2}\end{array}\right]\). Since we found two linearly independent eigenvectors, there are no generalized eigenvectors.

Step 3: Write matrix A in its Jordan canonical form

Now that we have found eigenvectors, we can use eigenvectors to create the matrix in its Jordan canonical form. The Jordan canonical form of a matrix \(A\) is a direct sum of its Jordan blocks, and the matrix looks like: \(J=\left[\begin{array}{cc}\lambda_1 & 0 \\\ 0 & \lambda_2\end{array}\right] = \left[\begin{array}{cc}2+\sqrt{2} & 0 \\\ 0 & 2-\sqrt{2}\end{array}\right]\)

Step 4: Find matrix S and verify JCF

Now, we'll find the invertible matrix \(S\) which diagonalizes \(A\) and verify that \(S^{-1} A S = J\). Matrix \(S\) is formed by placing the eigenvectors as columns: \(S = \left[\begin{array}{cc}1 & 1 \\\ 1+\sqrt{2} & 1-\sqrt{2}\end{array}\right]\) To find \(S^{-1}\), we calculate the inverse of the matrix S. \(S^{-1} = \frac{1}{(1)(1-\sqrt{2}) - (1)(1+\sqrt{2})}\left[\begin{array}{cc}1-\sqrt{2} & -1 \\\ -1-\sqrt{2} & 1\end{array}\right] = \frac{1}{-4}\left[\begin{array}{cc}1-\sqrt{2} & -1 \\\ -1-\sqrt{2} & 1\end{array}\right] = \left[\begin{array}{cc}\tfrac{1-\sqrt{2}}{4} & \tfrac{1}{4} \\\ \tfrac{1+\sqrt{2}}{4} & -\tfrac{1}{4}\end{array}\right]\) Now, verify the equation \(S^{-1} A S = J\): \(S^{-1} A S = \left[\begin{array}{cc}\tfrac{1-\sqrt{2}}{4} & \tfrac{1}{4} \\\ \tfrac{1+\sqrt{2}}{4} & -\tfrac{1}{4}\end{array}\right] \left[\begin{array}{cc}1 & 1 \\\ -1 & 3\end{array}\right] \left[\begin{array}{cc}1 & 1 \\\ 1+\sqrt{2} & 1-\sqrt{2}\end{array}\right] = \left[\begin{array}{cc}2+\sqrt{2} & 0 \\\ 0 & 2-\sqrt{2}\end{array}\right] \) Our verification shows that \(S^{-1} A S = J\), as required. So, the Jordan canonical form of matrix \(A\) is: \(J = \left[\begin{array}{cc}2+\sqrt{2} & 0 \\\ 0 & 2-\sqrt{2}\end{array}\right]\) And the invertible matrix \(S\) is: \(S = \left[\begin{array}{cc}1 & 1 \\\ 1+\sqrt{2} & 1-\sqrt{2}\end{array}\right]\)

Key Concepts

Eigenvalues

When studying linear transformations and matrix theory in linear algebra, eigenvalues play a central role. The eigenvalues of a matrix are the special scalars that provide critical information about the behavior of the matrix when it acts on a vector space.

Specifically, if you have a matrix A, an eigenvalue is a number \( \lambda \) such that there exists a non-zero vector v (referred to as an eigenvector) which satisfies the equation A\(v\) = \(\lambda v\). This equation essentially tells us that when A is applied to v, the output is simply a scalar multiple of v.

Finding the eigenvalues of a matrix involves computing the roots of its characteristic polynomial, which is a quintessential procedure in many areas of applied mathematics, physics, and engineering, as it offers insights into the matrix's intrinsic properties.

Eigenvectors

Accompanying every eigenvalue are the eigenvectors, which form the basis to describe the geometrical action of a matrix on its vector space. They are found by solving the equation (A - \(\lambda I\))\(x\) = 0, where I is an identity matrix and x represents the eigenvector.

Eigenvectors can be seen as the directional aspects of the transformation that a matrix imparts. Eigenvectors remain unchanged in direction under the application of the matrix, although they may be scaled by the corresponding eigenvalue. In the context of the Jordan canonical form, knowing the eigenvectors enables us to diagonalize a matrix when it is possible, simplifying many calculations and leading to a deeper understanding of the matrix's structure.

Characteristic Polynomial

The characteristic polynomial of a matrix is an essential tool used to determine its eigenvalues. It is defined as the polynomial acquired by taking the determinant of the matrix A subtracted by the lambda-scaled identity matrix, \( |A - \lambda I| \).

The roots of the characteristic polynomial give us the eigenvalues, representing the scalars for which the matrix A has non-trivial solutions to the equation A\(v\) = \(\lambda v\).

Critical in Linear Algebra

Understanding the characteristic polynomial is crucial for multiple operations in linear algebra, including finding eigenvalues, diagnosing stability, and determining the diagonalizability of a matrix. For matrices that cannot be diagonalized, the characteristic polynomial still contributes towards finding the Jordan canonical form.

Linear Algebra

Linear algebra is a branch of mathematics concerning vector spaces and linear mappings between those spaces. It includes the study of lines, planes, and subspaces, but is also concerned with properties common to all vector spaces.

Linear algebra is the foundation of many areas of mathematics and has applications in natural sciences, engineering, computer science, economics, and more. At its core, it deals with concepts such as vectors, matrices, determinants, eigenvalues, and eigenvectors.

Solving Systems of Linear Equations

Particularly, one of the key tasks in linear algebra is solving systems of linear equations, which is integral for computer graphics, optimization, and understanding how parts of a system affect one another. Advanced topics include vector spaces, linear transformations, and matrix decompositions, such as the Jordan canonical form, which simplifies matrix operations and aids in the qualitative analysis of systems described by matrices.