Problem 22 - 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. $$ \left(\begin{array}{lll} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 1 \end{array}\right) $$

Step-by-Step Solution

Answer

Eigenvalues are 0, 0, and 1. Eigenvectors: any vector in the form \( \begin{bmatrix} x \\ y \\ 0 \end{bmatrix} \) for \( \lambda = 0 \) and \( \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix} \) for \( \lambda = 1 \). Matrix is singular.

Step 1: Write Down the Matrix

The matrix given is a 3x3 matrix: \[ A = \begin{bmatrix} 0 & 0 & 0 \ 0 & 0 & 0 \ 0 & 0 & 1 \end{bmatrix} \]. We will find the eigenvalues and eigenvectors of this matrix.

Step 2: Finding Eigenvalues

To find the eigenvalues, we need to solve the characteristic equation \( \det(A - \lambda I) = 0 \). Here, \( I \) is the identity matrix. Thus, \[ A - \lambda I = \begin{bmatrix} -\lambda & 0 & 0 \ 0 & -\lambda & 0 \ 0 & 0 & 1-\lambda \end{bmatrix} \].\ The determinant is \( (-\lambda)(-\lambda)(1-\lambda) \). Solving \( \lambda^2(1-\lambda) = 0 \), we find the eigenvalues: \( \lambda_1 = 0, \lambda_2 = 0, \lambda_3 = 1 \).

Step 3: Finding Eigenvectors

For \( \lambda_1 = 0 \), we solve \( (A - 0I) \mathbf{v} = 0 \). Thus, \( A\mathbf{v} = 0 \), where \( A = \begin{bmatrix} 0 & 0 & 0 \ 0 & 0 & 0 \ 0 & 0 & 1 \end{bmatrix} \). This leads to the system: \( 0v_1 = 0, 0v_2 = 0, 0v_3 = 0 \). So the eigenvectors form a subspace generated by \( \begin{bmatrix} 1 \ 0 \ 0 \end{bmatrix}, \begin{bmatrix} 0 \ 1 \ 0 \end{bmatrix} \).For \( \lambda_3 = 1 \), we solve \( (A - I) \mathbf{v} = 0 \), which results in the system \( -v_1 = 0, -v_2 = 0, 0 = 0 \). This gives the eigenvector \( \begin{bmatrix} 0 \ 0 \ 1 \end{bmatrix} \).

Step 4: Determine if the Matrix is Singular or Nonsingular

According to Theorem 8.8.2, a matrix is singular if it has a zero eigenvalue. Since \( \lambda_1 = 0 \) and \( \lambda_2 = 0 \), the matrix \( A \) is singular.

Key Concepts

Singular Matrix

A matrix is termed as 'singular' if it does not have an inverse. This property is particularly important in various calculations in linear algebra, such as solving linear equations, where a non-invertible matrix can indicate that some solutions may not be unique or even exist.

• A matrix is singular if it has at least one eigenvalue that is zero.
• This indicates that the determinant of the matrix is zero, affirming singularity.

In the exercise provided, the matrix has eigenvalues including zero (\( \lambda_1 = 0 \) and \( \lambda_2 = 0 \)). Thus, by theorem and definition, such a matrix is singular, since it lacks a full rank and has reduced dimensionality of the output space relative to its input space.

Characteristic Equation

The characteristic equation is a pivotal concept in linear algebra, used to find the eigenvalues of a matrix. It forms by equating the determinant of the matrix minus a scalar multiple of the identity matrix to zero. Essentially, solving the characteristic equation allows you to discover values of \( \lambda \) that make the matrix \( (A - \lambda I) \) singular.For a matrix \( A \), the characteristic equation is defined as \[ \det(A - \lambda I) = 0 \]where \( I \) is the identity matrix and \( \lambda \) represents the eigenvalues. In our step-by-step solution, solving \[ \begin{bmatrix} -\lambda & 0 & 0 \ 0 & -\lambda & 0 \ 0 & 0 & 1-\lambda \ \end{bmatrix} \]leads to the equation \( \lambda^2(1-\lambda) = 0 \), yielding eigenvalues \( \lambda_1 = 0, \lambda_2 = 0, \lambda_3 = 1 \). This equation bridges the connection between matrix algebra and the properties of linear transformations.

Determinant of a Matrix

The determinant of a matrix is a scalar that provides insightful information about the matrix. In particular, it helps in understanding properties such as singularity, invertibility, and volume scaling in transformations.

• A non-zero determinant suggests that the matrix is nonsingular and invertible.
• A zero determinant indicates singularity, confirming that the matrix cannot be inverted.

For a matrix \( A \), the determinant is calculated by expanding \( A - \lambda I \). In the problem matrix, \[ \det \begin{bmatrix} -\lambda & 0 & 0 \ 0 & -\lambda & 0 \ 0 & 0 & 1-\lambda \ \end{bmatrix} = (-\lambda)(-\lambda)(1-\lambda) = \lambda^2(1-\lambda) \]results in zero and nonzero values as \( \lambda \) varies, confirming the presence of zero eigenvalues and thus, the singular nature of \( A \).

Linear Algebra Concepts

Linear algebra concepts form the backbone of understanding how matrices operate, transform spaces, and solve systems of equations effectively. Two critical components featured in the problem are eigenvalues and eigenvectors.

• Eigenvalues are scalars that illuminate crucial aspects of transformation operations represented by a matrix.
• Eigenvectors, on the other hand, signify directions that remain invariant under the transformation aside from scaling by their respective eigenvalues.

The exercise involves searching for these significant values to comprehend the transformation properties of a given matrix thoroughly. By harnessing these insights, one can determine paths where certain operations remain invariant or are informatively altered by the matrix's inherent characteristics.