Problem 6 - 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. $$A=\left[\begin{array}{rrr} 3 & -4 & -1 \\ 0 & -1 & -1 \\ 0 & -4 & 2 \end{array}\right]$$

Step-by-Step Solution

Answer

The eigenvalues of the matrix \(A\) are \(\lambda_1 = 3\), \(\lambda_2 = -1\), and \(\lambda_3 = 2\), each with a multiplicity of 1. The eigenspaces and their dimensions for these eigenvalues are: - For \(\lambda_1 = 3\), eigenspace basis: \(\left\{ \begin{bmatrix}0\\-1\\1\end{bmatrix} \right\}\), dimension: 1 - For \(\lambda_2 = -1\), eigenspace basis: \(\left\{ \begin{bmatrix}1\\1\\0\end{bmatrix} \right\}\), dimension: 1 - For \(\lambda_3 = 2\), eigenspace basis: \(\left\{ \begin{bmatrix}4\\1\\0\end{bmatrix} \right\}\), dimension: 1 Since the sum of the dimensions of the eigenspaces is equal to the dimension of the matrix (3), the matrix \(A\) is nondefective.

Step 1: Find the eigenvalues from the characteristic equation

To find the eigenvalues of the matrix \(A\), we need to calculate the determinant of \((A-\lambda I\)) for the given matrix and equate it to 0, where \(I\) is the identity matrix and \(\lambda\) is the eigenvalue. \((A-\lambda I)=\begin{bmatrix}3-\lambda & -4 & -1\\0 & -1-\lambda & -1\\0 & -4 & 2-\lambda \end{bmatrix}\) We now calculate the determinant: \(|A-\lambda I| = (3-\lambda)[(-1-\lambda)(2-\lambda)+4] \) Now, we can simplify and find the roots. \( (3-\lambda)[\lambda^2-\lambda+2] = 0\) The roots of this equation are the eigenvalues. Upon solving this equation, we get 3 eigenvalues: \(\lambda_1 = 3\), \(\lambda_2 = -1\), and \(\lambda_3 = 2\).

Step 2: Find the eigenvectors corresponding to each eigenvalue

Now, we will find the eigenvectors corresponding to each eigenvalue by solving the following equation: \((A-\lambda I)X=0\). A. For \(\lambda_1 = 3\), the equation becomes: \((A-3I)X = \begin{bmatrix}0 & -4 & -1\\0 & -4 & -1\\0 & -4 & -1 \end{bmatrix}X = 0\) Row reduce to echelon form: \(\begin{bmatrix}1 & 1 & 1\\0 & 1 & 1\\0 & -2 & -1\end{bmatrix}\xrightarrow[Row \thinspace 3 = -2 \times Row \thinspace 2 + Row \thinspace 3]{Row\thinspace 1 = -Row \thinspace 2 + Row \thinspace 1} \begin{bmatrix}1 & 0 & 0\\0 & 1 & 1\\0 & 0 & 0\end{bmatrix}\) The eigenspace for \(\lambda_1 = \{X \thinspace | \thinspace X=a\begin{bmatrix}0\\-1\\1\end{bmatrix},\thinspace a\in \mathbb{R}\}\). Hence, the basis for this eigenspace is \(\{ \begin{bmatrix}0\\-1\\1\end{bmatrix} \}\) B. For \(\lambda_2 = -1\), the equation becomes: \((A+I)X = \begin{bmatrix}4 & -4 & -1\\0 & 0 & -1\\0 & -4 & 3 \end{bmatrix}X = 0\) Row reduce to echelon form: \(\begin{bmatrix}1 & -1 & -1/4\\0 & -4 & 3\\0 & 0 & -1\end{bmatrix}\) The eigenspace for \(\lambda_2 = \{X \thinspace | \thinspace X=a\begin{bmatrix}1\\1\\0\end{bmatrix},\thinspace a\in \mathbb{R}\}\). Hence, the basis for this eigenspace is \(\{ \begin{bmatrix}1\\1\\0\end{bmatrix} \}\) C. For \(\lambda_3 = 2\), the equation becomes: \((A-2I)X = \begin{bmatrix}1 & -4 & -1\\0 & -3 & -1\\0 & -4 & 0 \end{bmatrix}X = 0\) Row reduce to echelon form: \(\begin{bmatrix}1 & -4 & -1\\0 & -4 & 0\\0 & 0 & -1\end{bmatrix}\) The eigenspace for \(\lambda_3 = \{X \thinspace | \thinspace X=a\begin{bmatrix}4\\1\\0\end{bmatrix},\thinspace a\in \mathbb{R}\}\) . Hence, the basis for this eigenspace is \(\{ \begin{bmatrix}4\\1\\0\end{bmatrix} \}\)

Step 3: Determine the dimensions of each eigenspace and defective or nondefective matrix

