(I) Introduction and Complete Eigenvalues

Previously, we saw that if an \( n \times n \) coefficient matrix \( \mathbf{A} \) has \( n \) distinct eigenvalues \( \lambda_1, \dots, \lambda_n \), the general solution is:

\[ \vec{X}(t) = C_1\vec{v}_1e^{\lambda_1 t} + \dots + C_n\vec{v}_ne^{\lambda_n t} \]

We now consider the case when the characteristic equation \( \det(\mathbf{A} - \lambda \mathbf{I}) = 0 \) has at least one repeated root.

Complete Eigenvalues

An eigenvalue \( \lambda \) of multiplicity \( k \) is complete if it has \( k \) linearly independent associated eigenvectors.

If all eigenvalues of \( \mathbf{A} \) are complete, there exists a complete set of \( n \) linearly independent eigenvectors \( \vec{v}_1, \dots, \vec{v}_n \) corresponding to \( \lambda_1 , \dots, \lambda_n \) (each repeated with its multiplicity), and the general solution remains:

\[ \vec{X}(t) = C_1\vec{v}_1e^{\lambda_1 t} + \dots + C_n\vec{v}_ne^{\lambda_n t} \]
Example 1: Complete Multiplicity

Find the general solution of:

\[ \vec{X}' = \begin{bmatrix} 9 & 4 & 0 \\ -6 & -1 & 0 \\ 6 & 4 & 3 \end{bmatrix} \vec{X} \]

Step 1: Characteristic Equation.

\[ \det(\mathbf{A} - \lambda \mathbf{I}) = (3-\lambda) \begin{vmatrix} 9-\lambda & 4 \\ -6 & -1-\lambda \end{vmatrix} = -(\lambda-3)^2(\lambda-5) = 0 \]

Eigenvalues: \( \lambda_1 = 5 \) (multiplicity 1) and \( \lambda_2 = 3 \) (multiplicity 2).

Step 2: Find Eigenvectors.

  • For \( \lambda_1 = 5 \): \( (\mathbf{A} - 5\mathbf{I})\vec{v} = \vec{0} \implies \begin{bmatrix} 4 & 4 & 0 \\ -6 & -6 & 0 \\ 6 & 4 & -2 \end{bmatrix} \sim \begin{bmatrix} 1 & 0 & -1 \\ 0 & 1 & 1 \\ 0 & 0 & 0 \end{bmatrix} \implies \vec{v}_1 = \begin{bmatrix} 1 \\ -1 \\ 1 \end{bmatrix} \).
  • For \( \lambda_2 = 3 \): \( (\mathbf{A} - 3\mathbf{I})\vec{v} = \vec{0} \implies \begin{bmatrix} 6 & 4 & 0 \\ -6 & -4 & 0 \\ 6 & 4 & 0 \end{bmatrix} \sim \begin{bmatrix} 3 & 2 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} \implies 3a + 2b = 0 \).

This gives two linearly independent eigenvectors for \( \lambda = 3 \):

\[ \vec{v}_2 = \begin{bmatrix} 2 \\ -3 \\ 0 \end{bmatrix} \quad \text{and} \quad \vec{v}_3 = \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix} \]

General Solution:

\[ \vec{X}(t) = C_1 \begin{bmatrix} 1 \\ -1 \\ 1 \end{bmatrix} e^{5t} + C_2 \begin{bmatrix} 2 \\ -3 \\ 0 \end{bmatrix} e^{3t} + C_3 \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix} e^{3t} \]

(II) Defective Eigenvalues

An eigenvalue of multiplicity \( k \) is defective if it has fewer than \( k \) linearly independent associated eigenvectors. The difference \( d = k - p \) (where \( p \) is the number of independent eigenvectors) is called the defect.

