Problem 11 - 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. $$A=\left[\begin{array}{lll} 2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2 \end{array}\right]$$

Step-by-Step Solution

Answer

The eigenvalue of matrix A is \(\lambda = 2\), with a multiplicity of 3. The basis of the eigenspace corresponding to this eigenvalue is given by: \(\left\{ \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix}, \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix}, \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix} \right\}\), with a dimension of 3. As the sum of the dimensions of the eigenspaces is equal to the size of the matrix, matrix A is nondefective.

Step 1: Find eigenvalues

We begin by finding the eigenvalues of matrix A. We do this by solving the characteristic equation given by: \[ det(A - \lambda I) = 0 \] where \(\lambda\) is the eigenvalue and \(I\) is the identity matrix. For the given matrix A, the characteristic equation is: \[ \begin{vmatrix} 2-\lambda & 0 & 0 \\ 0 & 2-\lambda & 0 \\ 0 & 0 & 2-\lambda \end{vmatrix} = 0 \] Solving for \(\lambda\), we get one eigenvalue: \((2-\lambda)^3 = 0 \implies \lambda = 2\)

Step 2: Find eigenvectors associated with each eigenvalue

Next, we find the eigenvectors associated with the eigenvalue \(\lambda = 2\). To do this, we solve the equation: \((A - \lambda I) x = 0\) For \(\lambda = 2\), we get: \[\begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} x = 0\] Since there are no constraints on the values of \(x_1\), \(x_2\), and \(x_3\), all non-zero combinations of these values will form an eigenvector associated with the eigenvalue \(\lambda = 2\).

Step 3: Find basis and dimension of each eigenspace

To find the basis and dimension of each eigenspace, we use the eigenvectors associated with each eigenvalue. For the eigenvalue \(\lambda = 2\), any eigenvector will be of the form: \(x = \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix}\) We can choose three linearly independent eigenvectors for the eigenvalue \(\lambda = 2\) to form a basis for the eigenspace: \(\text{Basis} = \{ \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix}, \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix}, \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix} \}\) Since there are three linearly independent eigenvectors in the eigenspace, the dimension of the eigenspace is 3.

Step 4: Determine if the matrix is defective or nondefective

A matrix is said to be defective if the sum of the dimensions of its eigenspaces is less than the size of the matrix. In this case, the sum of the dimensions of the eigenspaces is 3, which is equal to the size of the matrix (3x3). Therefore, the matrix A is nondefective.

Key Concepts

Matrix Diagonalization

Matrix diagonalization is a process that transforms a matrix into a diagonal form. This makes complex operations easier, just like how working with simple, straightforward numbers is easier than dealing with a mess of equations. Diagonalizing a matrix involves finding its eigenvectors and eigenvalues, which are critical to the process. When a matrix is diagonalized, we can express it as \[ A = PDP^{-1} \] where \(P\) is a matrix formed of the eigenvectors, \(D\) is a diagonal matrix made up of the eigenvalues, and \(P^{-1}\) is the inverse of \(P\). This structure allows performing calculations such as matrix powers efficiently. To diagonalize a matrix, its eigenvalues should lead to enough linearly independent eigenvectors to fill the eigenspace. In other words, for an \(n \times n\) matrix, you need \(n\) linearly independent eigenvectors to ensure diagonalizability. This is precisely what happened in our example matrix, where we had three independent eigenvectors.

Defective and Nondefective Matrices

The terms defective and nondefective describe the completeness of eigenvectors in a matrix. This concept is crucial in understanding whether a matrix can be diagonalized. A nondefective matrix has a complete set of linearly independent eigenvectors that match its order. In our example, since all three dimensions of the eigenspace matched the size of the matrix (3x3), matrix A is nondefective. A defective matrix, however, lacks sufficient independent eigenvectors to fully diagonalize. This scenario arises when some eigenvalues do not provide the necessary eigenvectors. Without enough eigenvectors, the transformation to a diagonal matrix is impossible. Here's how you can identify defective matrices:

• If the sum of the dimensions of all eigenspaces is less than the order of the matrix, it is defective.
• Ensure you verify the linear independence of the derived eigenvectors to confirm the structure.

Recognizing these matrix traits is key for effective mathematical modeling and operations.

Linear Algebra Concepts

Linear algebra is the mathematical discipline that deals with vectors and transformations, foundational to understanding many scientific fields. At its core, it involves systems of linear equations, vector spaces, and matrices.Key elements of linear algebra relevant to our matrix problem include:

• Vectors: Objects unpacked into components in directions, useful in representing spatial relationships.
• Matrices: Arrays of numbers shaping systems of equations, transformations and having applications in statistics, physics, and computer science.
• Eigenvalues and Eigenvectors: These originate from the characteristic equation \(det(A - \lambda I) = 0\), representing the underlying features of the matrix transformations. Eigenvectors remain unchanged in direction after such transformations, scaled by the eigenvalue.

These concepts are integral to learning linear algebra and understanding how they fit into matrix operations, like finding if a matrix is defective or nondefective, highlights their practical role in problem-solving.