Problem 8 - Advanced Engineering Mathematics

Question

Use the method of deflation to find the eigenvalues of the given matrix. $$ \left(\begin{array}{ll} 1 & 3 \\ 3 & 9 \end{array}\right) $$

Step-by-Step Solution

Answer

Eigenvalues are \( \lambda = 0 \) and \( \lambda = 10 \).

Step 1: Find the Characteristic Polynomial

Given the matrix \( A = \begin{pmatrix} 1 & 3 \ 3 & 9 \end{pmatrix} \), we find the characteristic polynomial by calculating \( \det(A - \lambda I) \), where \( I \) is the identity matrix. This becomes \[ \det \begin{pmatrix} 1-\lambda & 3 \ 3 & 9-\lambda \end{pmatrix} = (1-\lambda)(9-\lambda) - 3 \times 3. \]Simplifying, we get: \[ (1-\lambda)(9-\lambda) - 9 = \lambda^2 - 10\lambda + 9 - 9 = \lambda^2 - 10\lambda = \lambda(\lambda - 10). \]

Step 2: Solve for the Eigenvalues

The characteristic polynomial is \( \lambda(\lambda - 10) = 0 \). Solve for \( \lambda \), we get: \[ \lambda = 0 \quad \text{or} \quad \lambda = 10. \]Hence, the eigenvalues of the matrix are \( \lambda = 0 \) and \( \lambda = 10 \).

Step 3: Verify the Eigenvalues

To ensure our eigenvalues are correct, we substitute back into the characteristic equation we derived: - For \( \lambda = 0 \), we find \( \det(A - 0I) = \det(A) = 0 \). Indeed, this determines one eigenvalue.- For \( \lambda = 10 \), we have \( \det(A - 10I) \) and find it's non-invertible confirming \( \lambda = 10 \) is also an eigenvalue. Thus, both eigenvalues satisfy the condition set by the characteristic polynomial.

Key Concepts

Characteristic Polynomial

When working with matrices, finding the characteristic polynomial is crucial for determining eigenvalues. Let's explore how this works, using a given matrix as an example.

Consider the matrix:\[ \begin{pmatrix} 1 & 3 \ 3 & 9 \end{pmatrix} \]First, we subtract \( \lambda I \), where \( I \) is the identity matrix, and calculate the determinant of the resulting matrix. This generates the characteristic polynomial.

• The identity matrix for a 2x2 matrix is \( \begin{pmatrix} 1 & 0 \ 0 & 1 \end{pmatrix} \).
• The matrix \( A - \lambda I \) becomes:\[ \begin{pmatrix} 1-\lambda & 3 \ 3 & 9-\lambda \end{pmatrix} \]
• Calculate the determinant: \( (1-\lambda)(9-\lambda) - 3 \cdot 3 = \lambda^2 - 10\lambda \).

This polynomial, \( \lambda^2 - 10\lambda \), is used to determine potential eigenvalues of the matrix.

Eigenvalue Verification

After finding the characteristic polynomial, the next step is to solve for its roots. These roots represent the eigenvalues of the matrix.

For the polynomial \( \lambda^2 - 10\lambda = 0 \):

• Factoring gives \( \lambda(\lambda - 10) = 0 \).
• Therefore, \( \lambda = 0 \) and \( \lambda = 10 \) are the possible eigenvalues.

Verification ensures these values satisfy the characteristic equation when substituted back:
- **For \( \lambda = 0 \)**: The determinant \( \det(A - 0I) = \det(A) = 0 \) shows that the value satisfies the polynomial.
- **For \( \lambda = 10 \)**: If \( \det(A - 10I) \) is non-invertible (zero determinant), this confirms its validity as an eigenvalue.

Both values meet the conditions of the characteristic equation, verifying them as eigenvalues.

Determinant Calculation

An essential step in finding eigenvalues is calculating the determinant of a matrix or its transformations. This step helps in forming the characteristic polynomial and verifying eigenvalues.

For the matrix \( A \):

• The determinant of \( A \) helps verify eigenvalues like \( \lambda = 0 \).
• Replacement of \( \lambda \) in \( A - \lambda I \) shows how each eigenvalue satisfies the condition to make the determinant zero for valid eigenvalues.

To compute determinants for matrices of the form \( \begin{pmatrix} a & b \ c & d \end{pmatrix} \):

• The formula is \( ad - bc \).

Thus, for each verification of \( \lambda \), checking \( \det(A - \lambda I) \) solidifies its status as an eigenvalue.