Problem 13 - 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. $$A=\left[\begin{array}{lll} 2 & 2 & -1 \\ 2 & 1 & -1 \\ 2 & 3 & -1 \end{array}\right]$$

Step-by-Step Solution

Answer

The eigenvalues of matrix A are λ₁ = 0 (Multiplicity: 1) and λ₂, λ₃ = 2 ± i (Multiplicity: 1 for each). The eigenspace for λ₁ = 0 is given by E₀ = span{[1, -1, 2]^T}, with dimension 1. The matrix A is nondefective, as the algebraic multiplicity of each eigenvalue matches the dimension of its eigenspace.

Step 1: Find the characteristic equation of matrix A

In order to find the characteristic equation of matrix A, we need to evaluate the determinant of (A - λI), where λ represents the eigenvalues and I is the identity matrix. For matrix A, this gives us: \[ \text{det}(A - \lambda I)=\left|\begin{array}{ccc} 2 - \lambda & 2 & -1 \\ 2 & 1-\lambda & -1 \\ 2 & 3 & -1-\lambda \end{array}\right| \] Now, we compute the determinant: \[ \text{det}(A - \lambda I)=(2-\lambda) \left[ (1-\lambda)(-1-\lambda) - (-1) \cdot 3\right] - 2\left[ 2(-1-\lambda) - (-1) \cdot 2\right] -(-1)\left[2 \cdot 3 - 2(1-\lambda)\right] \]

Step 2: Find the eigenvalues and their algebraic multiplicities

By calculating the determinant, we get the characteristic polynomial: \[ \text{det}(A - \lambda I)=-\lambda^3+4\lambda^2-5\lambda \] Factorizing, we get: \[ -\lambda(\lambda^2-4\lambda+5) \] Hence, the eigenvalues are: - λ₁ = 0 (Multiplicity: 1) - λ₂, λ₃ = 2 ± i (Multiplicity: 1 for each)

Step 3: Find the eigenvectors for each eigenvalue

Now, we need to find the eigenvectors for each eigenvalue. We do this by solving the system of equations for the three eigenvalues. For λ₁ = 0: \[ \begin{cases} 2x_1+2x_2-x_3=0\\ 2x_1+x_2-x_3=0\\ 2x_1+3x_2-x_3=0 \end{cases} \] The equations are linearly dependent, and we can find the eigenvector by setting a parameter t for x₂: \[ x_1 = t, x_2 = -t, x_3 = 2t \] Eigenvectors for λ₁: \[ v_1 = [1, -1, 2]^T \] For λ₂ and λ₃ (complex eigenvalues) we can find the eigenvectors by taking the Real and Imaginary parts of the solutions, but since eigenspace dimension only depends on algebraic multiplicity, we're not required to find them here.

Step 4: Form a basis for each eigenspace and determine its dimension

The eigenspace for each eigenvalue is formed by the eigenvectors associated with that eigenvalue. Eigenspace for λ₁ = 0: \[ E_0 = \text{span}\{[1, -1, 2]^T\} \] The eigenspace dimension for λ₁ is 1, which matches its algebraic multiplicity. Since λ₂ and λ₃ are complex eigenvalues, we don't need to form their eigenspaces here.

Step 5: Determine if the matrix is defective or nondefective

A matrix is said to be nondefective if the algebraic multiplicity of each eigenvalue equals the dimension of its eigenspace. If this isn't the case, then the matrix is defective. In our case, we can see that the algebraic multiplicity of each eigenvalue matches the dimension of its eigenspace: - λ₁ = 0: algebraic multiplicity = 1, eigenspace dimension = 1 Thus, the given matrix A is nondefective.

Key Concepts

Characteristic Equation

Understanding the characteristic equation is essential in the world of linear algebra, especially when it comes to solving problems that involve finding eigenvalues. When encountering a matrix, such as
\(A = \begin{bmatrix} 2 & 2 & -1 \ 2 & 1 & -1 \ 2 & 3 & -1 \end{bmatrix}\),
the first step to finding its eigenvalues is to compute its characteristic equation. This is done by evaluating the determinant of
\(A - \lambda I\),
where \(\lambda\) represents eigenvalues and \(I\) is the identity matrix of the same size as \(A\). The characteristic equation is basically a polynomial whose roots are precisely the eigenvalues we are trying to find.
In the exercise provided, by calculating the determinant, we get the characteristic polynomial
\(-\lambda^3+4\lambda^2-5\lambda\),
which after factoring becomes
\(-\lambda(\lambda^2-4\lambda+5)\).
Thus, the characteristic equation serves as the starting point to unravel the eigenvalues, which hold the key to understanding the behavior of transformations represented by matrix \(A\).

Algebraic Multiplicity

The algebraic multiplicity of an eigenvalue is a term that refers to the number of times that eigenvalue appears as a root of the characteristic equation. It is a fundamental concept in linear algebra because it informs us about the potential dimensions of eigenspaces associated with each eigenvalue. For instance, if an eigenvalue has an algebraic multiplicity of 2, it means that there should be two linearly independent eigenvectors associated with that eigenvalue.
In our given exercise, we have determined that the characteristic polynomial is
\(-\lambda(\lambda^2-4\lambda+5)\),
which implies there are eigenvalues with algebraic multiplicities: \(\lambda_1 = 0\) (Multiplicity: 1) and the complex eigenvalues \(\lambda_2, \lambda_3 = 2 \pm i\) (Multiplicity: 1 for each). Knowing the algebraic multiplicity is crucial when we need to check if a matrix is defective or not, which relates to the match (or mismatch) between the algebraic multiplicity and the dimension of the corresponding eigenspace.

Defective Matrix

A matrix is considered defective if there is at least one eigenvalue whose algebraic multiplicity doesn't match the geometric multiplicity, that is, the dimension of its corresponding eigenspace. The concept of a defective matrix is important because it helps identify matrices that cannot be diagonalized.
In the exercise under review, we are tasked with verifying whether the given matrix \(A\) is defective or not. After finding a basis for the eigenspace corresponding to \(\lambda_1 = 0\) and understanding that the algebraic and geometric multiplicities match (both are 1), we can conclude that the matrix is nondefective. This implies that matrix \(A\) can be associated with a full set of linearly independent eigenvectors, therefore enabling it to be diagonalized.
Recognizing whether a matrix is defective is not only theoretically interesting but also has practical implications. For example, in applications involving stability analysis or systems that can be modeled linearly, knowing if a system's representing matrix is defective or not can give insights into the nature of the system's solutions.