Problem 24 - Advanced Engineering Mathematics

Question

In Problems 23-26, find the eigenvalues and eigenvectors of the given nonsingular matrix \(\mathbf{A}\). Then without finding \(\mathbf{A}^{-1}\), find its eigenvalues and corresponding eigenvectors. $$ \mathbf{A}=\left(\begin{array}{rr} 4 & 2 \\ 7 & -1 \end{array}\right) $$

Step-by-Step Solution

Answer

Eigenvalues of \( \mathbf{A} \) are 6 and -3; corresponding eigenvectors are \( \begin{pmatrix} 1 \\ 1 \end{pmatrix} \) and \( \begin{pmatrix} 2 \\ -7 \end{pmatrix} \). Eigenvalues of \( \mathbf{A}^{-1} \) are \( \frac{1}{6} \) and \( -\frac{1}{3} \).

Step 1: Define the Characteristic Equation

First, we need to determine the characteristic equation, which is derived from the matrix \( \mathbf{A} \) and an identity matrix \( \mathbf{I} \). This is given by the equation \( \det(\mathbf{A} - \lambda \mathbf{I}) = 0 \), where \( \lambda \) represents the eigenvalues.

Step 2: Compute the Determinant

Write the characterstic equation: \( \mathbf{A} - \lambda \mathbf{I} = \begin{pmatrix} 4 - \lambda & 2 \ 7 & -1 - \lambda \end{pmatrix} \). Calculate the determinant: \((4-\lambda)(-1-\lambda) - (2)(7) = (\lambda^2 - 3\lambda - 18) = 0\).

Step 3: Solve the Characteristic Equation

To find eigenvalues \( \lambda \), solve the quadratic equation: \( \lambda^2 - 3\lambda - 18 = 0 \). Factor the equation: \( \lambda^2 - 3\lambda - 18 = (\lambda - 6)(\lambda + 3) = 0 \), giving solutions \( \lambda_1 = 6 \) and \( \lambda_2 = -3 \).

Step 4: Find the Eigenvectors for \( \lambda_1 = 6 \)

Substitute \( \lambda_1 = 6 \) into \( \mathbf{A} - \lambda \mathbf{I} \): \( \begin{pmatrix} 4-6 & 2 \ 7 & -1-6 \end{pmatrix} = \begin{pmatrix} -2 & 2 \ 7 & -7 \end{pmatrix} \). Solving for \( \mathbf{x} = \begin{pmatrix} x_1 \ x_2 \end{pmatrix} \) in the equation \( \begin{pmatrix} -2 & 2 \ 7 & -7 \end{pmatrix} \begin{pmatrix} x_1 \ x_2 \end{pmatrix} = \begin{pmatrix} 0 \ 0 \end{pmatrix} \), leads to \( x_1 = x_2 \). Hence, an eigenvector is \( \begin{pmatrix} 1 \ 1 \end{pmatrix} \).

Step 5: Find the Eigenvectors for \( \lambda_2 = -3 \)

Substitute \( \lambda_2 = -3 \) into \( \mathbf{A} - \lambda \mathbf{I} \): \( \begin{pmatrix} 4+3 & 2 \ 7 & -1+3 \end{pmatrix} = \begin{pmatrix} 7 & 2 \ 7 & 2 \end{pmatrix} \). Solving \( \begin{pmatrix} 7 & 2 \ 7 & 2 \end{pmatrix} \begin{pmatrix} x_1 \ x_2 \end{pmatrix} = \begin{pmatrix} 0 \ 0 \end{pmatrix} \), results in \( 7x_1 = -2x_2 \). Thus, a valid eigenvector is \( \begin{pmatrix} 2 \ -7 \end{pmatrix} \).

Step 6: Utilize the Inverse Property for Eigenvalues

For a nonsingular matrix \( \mathbf{A} \), if \( \lambda \) is an eigenvalue, then \( \frac{1}{\lambda} \) is the eigenvalue of \( \mathbf{A}^{-1} \). Hence, the eigenvalues of \( \mathbf{A}^{-1} \) are \( \frac{1}{6} \) and \( \frac{1}{-3} \). Eigenvectors for \( \mathbf{A}^{-1} \) remain the same as those of \( \mathbf{A} \), which are \( \begin{pmatrix} 1 \ 1 \end{pmatrix} \) and \( \begin{pmatrix} 2 \ -7 \end{pmatrix} \).

Key Concepts

Characteristic Equation

The characteristic equation is essential when finding eigenvalues of a matrix. It is derived from the equation \( ext{det}(\mathbf{A} - \lambda \mathbf{I}) = 0\), where \(\mathbf{A}\) is your matrix and \(\lambda\) represents the eigenvalues. This equation helps us find the values of \(\lambda\) that satisfy the equation, revealing the matrix's eigenvalues. By replacing \(\mathbf{A}\) with the specific matrix from our exercise and subtracting \(\lambda I\), we form a new matrix whose determinant must equal zero. Solving this equation, using methods such as expansion by minors, is crucial in locating these special \(\lambda\) values.

Nonsingular Matrix

A nonsingular matrix, also known as an invertible matrix, has several important properties. One key feature is that it has a determinate different from zero, which corresponds to having full rank. This means there's no row or column which is completely zero. This non-zero determinant indicates that the matrix has no zero eigenvalues, ensuring all eigenvalues are non-zero. Understanding this is critical because non-zero eigenvalues allow the matrix to be inverted. Without this property, calculating the inverse matrix and finding its eigenvalues would be impossible.

Matrix Determinant

The determinant of a matrix is a special number that can be calculated from its elements, often symbolized by \(\text{det}(\mathbf{A})\). It's more than just a number; it tells us if a matrix is invertible and even the volume transformation it applies in space. When the determinant is zero, it's a clue that the matrix is singular and can't be inverted. However, if it's non-zero (as in our exercise), the matrix is nonsingular and invertible. In our characteristic equation, calculating the determinant helps find the specific eigenvalues by setting the whole expression to zero.

Inverse Matrix Eigenvalues

For any nonsingular matrix, you don't need to compute its inverse to know the eigenvalues of that inverse. This is because if \(\lambda\) is an eigenvalue of the original matrix \(\mathbf{A}\), then its inverse matrix \(\mathbf{A}^{-1}\) will have eigenvalues that are \(\frac{1}{\lambda}\). This property saves a lot of computational work since calculating the matrix's inverse can be complex and time-consuming. In our exercise, since the eigenvalues of \(\mathbf{A}\) were found to be 6 and -3, this means the eigenvalues of \(\mathbf{A}^{-1}\) are \(\frac{1}{6}\) and \(-\frac{1}{3}\), respectively. Remarkably, the eigenvectors for both matrices remain the same, highlighting the beauty of linear algebra properties.