Problem 48 - Differential Equations and Linear Algebra

Question

Use some form of technology to determine the eigenvalues and eigenvectors of \(A\) in the following manner: (1) Form the matrix \(A-\lambda I.\) (2) Solve the characteristic equation \(\operatorname{det}(A-\lambda I)=0\) to determine the eigenvalues of \(A.\) (3) For each eigenvalue \(\lambda_{i}\) found in \((2),\) solve the system \(\left(A-\lambda_{i} I\right) \mathbf{v}=\mathbf{0}\) to determine the eigenvectors of \(A.\) $$\diamond A=\left[\begin{array}{rrr}1 & 1 & 1 \\\3 & -1 & 2 \\\3 & 1 & 4\end{array}\right]$$

Step-by-Step Solution

Answer

The eigenvalues of the matrix A are \(\lambda_1 = 1\), \(\lambda_2 = 2\), and \(\lambda_3 = 3\). The corresponding eigenvectors are \(\mathbf{v}_1 = \begin{bmatrix}1 \\ -1 \\ 1 \end{bmatrix}\), \(\mathbf{v}_2 = \begin{bmatrix}1 \\ -2 \\ 1 \end{bmatrix}\), and \(\mathbf{v}_3 = \begin{bmatrix}-1 \\ 0 \\ 1 \end{bmatrix}\).

Step 1: Form the matrix A - λI

To find the matrix \(A - \lambda I\), we need to subtract \(\lambda\) from the diagonal elements of matrix A and get: $$ A-\lambda I = \left[\begin{array}{ccc}1-\lambda & 1 & 1 \\ 3 & -1-\lambda & 2 \\ 3 & 1 & 4-\lambda \end{array}\right] $$ This is the matrix we will work with to find eigenvalues and eigenvectors.

Step 2: Solve the characteristic equation

Now, we will find the determinant of the matrix \(A - \lambda I\) to obtain the characteristic equation: $$ \operatorname{det}(A-\lambda I) = \begin{vmatrix}1-\lambda & 1 & 1 \\ 3 & -1-\lambda & 2 \\ 3 & 1 & 4-\lambda \end{vmatrix} $$ Compute the determinant: $$ \begin{aligned} \operatorname{det}(A-\lambda I) &= (1-\lambda) \begin{vmatrix}-1-\lambda & 2 \\ 1 & 4-\lambda \end{vmatrix} - 1\begin{vmatrix}3 & 2 \\ 3 & 4-\lambda \end{vmatrix} + 1\begin{vmatrix}3 & -1-\lambda \\ 3 & 1 \end{vmatrix} \\ &= (1-\lambda)((-1-\lambda)(4-\lambda)-(2)(1)) - (3(4-\lambda)-2(3))+(3(1)-(-1-\lambda)(3)) \end{aligned} $$ When the above expression is simplified, we get: $$ \operatorname{det}(A-\lambda I) = -\lambda^3 + 4\lambda^2 - \lambda - 6 $$ Now, we need to find the roots of this polynomial equation to get the eigenvalues: $$ -\lambda^3 + 4\lambda^2 - \lambda - 6 = 0 $$ Using any technology or method to find the roots, we obtain the eigenvalues as \(\lambda_1 = 1\), \(\lambda_2 = 2\), and\(\lambda_3 = 3\).