For each eigenvalue, the dimension of the eigenspace is equal to the number of eigenvectors in its basis. - For \(\lambda_1 = 3\), the dimension of the eigenspace is 1 (basis: \(\{ \begin{bmatrix}0\\-1\\1\end{bmatrix} \}\)). - For \(\lambda_2 = -1\), the dimension of the eigenspace is 1 (basis: \(\{ \begin{bmatrix}1\\1\\0\end{bmatrix} \}\)). - For \(\lambda_3 = 2\), the dimension of the eigenspace is 1 (basis: \(\{ \begin{bmatrix}4\\1\\0\end{bmatrix} \}\)). Since the sum of the dimensions of the eigenspaces is equal to the dimension of the matrix (which is 3 in this case), the matrix \(A\) is nondefective.

Key Concepts

Eigenspace

The eigenspace of a given eigenvalue is essentially the collection of all eigenvectors associated with that eigenvalue, combined with the zero vector. When we solve for the eigenvectors of a matrix, we are really pinpointing directions in the vector space that don’t change direction during the transformation represented by that matrix. Consider it like a vector capsule for each eigenvalue.

Each eigenspace can be seen as a subspace of the original vector space. This means it can contain multiple eigenvectors, whose linear combinations also reside within this space. For instance, for our matrix, we identified the eigenspace for

• the eigenvalue \(\lambda_1 = 3\): any multiple of \(\begin{bmatrix}0\-1\1\end{bmatrix}\)
• the eigenvalue \(\lambda_2 = -1\): any multiple of \(\begin{bmatrix}1\1\0\end{bmatrix}\)
• the eigenvalue \(\lambda_3 = 2\): any multiple of \(\begin{bmatrix}4\1\0\end{bmatrix}\)

This explains how each specific vector points in the invariant direction that the transformation stabilizes. Understanding eigenspaces gives a clear geometric interpretation of linear transformations.

Defective Matrix

A defective matrix is one that doesn't have enough linearly independent eigenvectors to form a complete basis for its space. This occurs when the algebraic multiplicity of an eigenvalue doesn't match its geometric multiplicity. In simpler terms, it means you don't have enough eigenvectors as you'd expect given the size or dimensions of the matrix.

In the context of our exercise, the matrix A is nondefective, which means:

• Each eigenvalue of matrix A corresponds to a sufficient number of linearly independent eigenvectors.
• The sum of the dimensions of the eigenspaces adds up to the matrix's dimension.

This is a desirable property because it allows the matrix to be diagonalized, simplifying computations significantly when applying the matrix to other vectors or transformations. Knowing whether a matrix is defective or not is crucial when performing eigenvalue analysis, especially in practical applications where diagonalization is necessary.

Characteristic Equation

The characteristic equation is a fundamental equation that emerges from setting the determinant of \((A - \lambda I)\) equal to zero. It is of the form \[ det(A - \lambda I) = 0 \]This equation is instrumental because it reveals the eigenvalues of a matrix, each serving as the root of the equation.

Solving the characteristic equation gives us essential information regarding the behavior and properties of the matrix. For matrix A, by solving the characteristic equation derived from its determinant, we obtain the eigenvalues \(\lambda_1 = 3\), \(\lambda_2 = -1\), and \(\lambda_3 = 2\). Each eigenvalue represents a scalar indicating how much its associated eigenvector is stretched or shrunk during a transformation.

• It’s essentially the "DNA" of a matrix, giving structural insights.
• Understanding and computing it accurately sets the stage for further calculations.

In eigenvalue analysis, deriving and solving the characteristic equation isn’t just an academic exercise; it’s a step with real, tangible payoff in understanding a matrix's transformational characteristics.

Eigenvectors

Eigenvectors are special vectors associated with matrices that remain in the same direction after a transformation, albeit sometimes reversed or scaled by a factor—the corresponding eigenvalue. Finding eigenvectors is essential as they represent the axes of transformation for a given matrix.

The process to solve eigenvectors follows after identifying the eigenvalues from the characteristic equation. You substitute each eigenvalue back into the equation \((A - \lambda I)X = 0\) and solve for the vector \(X\). Each eigenvector helps describe the transformation brought about by the matrix, as it maintains its direction throughout the operation.

• For \(\lambda_1 = 3\), the eigenvector is \(\begin{bmatrix}0\-1\1\end{bmatrix}\)
• For \(\lambda_2 = -1\), the eigenvector is \(\begin{bmatrix}1\1\0\end{bmatrix}\)
• For \(\lambda_3 = 2\), the eigenvector is \(\begin{bmatrix}4\1\0\end{bmatrix}\)

Understanding and computing eigenvectors allows for deeper insights into the inherent symmetrical properties of the matrix. They simplify complex matrix operations by transforming them into more manageable diagonal forms.