Problem 10 - Advanced Engineering Mathematics

Question

In Problems 7-22, find the eigenvalues and eigenvectors of the given matrix. Using Theorem 8.8.2 or (6), state whether the matrix is singular or nonsingular. $$ \left(\begin{array}{ll} 1 & 1 \\ \frac{1}{4} & 1 \end{array}\right) $$

Step-by-Step Solution

Answer

The eigenvalues are \( \frac{3}{2} \) and \( \frac{1}{2} \). It is nonsingular.

Step 1: Setup the Characteristic Equation

Given the matrix \( A = \begin{pmatrix} 1 & 1 \ \frac{1}{4} & 1 \end{pmatrix} \), we find the eigenvalues by solving the characteristic equation \( \det(A - \lambda I) = 0 \). First, write \( A - \lambda I = \begin{pmatrix} 1-\lambda & 1 \ \frac{1}{4} & 1-\lambda \end{pmatrix} \).

Step 2: Compute the Determinant

Calculate the determinant of \( A - \lambda I \):\[\det(A - \lambda I) = (1-\lambda)(1-\lambda) - \left( \frac{1}{4} \times 1 \right) = (1-\lambda)^2 - \frac{1}{4}.\]

Step 3: Solve the Characteristic Equation

Set the determinant equal to zero to find the eigenvalues:\[(1-\lambda)^2 - \frac{1}{4} = 0\]Simplify and solve for \( \lambda \):\[(1-\lambda)^2 = \frac{1}{4}\]Taking the square root, we find:\[1-\lambda = \pm \frac{1}{2}\] which gives\[\lambda = \frac{3}{2}, \quad \lambda = \frac{1}{2}.\]

Step 4: Find Eigenvectors for \( \lambda = \frac{3}{2} \)

Substitute \( \lambda = \frac{3}{2} \) back into \( A - \lambda I \):\[ A - \lambda I = \begin{pmatrix} 1-\frac{3}{2} & 1 \ \frac{1}{4} & 1-\frac{3}{2} \end{pmatrix} = \begin{pmatrix} -\frac{1}{2} & 1 \ \frac{1}{4} & -\frac{1}{2} \end{pmatrix}. \]Solve \( (A - \lambda I)\mathbf{x} = 0 \) using a vector \( \mathbf{x} = \begin{pmatrix} x_1 \ x_2 \end{pmatrix} \):\[\begin{pmatrix} -\frac{1}{2} & 1 \ \frac{1}{4} & -\frac{1}{2} \end{pmatrix} \begin{pmatrix} x_1 \ x_2 \end{pmatrix} = \begin{pmatrix} -\frac{1}{2}x_1 + x_2 \ \frac{1}{4}x_1 - \frac{1}{2}x_2 \end{pmatrix} = \begin{pmatrix} 0 \ 0 \end{pmatrix}.\]This gives two equations: - \(-\frac{1}{2}x_1 + x_2 = 0\)- \(\frac{1}{4}x_1 - \frac{1}{2}x_2 = 0\)Both simplify to \(x_2 = \frac{1}{2} x_1\), thus, an eigenvector is \( \begin{pmatrix} 1 \ \frac{1}{2} \end{pmatrix} \).

Step 5: Find Eigenvectors for \( \lambda = \frac{1}{2} \)

Substitute \( \lambda = \frac{1}{2} \) back:\[ A - \lambda I = \begin{pmatrix} 1-\frac{1}{2} & 1 \ \frac{1}{4} & 1-\frac{1}{2} \end{pmatrix} = \begin{pmatrix} \frac{1}{2} & 1 \ \frac{1}{4} & \frac{1}{2} \end{pmatrix}. \]Solve \( (A - \lambda I)\mathbf{x} = 0 \):\[\begin{pmatrix} \frac{1}{2} & 1 \ \frac{1}{4} & \frac{1}{2} \end{pmatrix} \begin{pmatrix} x_1 \ x_2 \end{pmatrix} = \begin{pmatrix} \frac{1}{2}x_1 + x_2 \ \frac{1}{4}x_1 + \frac{1}{2}x_2 \end{pmatrix} = \begin{pmatrix} 0 \ 0 \end{pmatrix}.\]This gives:- \(\frac{1}{2}x_1 + x_2 = 0\)- \(\frac{1}{4}x_1 + \frac{1}{2}x_2 = 0\)Both imply \(x_2 = -\frac{1}{2}x_1\), thus an eigenvector is \( \begin{pmatrix} 1 \ -\frac{1}{2} \end{pmatrix} \).

Step 6: Determine if the Matrix is Singular or Nonsingular

The matrix is nonsingular because none of the eigenvalues of the matrix are zero.

Key Concepts

Characteristic Equation

When dealing with eigenvalues and eigenvectors, the characteristic equation is your starting point. It is created from a square matrix, like the one in our exercise: \(A = \begin{pmatrix} 1 & 1 \ \frac{1}{4} & 1 \end{pmatrix}\). To find this equation, we subtract \(\lambda I\) from \(A\), where \(I\) is the identity matrix, and set the determinant of this new matrix to zero:
\[\det(A - \lambda I) = 0\]This process leads us to the equation \((1-\lambda)^2 - \frac{1}{4} = 0\).
After solving, we identify the values of \(\lambda\), known as eigenvalues. For the matrix in our problem, the eigenvalues are \(\lambda = \frac{3}{2}\) and \(\lambda = \frac{1}{2}\).
By finding these, we begin to uncover key properties about the matrix, such as dimensions of eigenspaces and matrix diagonalizability.

Singular and Nonsingular Matrices

Understanding whether a matrix is singular or nonsingular is crucial. A matrix is called **singular** if it has no inverse, while it is **nonsingular** if an inverse exists.
Typically, a matrix becomes singular if at least one of its eigenvalues is zero. On the flip side, if all eigenvalues are non-zero, the matrix is nonsingular.

• Singular matrices have a determinant of zero.
• Nonsingular matrices have a non-zero determinant.

In the provided exercise, the matrix is nonsingular since both eigenvalues (\(\lambda = \frac{3}{2}, \frac{1}{2}\)) are not zero.
As such, an inverse matrix can be computed, highlighting its nonsingular nature.

Determinant of a Matrix

The determinant of a matrix is a numerical value that offers insights into several matrix properties, such as invertibility, eigenvalues, and rank. For a 2x2 matrix, the determinant is composed using \[\det\begin{pmatrix} a & b \ c & d \end{pmatrix} = ad - bc\]In the exercise's matrix, we generated the determinant for \(A - \lambda I\):
\((1-\lambda)^2 - \frac{1}{4}\).This result plays a vital role in forming the characteristic equation by equating it to zero to find eigenvalues.
The determinant also helps determine a matrix's singular status:

• Non-zero determinant: matrix is nonsingular.
• Zero determinant: matrix is singular.

So, fully grasping how to compute and interpret determinants is fundamental in linear algebra and helps in solving matrix-related problems efficiently.