Defective Eigenvalue of Multiplicity 2

If there is only one linearly independent eigenvector \( \vec{v}_1 \), we cannot use the standard form. Instead, we find a second solution of the form:

\[ \vec{X}_2(t) = (\vec{v}_1 t + \vec{v}_2)e^{\lambda t} \]

Substituting into \( \vec{X}' = \mathbf{A}\vec{X} \) leads to the following relations:

  1. \( (\mathbf{A} - \lambda \mathbf{I})\vec{v}_1 = \vec{0} \) (\( \vec{v}_1 \) is an eigenvector)
  2. \( (\mathbf{A} - \lambda \mathbf{I})\vec{v}_2 = \vec{v}_1 \) (\( \vec{v}_2 \) is a generalized eigenvector)

Algorithm for Multiplicity 2

  1. Find a non-zero solution \( \vec{v}_2 \) to \( (\mathbf{A} - \lambda \mathbf{I})^2 \vec{v}_2 = \vec{0} \) such that \( (\mathbf{A} - \lambda \mathbf{I})\vec{v}_2 = \vec{v}_1 \neq \vec{0} \).
  2. Construct the general solution: \[ \vec{X}(t) = C_1 \vec{v}_1 e^{\lambda t} + C_2 (\vec{v}_1 t + \vec{v}_2)e^{\lambda t} \]
Example 2

Find the general solution of: \( \vec{X}' = \begin{bmatrix} 1 & -3 \\ 3 & 7 \end{bmatrix} \vec{X} \).

Step 1: Eigenvalues.

\[ \det(\mathbf{A} - \lambda \mathbf{I}) = \lambda^2 - 8\lambda + 16 = (\lambda - 4)^2 = 0 \implies \lambda = 4 \text{ (multiplicity 2)} \]

Step 2: Find Generalized Eigenvectors.

\[ \mathbf{A} - 4\mathbf{I} = \begin{bmatrix} -3 & -3 \\ 3 & 3 \end{bmatrix}, \quad (\mathbf{A} - 4\mathbf{I})^2 = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} \]

Pick \( \vec{v}_2 = \begin{bmatrix} 1 \\ 0 \end{bmatrix} \). Then \( \vec{v}_1 = (\mathbf{A} - 4\mathbf{I})\vec{v}_2 = \begin{bmatrix} -3 \\ 3 \end{bmatrix} \neq \vec{0} \).

General Solution:

\[ \vec{X}(t) = C_1 \begin{bmatrix} -3 \\ 3 \end{bmatrix} e^{4t} + C_2 \left( \begin{bmatrix} -3 \\ 3 \end{bmatrix} t + \begin{bmatrix} 1 \\ 0 \end{bmatrix} \right) e^{4t} \]
A phase portrait plot for an improper node (degenerate node) system. The origin (0,0) is an unstable source. Trajectories curve away from the origin and are all tangent to the dashed eigenvector line y = -x. The arrows on the colored solution curves indicate outward motion as time increases.
Figure: Phase Portrait for a Defective System (Improper Node)

The origin is called an improper nodal source. Trajectories become tangent to the line \( x_1 = -x_2 \) as \( t \to \infty \).

(III) Generalized Eigenvectors

A rank \( r \) generalized eigenvector associated with eigenvalue \( \lambda \) is a vector \( \vec{u} \) such that:

\[ (\mathbf{A} - \lambda \mathbf{I})^r \vec{u} = \vec{0} \quad \text{but} \quad (\mathbf{A} - \lambda \mathbf{I})^{r-1} \vec{u} \neq \vec{0} \]

Algorithm for Length 3 Chain

  1. Find a rank 3 generalized eigenvector \( \vec{v}_3 \) such that \( (\mathbf{A} - \lambda \mathbf{I})^3 \vec{v}_3 = \vec{0} \).
  2. Define \( \vec{v}_2 = (\mathbf{A} - \lambda \mathbf{I})\vec{v}_3 \).
  3. Define \( \vec{v}_1 = (\mathbf{A} - \lambda \mathbf{I})\vec{v}_2 \).
  4. The solutions are:
    \( \vec{X}_1(t) = \vec{v}_1 e^{\lambda t} \)
    \( \vec{X}_2(t) = (\vec{v}_1 t + \vec{v}_2) e^{\lambda t} \)
    \( \vec{X}_3(t) = \left( \frac{1}{2}\vec{v}_1 t^2 + \vec{v}_2 t + \vec{v}_3 \right) e^{\lambda t} \)