Problem 20 - Differential Equations and Linear Algebra

Question

Determine whether the given matrix is defective or nondefective. $$\begin{aligned} &A=\left[\begin{array}{rrr} 1 & -3 & 1 \\ -1 & -1 & 1 \\ -1 & -3 & 3 \end{array}\right]\\\ &\text { characteristic polynomial } p(\lambda)=-(\lambda-2)^{2}(\lambda+1) \end{aligned}$$

Step-by-Step Solution

Answer

The matrix \(A\) is defective because the eigenvalue \(\lambda_1=2\) has algebraic multiplicity 2, but the corresponding eigenspace has dimension 1, which is less than its algebraic multiplicity.

Step 1: Identify the eigenvalues

The given characteristic polynomial is \(p(\lambda)=-(\lambda-2)^{2}(\lambda+1)\). Setting this polynomial equal to zero, we find the eigenvalues: \((\lambda-2)^{2}(\lambda+1) = 0\) This equation has two roots: \(\lambda_1 = 2\) and \(\lambda_2 = -1\). The algebraic multiplicity of the eigenvalue 2 is 2, and the algebraic multiplicity of the eigenvalue -1 is 1.

Step 2: Compute the eigenspaces

Now, we need to compute the eigenspaces for each eigenvalue: 1. For \(\lambda_1 = 2\), we want to find the eigenvectors that satisfy \((A-\lambda I)x = 0\). Calculate the matrix equation for \(\lambda = 2\): \((A-2I) = \begin{bmatrix} 1 & -3 & 1 \\ -1 & -1 & 1 \\ -1 & -3 & 3 \\ \end{bmatrix} - 2\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{bmatrix}= \begin{bmatrix} -1 & -3 & 1 \\ -1 & -3 & 1 \\ -1 & -3 & 1 \\ \end{bmatrix}\) Now, solve \((A-2I)x = 0\): \(\begin{bmatrix} -1 & -3 & 1 \\ -1 & -3 & 1 \\ -1 & -3 & 1 \\ \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ x_3 \\ \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \\ \end{bmatrix}\) From this equation, we can see that the eigenspace for \(\lambda = 2\) contains only one linearly independent vector, which is the eigenvector for this eigenvalue: \(\begin{bmatrix} 2 \\ 1 \\ 0 \\ \end{bmatrix}\) 2. For \(\lambda_2 = -1\), calculate the matrix equation for \(\lambda = -1\): \((A+I) = \begin{bmatrix} 1 & -3 & 1 \\ -1 & -1 & 1 \\ -1 & -3 & 3 \\ \end{bmatrix} + \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{bmatrix}=\begin{bmatrix} 2 & -3 & 1 \\ -1 & 0 & 1 \\ -1 & -3 & 4 \\ \end{bmatrix}\) Now, solve \((A+I)x = 0\): \(\begin{bmatrix} 2 & -3 & 1 \\ -1 & 0 & 1 \\ -1 & -3 & 4 \\ \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ x_3 \\ \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \\ \end{bmatrix}\) The eigenspace for \(\lambda = -1\) contains only one linearly independent vector, which is the eigenvector for this eigenvalue: \(\begin{bmatrix} 1 \\ 1 \\ 1 \\ \end{bmatrix}\)

Step 3: Determine if the matrix is defective or nondefective

Now we can evaluate if our matrix is defective or nondefective based on the dimensions of the eigenspaces we found. For \(\lambda_1=2\), we have an algebraic multiplicity of 2, but the eigenspace only has dimension 1. This implies that the matrix is defective because the eigenspace corresponding to \(\lambda_1=2\) doesn't have the same dimension as its algebraic multiplicity.

Key Concepts

Eigenvalues

Eigenvalues are a fundamental concept in linear algebra, associated with matrices. Simply put, they are special numbers that reveal important properties of a matrix. Eigenvalues are derived from the characteristic polynomial of a matrix. In the context of transformations, an eigenvalue illustrates how much the transformation stretches or compresses a vector along its associated eigenvector direction. Thus, eigenvalues are pivotal in determining the behavior of a matrix under transformation.

When encountering a matrix, we often set its characteristic polynomial to zero to find its eigenvalues. In our given exercise, the characteristic polynomial is \(p(\lambda)=-(\lambda-2)^{2}(\lambda+1)\). Solving \((\lambda-2)^{2}(\lambda+1) = 0\) yields the eigenvalues \(2\) and \(-1\). These eigenvalues help describe how the matrix behaves and interact with its eigenspaces. By revealing the multiplicities of these roots, they also help determine whether the matrix is defective.

Characteristic Polynomial

The characteristic polynomial is an essential tool used to find the eigenvalues of a matrix. It is derived from the determinant of the matrix \(A\) subtracted by \(\lambda\) times the identity matrix. Mathematically, this is expressed as \( \det (A - \lambda I) \). The roots of this polynomial are the eigenvalues of the matrix.

For example, in our exercise, the characteristic polynomial is given as \(-\lambda^3 + 3\lambda^2 - 4\lambda + 2\) which simplifies to \(-(\lambda-2)^{2}(\lambda+1)\). Solving \(p(\lambda) = 0\) provides the eigenvalues \(\lambda = 2\) and \(\lambda = -1\). The nature of this polynomial, particularly its factors and roots, also reveals vital information about the matrix's properties, such as whether the matrix is defective or nondefective.

• The factors of the polynomial indicate potential multiplicities of the roots.
• Solving \(p(\lambda) = 0\) yields the eigenvalues.
• The structure reveals algebraic multiplicities, which are crucial for understanding the matrix's behavior.

Eigenspaces

Eigenspaces are collections or sets of eigenvectors corresponding to a particular eigenvalue, including the zero vector. For each eigenvalue, there's an associated eigenspace that describes all vectors that remain unchanged, except for a scalar multiple, when a transformation described by a matrix is applied.

To find an eigenspace, we solve \((A - \lambda I)x = 0\) for each eigenvalue. The solution yields the linearly independent vectors forming the basis of the eigenspace. In our exercise:

• For \(\lambda = 2\), we found the eigenspace consisting of eigenvectors that satisfy \((A-2I)x = 0\). This yielded a single linearly independent vector \([2, 1, 0]^T\).
• For \(\lambda = -1\), solving \((A+I)x = 0\) provided another single linearly independent vector \([1, 1, 1]^T\).

Eigenspaces provide comprehensive insights into how matrices transform vectors. Their dimensionality is essential for determining if a matrix is defective, revealing whether there's a sufficient number of linearly independent vectors for each eigenvalue.

Algebraic Multiplicity

Algebraic multiplicity of an eigenvalue is the number of times that eigenvalue appears as a root of the characteristic polynomial. Each eigenvalue of a matrix can have a different algebraic multiplicity, indicating how many dimensions their corresponding eigenspaces should ideally span.

In our exercise, the eigenvalue \(\lambda = 2\) has an algebraic multiplicity of 2, and \(\lambda = -1\) has an algebraic multiplicity of 1. This suggests that ideally, the eigenspace related to \(\lambda = 2\) should span two dimensions, presenting two linearly independent eigenvectors.

• Algebraic multiplicity is derived from the power of the factor associated with that eigenvalue in the characteristic polynomial.
• If an eigenspace has fewer linearly independent vectors than the algebraic multiplicity suggests, then the matrix may be defective.

Therefore, algebraic multiplicity is intrinsic to assessing whether a matrix is defective or nondefective, based on how well the eigenspaces fulfill these multiplicities.