Problem 18 - Differential Equations and Linear Algebra

Question

Determine whether the given matrix is defective or nondefective. $$A=\left[\begin{array}{rr} 6 & 5 \\ -5 & -4 \end{array}\right]$$

Step-by-Step Solution

Answer

The eigenvalues of the matrix A are λ1 = 3 and λ2 = -3. Since we found two distinct eigenvalues with their corresponding eigenvectors, the sum of their geometric multiplicities is equal to the matrix dimension, which is 2. Hence, the given matrix A is nondefective.

Step 1: Find the eigenvalues of the matrix A

To find the eigenvalues, we need to solve the characteristic equation given by det(A - λI), where λ is the eigenvalue and I is the identity matrix of the same size as A. For matrix A, the characteristic equation becomes: \[ \left|\begin{array}{cc} 6-\lambda & 5 \\ -5 & -4-\lambda \end{array}\right| = (6-\lambda)((-4)-\lambda) - (5)(-5) = \lambda^2 - 2\lambda - 9 \] Now, we need to find the roots of the above polynomial: \( \lambda^2 - 2\lambda - 9 = (\lambda-3)(\lambda+3) = 0 \) So, the eigenvalues (λ) are λ1 = 3 and λ2 = -3.

Step 2: Find the eigenvectors corresponding to each eigenvalue

To find the corresponding eigenvectors for each eigenvalue, we need to solve the equation (A - λI)v = 0, where λ is the eigenvalue and v is the eigenvector. For λ1 = 3 : \( (A - λ_1I)v = \left[\begin{array}{cc} 6-3 & 5 \\ -5 & -4-3 \end{array}\right]v = \left[\begin{array}{cc} 3 & 5 \\ -5 & -7 \end{array}\right]v = 0\) Row reducing the above matrix, we get: \( \left[\begin{array}{cc|c} 1 & \dfrac{5}{3} & 0 \\ 0 & 0 & 0 \end{array}\right] \) So v = k\(\left[\begin{array}{c} - 5/3 \\ 1 \end{array}\right]\), where k is a scalar constant. For λ2 = -3 : \( (A - λ_2I)v = \left[\begin{array}{cc} 6-(-3) & 5 \\ -5 & -4-(-3) \end{array}\right]v = \left[\begin{array}{cc} 9 & 5 \\ -5 & 1 \end{array}\right]v = 0\) Row reducing the above matrix, we get: \( \left[\begin{array}{cc|c} 1 & \dfrac{5}{9} & 0 \\ 0 & 0 & 0 \end{array}\right] \) So v = k\(\left[\begin{array}{c} -\dfrac{5}{9} \\ 1 \end{array}\right]\), where k is a scalar constant.

Step 3: Check if the matrix is defective or nondefective

The matrix dimension is 2. We found two distinct eigenvalues, λ1 = 3 and λ2 = -3, each with an eigenvector. Thus, the sum of the geometric multiplicities of the eigenvalues is also 2. Since the sum of the geometric multiplicities of the eigenvalues is equal to the matrix dimension, the matrix A is nondefective.

Key Concepts

Eigenvalues

In linear algebra, eigenvalues are special scalars associated with a square matrix. They provide valuable information about the linear transformations represented by the matrix. To find the eigenvalues, we solve the characteristic equation, which is derived from setting the determinant of \( A - \lambda I \) equal to zero. Here, \( \lambda \) represents an eigenvalue, \( A \) is the given matrix, and \( I \) is the identity matrix.

• The characteristic equation for a 2x2 matrix \( A \) can be expressed as: \[ \left| \begin{array}{cc} a-\lambda & b \ c & d-\lambda \end{array} \right| = 0 \]
• When expanded, it forms a quadratic equation: \[ \lambda^2 - (a+d)\lambda + (ad-cb) = 0 \]

In our example, after solving the characteristic equation \( \lambda^2 - 2\lambda - 9 = (\lambda - 3)(\lambda + 3) = 0 \), the eigenvalues are found to be \( \lambda_1 = 3 \) and \( \lambda_2 = -3 \). These values indicate the scalings of the vectors in the direction of the eigenvectors associated with those eigenvalues.

Eigenvectors

Eigenvectors provide the directions along which a linear transformation acts by only scaling. Once an eigenvalue is known, we can identify the corresponding eigenvectors by solving the equation \( (A - \lambda I)v = 0 \).

• For each eigenvalue \( \lambda \), substitute it into \( A - \lambda I \) and solve for vector \( v \).
• The solutions for \( v \) are scaled versions of the eigenvectors for that \( \lambda \).

In our matrix \( A \), once we substituted \( \lambda_1 = 3 \) and \( \lambda_2 = -3 \), we found corresponding eigenvectors:

• For \( \lambda_1 = 3 \), the eigenvector is \( v = k\left[\begin{array}{c} -\frac{5}{3} \ 1 \end{array}\right] \)
• For \( \lambda_2 = -3 \), the eigenvector is \( v = k\left[\begin{array}{c} -\frac{5}{9} \ 1 \end{array}\right] \)

These eigenvectors reveal how the matrix linearly transforms space, by stretching or contracting along specific directions.

Characteristic Equation

The characteristic equation is a crucial concept in determining eigenvalues. It arises from evaluating the determinant of \( A - \lambda I \), set to zero. This equation allows us to identify eigenvalues, which are the roots of the equation.

• The process involves \( A - \lambda I \), where \( A \) is the original matrix and \( I \) is the identity matrix of the same order.
• For a matrix \( A \), the equation becomes \( det(A - \lambda I) = 0 \).
• Solving this yields the polynomial whose roots are the matrix's eigenvalues.

In this case, the determinant for \( A \) is set up as:\[\left| \begin{array}{cc} 6-\lambda & 5 \ -5 & -4-\lambda \end{array} \right| = (6-\lambda)((-4)-\lambda) - (5)(-5)\]Which simplifies and equates to:\[ \lambda^2 - 2\lambda - 9 = 0 \]This quadratic equation reveals our eigenvalues \( \lambda_1 = 3 \) and \( \lambda_2 = -3 \). Solving the characteristic equation is integral in understanding how matrices impact vector spaces.

Nondefective Matrix

A nondefective matrix is one where the algebraic multiplicity equals the geometric multiplicity of each eigenvalue, indicating it has a complete set of linearly independent eigenvectors. In simpler terms, it is a matrix whose eigenvectors form a basis for the space.

• The dimension of the matrix matches the number of distinct eigenvectors.
• For matrix \( A \), checking its nondefectiveness involves ensuring that each eigenvalue's geometric multiplicity equals its algebraic multiplicity.
• The sum of geometric multiplicities equates to the matrix's order, implying complete diagonalizability.

As shown in our exercise, matrix \( A \) has two eigenvalues, \( \lambda_1 = 3 \) and \( \lambda_2 = -3 \), each associated with one eigenvector, leading to a geometric multiplicity sum of 2, equal to the matrix's order. Thus, this verifies the matrix is nondefective, and it can be diagonalized effectively.