Step 3: Find the eigenvectors

Now, we will find the eigenvectors corresponding to each eigenvalue. 1. For \(\lambda_1 = 1\): $$ (A-\lambda_1 I)\mathbf{v} = \left[\begin{array}{ccc}0 & 1 & 1 \\ 3 & -2 & 2 \\ 3 & 1 & 3 \end{array}\right] \mathbf{v} = \mathbf{0} $$ Row reducing this matrix to its row echelon form, we get: $$ \left[\begin{array}{ccc}1 & -\frac{1}{3} & \frac{1}{3} \\ 0 & 1 & 1 \\ 0 & 0 & 0 \end{array}\right] $$ From the row-reduced matrix, we have the free variable, \(v_3\), and the following system of linear equations: $$ \begin{cases} v_1 -\frac{1}{3}v_2 + \frac{1}{3}v_3 = 0 \\ v_2 + v_3 = 0 \end{cases} $$ Using this free variable, we get the eigenvector corresponding to \(\lambda_1 = 1\) as: $$ \mathbf{v}_1 = \begin{bmatrix}1 \\ -1 \\ 1 \end{bmatrix} $$ 2. For \(\lambda_2 = 2\): $$ (A-\lambda_2 I)\mathbf{v} = \left[\begin{array}{ccc}-1 & 1 & 1 \\ 3 & -3 & 2 \\ 3 & 1 & 2 \end{array}\right] \mathbf{v} = \mathbf{0} $$ Row reducing this matrix to its row echelon form, we get: $$ \left[\begin{array}{ccc}1 & 0 & -1 \\ 0 & 1 & 2 \\ 0 & 0 & 0 \end{array}\right] $$ From the row-reduced matrix, we have the free variable, \(v_3\), and the following system of linear equations: $$ \begin{cases} v_1 - v_3 = 0 \\ v_2 + 2v_3 = 0 \end{cases} $$ Using this free variable, we get the eigenvector corresponding to \(\lambda_2 = 2\) as: $$ \mathbf{v}_2 = \begin{bmatrix}1 \\ -2 \\ 1 \end{bmatrix} $$ 3. For \(\lambda_3 = 3\): $$ (A-\lambda_3 I)\mathbf{v} = \left[\begin{array}{ccc}-2 & 1 & 1 \\ 3 & -4 & 2 \\ 3 & 1 & 1 \end{array}\right] \mathbf{v} = \mathbf{0} $$ Row reducing this matrix to its row echelon form, we get: $$ \left[\begin{array}{ccc}1 & 0 & 1 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \end{array}\right] $$ From the row-reduced matrix, we have the free variable, \(v_3\), and the following system of linear equations: $$ \begin{cases} v_1 + v_3 = 0 \\ v_2 = 0 \end{cases} $$ Using this free variable, we get the eigenvector corresponding to \(\lambda_3 = 3\) as: $$ \mathbf{v}_3 = \begin{bmatrix}-1 \\ 0 \\ 1 \end{bmatrix} $$ The eigenvalues and their corresponding eigenvectors for matrix A are: Eigenvalue \(\lambda_1 = 1\): \(\mathbf{v}_1 = \begin{bmatrix}1 \\ -1 \\ 1 \end{bmatrix}\) Eigenvalue \(\lambda_2 = 2\): \(\mathbf{v}_2 = \begin{bmatrix}1 \\ -2 \\ 1 \end{bmatrix}\) Eigenvalue \(\lambda_3 = 3\): \(\mathbf{v}_3 = \begin{bmatrix}-1 \\ 0 \\ 1 \end{bmatrix}\)

Key Concepts

Matrix Algebra

Matrix algebra is like a mathematical puzzle where we deal with arrays of numbers, called matrices. These matrices can be added, subtracted, multiplied, and more, which helps solve many real-world and science problems. In the given exercise, we work with a matrix \(A\) and use matrix operations to find its eigenvalues and eigenvectors.

Here’s a simple breakdown of some key matrix operations:

• Adding or Subtracting Matrices: This involves adding or subtracting corresponding elements of matrices of the same size.
• Multiplying Matrices: This is a bit trickier. You take rows from the first matrix and columns from the second, multiply each element, and sum them up.
• Identity Matrix: Denoted as \(I\), it works like "1" in multiplication. When you multiply any matrix with \(I\), it remains unchanged.

In matrix algebra, the concept of finding \( A - \lambda I \) requires subtracting \( \lambda \), an unknown eigenvalue, from the diagonal elements of matrix \(A\). This operation is crucial in transforming the matrix into a form that helps us discover eigenvalues and eigenvectors.

Characteristic Equation

The characteristic equation is an essential part of computing eigenvalues, a key property of matrices in linear algebra. To derive this equation, we take the determinant of the matrix \( A - \lambda I \) and set it to zero.

• Determinant: This is a scalar value that summarizes many properties of a matrix. For a 2x2 matrix, it's calculated as \(ad - bc\) for \(\begin{bmatrix}a & b \ c & d \end{bmatrix}\). For larger matrices, like the 3x3 matrix in our exercise, the determinant requires more intricate calculations involving multiplying elements and subtracting products.

Let's take a look at how this works. The equation \( \operatorname{det}(A - \lambda I) = 0 \) is derived to find eigenvalues by solving a polynomial equation.For our example matrix:
\[\operatorname{det}(A - \lambda I) = -\lambda^3 + 4\lambda^2 - \lambda - 6 = 0\]

The coefficients of this polynomial and its roots (solutions for \( \lambda \)) lead us to the eigenvalues of the matrix. Solving this polynomial reveals the eigenvalues, which are specific \( \lambda \) values showing how the matrix behaves under certain conditions.

Row Echelon Form

The row echelon form (REF) is a simplified version of a matrix used to solve systems of linear equations. This form makes it easy to backtrack and find solutions using simple substitution methods.

To convert a matrix to REF, we use elementary row operations

• Row Swapping: Changing the position of two rows.
• Row Scaling: Multiplying all elements of a row by a non-zero number.
• Row Addition: Adding or subtracting multiples of one row from another.

In the exercise, for each eigenvalue, the matrix \( (A - \lambda I) \mathbf{v} = \mathbf{0} \) is row-reduced to find the eigenvectors. This row-reduced form helps identify free variables and simplifies the system of linear equations. Each row often zeroes out the elements below a leading coefficient, making it easier to solve the equations.

By simplifying matrices to REF, the system becomes clear, aiding in finding the solutions quickly and accurately. From this form, eigenvectors are extracted which highlight the direction in the matrix's transformation related to each eigenvalue.