Problem 40 - Differential Equations and Linear Algebra

Question

Use some form of technology to determine the eigenvalues and a basis for each eigenspace of the given matrix. Hence, determine the dimension of each eigenspace and state whether the matrix is defective or nondefective. $$\diamond A=\left[\begin{array}{rrr} 1 & -3 & 3 \\ -1 & -2 & 3 \\ -1 & -3 & 4 \end{array}\right]$$

Step-by-Step Solution

Answer

The eigenvalues of the matrix A are \(\lambda=0\) and \(\lambda=3\) (with a double root). The eigenspace for eigenvalue \(\lambda=0\) has basis \(E_0 = \{ \begin{bmatrix} 3 \\ 1 \\ 1 \end{bmatrix} \}\) with dimension 1 and the eigenspace for eigenvalue \(\lambda=3\) has basis \(E_3 = \{ \begin{bmatrix} 3 \\ -3 \\ 3 \end{bmatrix}, \begin{bmatrix} 0 \\ 3 \\ 3 \end{bmatrix} \}\) with dimension 2. Since the dimensions of the eigenspaces are equal to the multiplicities of the corresponding eigenvalues, the 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, \(|A - \lambda I| = 0\), where \(\lambda\) are the eigenvalues. For matrix A, $$ A - \lambda I = \begin{bmatrix} 1-\lambda & -3 & 3 \\ -1 & -2-\lambda & 3 \\ -1 & -3 & 4-\lambda \end{bmatrix} $$ Now, we compute the determinant: $$ |A - \lambda I| = (1-\lambda)((-2-\lambda)(4-\lambda)-(-3\cdot3)) - (-3)((-1)(4-\lambda)-(-1\cdot(-3))) + 3((-1)(-3)-(-1\cdot(-2-\lambda))) = \lambda^3-3\lambda^2 $$ To find the eigenvalues, we set the determinant equal to zero and solve for \(\lambda\): $$ \lambda^3-3\lambda^2=0 $$

Step 2: Find a basis for each eigenspace

The eigenvalues are the roots of the characteristic equation. In this case, we have \(\lambda=0\) and \(\lambda=3\) (with a double root). For each eigenvalue, we need to find a basis for the corresponding eigenspace by solving the following equation: $$ (A-\lambda I)X=0 $$ For \(\lambda = 0\), $$ (A-0I)X = \begin{bmatrix} 1 & -3 & 3 \\ -1 & -2 & 3 \\ -1 & -3 & 4 \end{bmatrix}X = 0 $$ Solving this system, we find that the eigenspace for eigenvalue \(\lambda=0\) has basis: $$ E_0 = \left\{ \begin{bmatrix} 3 \\ 1 \\ 1 \end{bmatrix} \right\} $$ For \(\lambda = 3\), $$ (A-3I)X = \begin{bmatrix} -2 & -3 & 3 \\ -1 & -5 & 3 \\ -1 & -3 & 1 \end{bmatrix}X = 0 $$ Solving this system, we find that the eigenspace for eigenvalue \(\lambda=3\) has basis: $$ E_3 = \left\{ \begin{bmatrix} 3 \\ -3 \\ 3 \end{bmatrix}, \begin{bmatrix} 0 \\ 3 \\ 3 \end{bmatrix} \right\} $$

Step 3: Determine the dimension of each eigenspace

The dimension of each eigenspace is simply the number of vectors in their bases. For the eigenvalues, we found the following eigenspaces: - Eigenspace for \(\lambda = 0\): \(E_0 = \{ \begin{bmatrix} 3 \\ 1 \\ 1 \end{bmatrix} \}\), dimension = 1 - Eigenspace for \(\lambda = 3\): \(E_3 = \{ \begin{bmatrix} 3 \\ -3 \\ 3 \end{bmatrix}, \begin{bmatrix} 0 \\ 3 \\ 3 \end{bmatrix} \}\), dimension = 2

Step 4: Check if the matrix is defective or nondefective

A matrix is defective if the dimension of each eigenspace is less than the multiplicity of the corresponding eigenvalue. In this case, the eigenvalue \(\lambda=0\) has a multiplicity of 1 and its eigenspace has a dimension of 1, while the eigenvalue \(\lambda=3\) has a multiplicity of 2 and its eigenspace has a dimension of 2. Since the dimensions of the eigenspaces are equal to the multiplicities of the corresponding eigenvalues, the matrix A is nondefective.

Key Concepts

Understanding Eigenvalues

One of the fundamental concepts in linear algebra is that of eigenvalues. They are the special scalars associated with a square matrix. When a matrix is multiplied by its eigenvector (a non-zero vector), the product is the eigenvector scaled by the corresponding eigenvalue. This concept can be written as the equation,
\( A \vec{x} = \lambda \vec{x} \),
where \( A \) is the matrix in question, \( \vec{x} \) is the eigenvector, and \( \lambda \) is the eigenvalue.
Determining eigenvalues is done by solving the characteristic equation of the matrix, a step crucial in many applications in physics and engineering, such as stability analysis, vibration analysis, and quantum mechanics. For instance, in the example of matrix
\( A = \begin{bmatrix} 1 & -3 & 3 \ -1 & -2 & 3 \ -1 & -3 & 4 \end{bmatrix} \),
the eigenvalues found by setting the determinant of the matrix \( A - \lambda I \) to zero are the roots of the characteristic polynomial, providing us with critical information about the matrix's behavior.

Characterizing Defective Matrices

Defective matrices are ones that do not have a complete set of linearly independent eigenvectors. This occurs when the algebraic multiplicity, which is the number of times an eigenvalue appears as a root of the characteristic polynomial, is greater than the geometric multiplicity, the dimension of the corresponding eigenspace. Geometrically, this means there are 'missing' directions along which the matrix cannot be simply scaled, which poses challenges in certain matrix decompositions like the diagonalization process.
In the context of the provided matrix
\( A = \begin{bmatrix} 1 & -3 & 3 \ -1 & -2 & 3 \ -1 & -3 & 4 \end{bmatrix} \),
it was determined that the matrix is nondefective since each eigenvalue's algebraic multiplicity equals its eigenspace's dimension. Understanding the concept of defective matrices is pivotal as it influences the approach to solving systems of linear equations and the behavior of dynamical systems.

The Role of the Characteristic Equation

The characteristic equation is the equation derived from a square matrix that helps determine its eigenvalues. It's formulated as the determinant of \( A - \lambda I \), where \( I \) is the identity matrix, set to zero. Solving this equation yields the eigenvalues which are key in understanding the matrix's properties.
Specifically, for our matrix
\( A = \begin{bmatrix} 1 & -3 & 3 \ -1 & -2 & 3 \ -1 & -3 & 4 \end{bmatrix} \),
the characteristic equation resulted in a cubic polynomial
\( \lambda^3 - 3\lambda^2 = 0 \).
The solutions to this equation determine the scaling factors under which the matrix compressions will occur. The characteristic equation plays a critical role in not just computing eigenvalues, but also in determining matrix invertibility, stability analysis of equilibrium points in differential equations, and more generally, in the field of control theory.