Problem 7 - Advanced Engineering Mathematics

Question

In Problems 7-10, use the method of deflation to find the eigenvalues of the given matrix. $$ \left(\begin{array}{ll} 3 & 2 \\ 2 & 6 \end{array}\right) $$

Step-by-Step Solution

Answer

The eigenvalues are \( \lambda_1 = 7 \) and \( \lambda_2 = 2 \).

Step 1: Calculate the Characteristic Polynomial

To find the eigenvalues, we first need the characteristic equation. This is determined by calculating the determinant of the matrix \( A - \lambda I \), where \( A \) is your matrix and \( \lambda \) is a scalar (the eigenvalue we need to find). For matrix \( A \), \( \lambda I \) is \( \begin{pmatrix} 3-\lambda & 2 \ 2 & 6-\lambda \end{pmatrix} \). The characteristic polynomial \( p(\lambda) \) is obtained by calculating the determinant: \[\det(\begin{pmatrix} 3-\lambda & 2 \ 2 & 6-\lambda \end{pmatrix}) = (3-\lambda)(6-\lambda) - 2\cdot2 \]Simplifying this gives: \[(3-\lambda)(6-\lambda) - 4 = \lambda^2 - 9\lambda + 14.\]

Step 2: Solve the Characteristic Polynomial

Now, solve the quadratic characteristic polynomial \( \lambda^2 - 9\lambda + 14 = 0 \) to find \( \lambda \), the eigenvalues. Use the quadratic formula \( \lambda = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1 \), \( b = -9 \), and \( c = 14 \). Plug in these values:\[\lambda = \frac{-(-9) \pm \sqrt{(-9)^2 - 4\cdot1\cdot14}}{2\cdot1} = \frac{9 \pm \sqrt{81 - 56}}{2} = \frac{9 \pm \sqrt{25}}{2}.\]This simplifies to two solutions:\[\lambda_1 = \frac{9 + 5}{2} = 7, \quad \lambda_2 = \frac{9 - 5}{2} = 2.\]

Step 3: Confirm the Eigenvalues by Deflation

Once the eigenvalues are found, use deflation by factoring the characteristic polynomial. Since we found \( \lambda_1 = 7 \) and \( \lambda_2 = 2 \), the polynomial can be expressed as \( (\lambda - 7)(\lambda - 2) = 0 \).Factoring the polynomial in this manner confirms our solution, showing that the eigenvalues are consistent.

Key Concepts

Matrix Determinant

The matrix determinant is a crucial concept in linear algebra. It helps us understand the properties of a matrix, such as whether it's invertible or not. To determine the eigenvalues of a matrix, we often calculate the determinant of the matrix \( A - \lambda I \). Here, \( A \) is your matrix and \( \lambda \) is a scalar. For our exercise matrix, this resulted in a new matrix:

• \( \begin{pmatrix} 3-\lambda & 2 \ 2 & 6-\lambda \end{pmatrix} \)

The determinant of this matrix is calculated using the formula:

• \((a\cdot d - b\cdot c)\). For our matrix, the determinant becomes: \((3-\lambda)(6-\lambda) - 2\cdot2\).

Determinants are essential for finding eigenvalues because they provide the characteristic polynomial, which needs to be solved to find these values.

Characteristic Polynomial

The characteristic polynomial arises from the matrix determinant process and plays a vital role in finding eigenvalues.
Given matrix \( A \), the characteristic polynomial is computed as the determinant of \( A - \lambda I \). For our matrix example, the determinant led us to the characteristic polynomial:

• \( \lambda^2 - 9\lambda + 14 \)

This polynomial equation is essential because its roots are the eigenvalues of \( A \). Hence, solving the characteristic polynomial directly leads us to the eigenvalues.
It's a quadratic equation, indicating that there may be two eigenvalues, which we will find using the quadratic formula.

Quadratic Formula

To find the roots of the characteristic polynomial, we employ the quadratic formula. This formula offers a simple way to solve any quadratic equation in the form \( ax^2 + bx + c = 0 \).
For our specific polynomial \( \lambda^2 - 9\lambda + 14 = 0 \), the quadratic formula is applied as follows:

• \( \lambda = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)
• where \( a = 1 \), \( b = -9 \), \( c = 14 \)

Solving, we find:

• \( \lambda = \frac{9 \pm \sqrt{81 - 56}}{2} \)

This results in the eigenvalues \( \lambda_1 = 7 \) and \( \lambda_2 = 2 \). The quadratic formula helps simplify the process of solving second-degree polynomials, making it an indispensable tool in linear algebra.

Linear Algebra

Linear algebra is the backbone of understanding systems involving vectors and matrices. It covers a myriad of concepts, including eigenvalues and eigenvectors, which are critical for various applications in science and engineering.

• Understanding vectors and matrices
• Matrix operations
• Determinants and inverses

For example, in finding eigenvalues, linear algebra provides the tools and formulas needed, such as the characteristic polynomial and quadratic formula methods.
Linear algebra helps us discuss relationships like transformations, where matrices represent functions applied to vectors. It's foundational in fields like computer graphics, data science, and physics, explaining phenomena through mathematical models.

Deflation Method

The deflation method is a technique used to simplify the process of finding additional eigenvalues once one has been identified. It's particularly useful when dealing with polynomials of high degree, as it reduces the degree of the polynomial.
After identifying one or more eigenvalues using the characteristic polynomial, we can factor these out of the polynomial.

• For example, if an eigenvalue \( \lambda_1 = 7 \) is found, the characteristic polynomial \( \lambda^2 - 9\lambda + 14 \) can be factored to \( (\lambda - 7)(\lambda - 2) = 0 \).

This process confirms the solutions and makes it easier to handle the matrix with reduced complexity. The deflation method effectively simplifies the problem into more manageable parts, aiding in understanding the solution structure better.