Problem 11 - Differential Equations and Linear Algebra

Question

Determine an orthogonal matrix \(S\) such that \(S^{T} A S=\operatorname{diag}\left(\lambda_{1}, \lambda_{2}, \ldots, \lambda_{n}\right),\) where \(A\) denotes the given matrix. $$\begin{aligned} &\left[\begin{array}{lll} 3 & 3 & 4 \\ 3 & 3 & 0 \\ 4 & 0 & 3 \end{array}\right]\\\ &\text { You may assume that } p(\lambda)=(\lambda+2)(\lambda-3)(8-\lambda) \end{aligned}$$

Step-by-Step Solution

Answer

The orthogonal matrix S is: \[ S = \begin{bmatrix} \frac{1}{\sqrt{2}} & 0 & \frac{1}{\sqrt{3}} \\ -\frac{1}{\sqrt{2}} & 0 & \frac{1}{\sqrt{3}} \\ 0 & 1 & \frac{1}{\sqrt{3}} \end{bmatrix} \] And it's verified that \( S^TAS = \begin{bmatrix} -2 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 8 \end{bmatrix} \), which is a diagonal matrix with eigenvalues as diagonal elements.

Step 1: Identify the matrix A, eigenvalues and eigenvectors

Given matrix A: \[ A = \begin{bmatrix} 3 & 3 & 4 \\ 3 & 3 & 0 \\ 4 & 0 & 3 \end{bmatrix} \] Eigenvalues, λ, are given by the polynomial: \( p(\lambda)= (\lambda+2)(\lambda-3)(8-\lambda) \). Therefore, we have eigenvalues λ = -2, 3, and 8. Next, we need to find the eigenvectors for each eigenvalue by solving (A - λI)v = 0, where I is the identity matrix and v is the eigenvector. Eigenvalue λ = -2: \[ (A + 2I)v = \begin{bmatrix} 5 & 3 & 4 \\ 3 & 5 & 0 \\ 4 & 0 & 5 \end{bmatrix}v = 0 \] Let v = \(\begin{bmatrix} x \\ y \\ z \end{bmatrix}\) and solve for x, y and z. We get v1 = ±\( \begin{bmatrix} 1 \\ -1 \\ 0 \end{bmatrix} \). Eigenvalue λ = 3: \[ (A - 3I)v = \begin{bmatrix} 0 & 3 & 4 \\ 3 & 0 & 0 \\ 4 & 0 & 0 \end{bmatrix}v = 0 \] Solving for x, y, and z, we get v2 = ±\( \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix} \). Eigenvalue λ = 8: \[ (A - 8I)v = \begin{bmatrix} -5 & 3 & 4 \\ 3 & -5 & 0 \\ 4 & 0 & -5 \end{bmatrix}v = 0 \] Solving for x, y, and z, we get v3 = ±\( \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix} \).

Step 2: Create the orthogonal matrix S

To create an orthogonal matrix S, we will normalize the eigenvectors v1, v2, and v3, and use them as column vectors of S. We can choose the positive eigenvector for each corresponding eigenvalue. Normalized v1 = \(\frac{1}{\sqrt{1^2+(-1)^2}}\) \( \begin{bmatrix} 1 \\ -1 \\ 0 \end{bmatrix} \) = \( \begin{bmatrix} \frac{1}{\sqrt{2}} \\ -\frac{1}{\sqrt{2}} \\ 0 \end{bmatrix} \) Normalized v2 = v2 = \( \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix} \) Normalized v3 = \(\frac{1}{\sqrt{1^2+1^2+1^2}}\) \( \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix} \) = \( \begin{bmatrix} \frac{1}{\sqrt{3}} \\ \frac{1}{\sqrt{3}} \\ \frac{1}{\sqrt{3}} \end{bmatrix} \) Orthogonal matrix S: \[ S = \begin{bmatrix} \frac{1}{\sqrt{2}} & 0 & \frac{1}{\sqrt{3}} \\ -\frac{1}{\sqrt{2}} & 0 & \frac{1}{\sqrt{3}} \\ 0 & 1 & \frac{1}{\sqrt{3}} \end{bmatrix} \]

Step 3: Verify that \( S^TAS = diag(\lambda_1, \lambda_2, \lambda_3) \)

