Problem 4 - Differential Equations and Linear Algebra

Question

Determine the multiplicity of each eigenvalue and a basis for each eigenspace of the given matrix \(A\). Hence, determine the dimension of each eigenspace and state whether the matrix is defective or nondefective. $$A=\left[\begin{array}{rr} 1 & 2 \\ -2 & 5 \end{array}\right]$$

Step-by-Step Solution

Answer

Eigenvalue: \(λ = 3\) (multiplicity 1). Eigenspace: Span{\(\begin{bmatrix} 1 \\ 1 \end{bmatrix}\)}. Dimension of eigenspace: 1. Matrix A is nondefective.

Step 1: Find the Characteristic Polynomial of A

To find the characteristic polynomial of the given matrix A, we first take the determinant of (A - λI), where λ is an eigenvalue and I is the identity matrix. Let A = \(\begin{bmatrix} 1 & 2 \\ -2 & 5 \end{bmatrix}\), then (A - λI) = \(\begin{bmatrix} 1 - λ & 2 \\ -2 & 5 - λ \end{bmatrix}\) The determinant of (A - λI) is: det(A - λI) = (1 - λ)(5 - λ) - (-2)(2) = λ^2 - 6λ + 9 So, the characteristic polynomial of A is: λ^2 - 6λ + 9.

Step 2: Find the Eigenvalues of A

To find the eigenvalues of A, we need to solve the equation det(A - λI) = 0. The equation det(A - λI) = 0 can be written as: λ^2 - 6λ + 9 = 0 This is a quadratic equation, and its solution can be found using the quadratic formula: λ = ( -b ± sqrt(b^2 - 4ac) ) / 2a Here, a = 1, b = -6, and c = 9. λ = ( 6 ± sqrt((-6)^2 - 4(1)(9)) )/ (2) λ = ( 6 ± sqrt(36 - 36) )/ 2 λ = 6 / 2 λ = 3 So, the matrix A has only one eigenvalue: λ = 3.

Step 3: Determine the Eigenspace for Eigenvalue λ = 3

Now we find the null space of (A - λI). Let's first solve (A - λI): (A - λI) = \(\begin{bmatrix} 1 - 3 & 2 \\ -2 & 5 - 3 \end{bmatrix}\) = \(\begin{bmatrix} -2 & 2 \\ -2 & 2 \end{bmatrix}\) Now, we solve the homogeneous linear system (A - λI)x = 0: \(\begin{bmatrix} -2 & 2 \\ -2 & 2 \end{bmatrix}\) \(\begin{bmatrix} x_1 \\ x_2 \end{bmatrix}\) = \(\begin{bmatrix} 0 \\ 0 \end{bmatrix}\) Since the second row is identical to the first row, we can eliminate it: \(- 2x_1 + 2x_2 = 0\) Solving for x_1: \(x_1 = x_2\) Let x_2 = k, where k is any scalar. Then, x_1 = k as well. So the solution to the homogeneous linear system is: \(\begin{bmatrix} x_1 \\ x_2 \end{bmatrix}\) = k\(\begin{bmatrix} 1 \\ 1 \end{bmatrix}\)

Step 4: Determine the Dimension and Basis for Eigenspace

Since the eigenvector for λ = 3 is given by k\(\begin{bmatrix} 1 \\ 1 \end{bmatrix}\), the basis of the eigenspace associated with λ = 3 is given by: Basis = { \(\begin{bmatrix} 1 \\ 1 \end{bmatrix}\)} The dimension of this eigenspace is 1, as there is only one independent eigenvector.

Step 5: Defective or Nondefective Matrix

A matrix is considered defective if the dimension of an eigenspace is less than the multiplicity of its eigenvalue, otherwise, it is nondefective. In this case, the dimension of the eigenspace (1) is equal to the multiplicity of the eigenvalue λ = 3 (also 1), so the matrix A is nondefective. To summarize: Eigenvalue: λ = 3 (multiplicity 1) Eigenspace: Span{\(\begin{bmatrix} 1 \\ 1 \end{bmatrix}\)} Dimension of eigenspace: 1 Matrix A is nondefective.

Key Concepts

Eigenspace

An eigenspace is a special subspace formed by all eigenvectors corresponding to a particular eigenvalue in a matrix. When a matrix acts on its eigenspace, it stretches or squishes all vectors by a constant factor, which is the eigenvalue itself. Understanding eigenspaces is crucial because they provide insight into the fundamental properties of a matrix.
Consider the given matrix \(A\) and its eigenvalue \(\lambda = 3\). An eigenspace associated with this eigenvalue includes all vectors that, when multiplied by \(A\), give rise to a scalar multiple of themselves. In mathematical terms, if \(v\) is an eigenvector, then \(Av = \lambda v\).

• Find eigenvectors by solving \((A - \lambda I)x = 0\). This linear system gives solutions that form a basis for the eigenspace.
• In our example, the solution \( \textbf{x} = k\begin{bmatrix} 1 \ 1 \end{bmatrix} \) is an eigenvector that spans the eigenspace.

The solution implies all vectors in this space are scalar multiples of the basis vector \(\begin{bmatrix} 1 \ 1 \end{bmatrix}\). This set of vectors describes the direction in which the matrix transforms its eigenvectors by the eigenvalue \(3\).

Matrix Defectiveness

Matrix defectiveness refers to a property that determines how well a matrix can be diagonalized. A square matrix is termed defective if its eigenvalues do not have enough linearly independent eigenvectors to match their algebraic multiplicities. Otherwise, it is nondefective.
For the square matrix \(A\) provided in this example, we have calculated the multiplicity of the eigenvalue \(3\) to be 1 and the dimension of its corresponding eigenspace to also be 1. In this scenario:

• The matrix has a full complement of independent eigenvectors for its eigenvalue.
• This sufficiency means the matrix is *nondefective*.

Being nondefective is significant because it indicates that the matrix's dynamics can be fully captured and expressed using its eigenvectors and eigenvalues, aligning completely with its characteristic structure.

Characteristic Polynomial

The characteristic polynomial is derived from a matrix and is crucial for understanding a matrix's eigenvalues. It is obtained by computing the determinant of \((A - \lambda I)\), where \(\lambda\) serves as a placeholder for potential eigenvalues, and \(I\) is the identity matrix.
The provided matrix \(A\) yielded the characteristic polynomial \(\lambda^2 - 6\lambda + 9\). Each coefficient in the polynomial conveys essential information about the matrix's properties:

• The quadratic term (\(\lambda^2\)) signifies a square matrix
• The linear term (\