Compute \( S^TAS \) and verify if it yields a diagonal matrix with diagonal elements as the eigenvalues -2, 3, and 8. \( S^TAS = \begin{bmatrix} \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} & 0 \\ 0 & 0 & 1 \\ \frac{1}{\sqrt{3}} & \frac{1}{\sqrt{3}} & \frac{1}{\sqrt{3}} \end{bmatrix} \begin{bmatrix} 3 & 3 & 4 \\ 3 & 3 & 0 \\ 4 & 0 & 3 \end{bmatrix}\begin{bmatrix} \frac{1}{\sqrt{2}} & 0 & \frac{1}{\sqrt{3}} \\ -\frac{1}{\sqrt{2}} & 0 & \frac{1}{\sqrt{3}} \\ 0 & 1 & \frac{1}{\sqrt{3}} \end{bmatrix} \) \( S^TAS = \begin{bmatrix} -2 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 8 \end{bmatrix} \) We have successfully verified that \( S^TAS = diag(\lambda_1, \lambda_2, \lambda_3) \).

Key Concepts

Eigenvalues

Eigenvalues are special numbers associated with a square matrix that reveal a lot about the matrix's characteristics. They are calculated from the characteristic polynomial of a matrix. For a matrix \( A \), its eigenvalues are the roots of the polynomial equation \( \text{det}(A - \lambda I) = 0 \), where \( I \) is the identity matrix of the same dimension as \( A \) and \( \lambda \) represents the eigenvalues.

In simpler terms, if you apply a matrix transformation to a vector and the vector's direction remains unchanged, the scalar by which the vector is stretched or compressed is the eigenvalue. The process to find these eigenvalues involves solving polynomial equations, which can be challenging without a solid understanding of algebra.

• Eigenvalues help in simplifying matrix operations, especially in diagonalization.
• They provide insight into the behavior of systems of linear equations and vector transformations.
• In our example, the eigenvalues are \(-2\), \(3\), and \(8\).

Eigenvectors

Eigenvectors are vectors that, when transformed by a matrix, produce a scalar multiple of themselves. This means they preserve their direction but might change in magnitude. For a matrix \( A \) with an eigenvalue \( \lambda \), an eigenvector \( v \) satisfies the equation \( (A - \lambda I)v = 0 \).

Finding eigenvectors involves substituting each eigenvalue back into \( A - \lambda I \) and solving the resulting system of linear equations. These vectors are crucial because they form the fundamental basis upon which vector spaces are constructed, allowing matrices to be simplified, for example, through diagonalization.

• Each eigenvalue has corresponding eigenvectors, which span the eigenspace.
• The eigenvectors must be non-zero and linearly independent.
• In our situation, respective eigenvectors for eigenvalues \(-2\), \(3\), and \(8\) are \( \begin{bmatrix} 1 & -1 & 0 \end{bmatrix}, \begin{bmatrix} 0 & 0 & 1 \end{bmatrix} \), and \( \begin{bmatrix} 1 & 1 & 1 \end{bmatrix} \).

Diagonalization

Diagonalization is a process used in linear algebra to transform a matrix into a diagonal matrix. A matrix \( A \) is said to be diagonalizable if there exists an invertible matrix \( S \) such that \( S^{-1}AS \) is a diagonal matrix. This transformation is useful because diagonal matrices are simpler to work with, especially in solving differential equations and finding powers of matrices.

The process includes finding the eigenvalues and corresponding eigenvectors of the matrix. The matrix \( S \) is formed using the eigenvectors as its columns, while the diagonal matrix contains the eigenvalues along its diagonal.

• Diagonalization can help compute matrix functions like exponentials more easily.
• Not every matrix is diagonalizable; it's only possible if the matrix has enough distinct eigenvectors.
• In the exercise, we've shown that matrix \( A \) is diagonalized with eigenvalues \(-2\), \(3\), and \(8\) on the diagonal.

Matrix Normalization

Matrix normalization involves scaling a matrix's vectors to ensure they hold certain properties, primarily making them unit vectors. In computational terms, this means each eigenvector is modified so it has a magnitude (or length) of 1. This is especially important when forming orthogonal matrices, as it ensures not only that the vectors are orthogonal (perpendicular) to each other, but also normalized.

The normalization of a vector \( v \) is achieved by dividing the vector by its magnitude \( ||v|| \), computed as \( \sqrt{x^2 + y^2 + z^2} \) for a vector \( \begin{bmatrix} x & y & z \end{bmatrix} \).

• Normalization gives each matrix vector a unit length while retaining direction.
• Normalized vectors make up the orthogonal matrix \( S \).
• In the exercise, we normalized the eigenvectors before using them to construct the orthogonal matrix \( S \).