Chapter 2 • Theory & Derivations

Homogeneous Matrix Systems with Constant Coefficients & Generalized Eigenvectors

Algebraic spectral theory of first-order vector equations: real distinct eigenvalues, complex conjugate modes and rotation matrices, defective matrices, and chains of generalized eigenvectors.

§2.1 The Matrix System x' = Ax with Real Distinct Eigenvalues

1. Eigenpair Formulation

To solve the homogeneous system $\vec{x}' = \mathbf{A}\vec{x}$ where $\mathbf{A} \in \mathbb{R}^{n \times n}$ is a constant matrix, we seek exponential vector solutions of the form $\vec{x}(t) = e^{\lambda t} \vec{v}$. Substituting into the system:

$$\lambda e^{\lambda t} \vec{v} = \mathbf{A} e^{\lambda t} \vec{v} \implies (\mathbf{A} - \lambda \mathbf{I}) \vec{v} = \vec{0}$$

Non-trivial solutions $\vec{v} \ne \vec{0}$ exist if and only if $\lambda$ satisfies the characteristic equation $\det(\mathbf{A} - \lambda \mathbf{I}) = 0$.

If $\mathbf{A}$ possesses $n$ real distinct eigenvalues $\lambda_1, \lambda_2, \dots, \lambda_n$, the corresponding eigenvectors $\vec{v}_1, \vec{v}_2, \dots, \vec{v}_n$ are linearly independent in $\mathbb{R}^n$. The general solution is a linear superposition of the fundamental modes:

$$\vec{x}(t) = \sum_{k=1}^n c_k e^{\lambda_k t} \vec{v}_k = c_1 e^{\lambda_1 t} \vec{v}_1 + c_2 e^{\lambda_2 t} \vec{v}_2 + \dots + c_n e^{\lambda_n t} \vec{v}_n$$

§2.2 Complex Conjugate Eigenvalues, Elliptic Orbits & Spiral Stability

1. Real Solutions from Complex Eigenpairs

When the real matrix $\mathbf{A}$ possesses a complex conjugate pair of eigenvalues $\lambda = \alpha \pm i\beta$ ($\beta \ne 0$), the corresponding eigenvector is also complex: $\vec{v} = \vec{u} + i\vec{w}$ where $\vec{u}, \vec{w} \in \mathbb{R}^n$. The complex solution is:

$$\vec{z}(t) = e^{(\alpha + i\beta)t} (\vec{u} + i\vec{w}) = e^{\alpha t} (\cos\beta t + i\sin\beta t)(\vec{u} + i\vec{w})$$

Splitting $\vec{z}(t)$ into its real and imaginary parts $\vec{z}(t) = \vec{x}_1(t) + i\vec{x}_2(t)$:

$$\begin{aligned} \vec{x}_1(t) &= \text{Re}[\vec{z}(t)] = e^{\alpha t} (\vec{u} \cos\beta t - \vec{w} \sin\beta t) \\ \vec{x}_2(t) &= \text{Im}[\vec{z}(t)] = e^{\alpha t} (\vec{u} \sin\beta t + \vec{w} \cos\beta t) \end{aligned}$$

Theorem 2.1: The vector functions $\vec{x}_1(t)$ and $\vec{x}_2(t)$ are linearly independent real solutions of $\vec{x}' = \mathbf{A}\vec{x}$ on $\mathbb{R}$. Their span describes a 2D plane in which orbits rotate with angular frequency $\beta$ while expanding ($\alpha > 0$) or decaying ($\alpha < 0$) exponentially.

§2.3 Repeated Eigenvalues, Defective Matrices & Chains of Generalized Eigenvectors

1. Algebraic vs Geometric Multiplicity

Let $\lambda$ be an eigenvalue of $\mathbf{A}$ with algebraic multiplicity $m_a = k$ (meaning $(\lambda - \lambda_0)^k$ divides the characteristic polynomial). The geometric multiplicity $m_g = \dim \ker(\mathbf{A} - \lambda \mathbf{I})$ is the number of linearly independent eigenvectors associated with $\lambda$.

If $m_g < m_a$, the matrix $\mathbf{A}$ is called defective. In this case, ordinary eigenvectors cannot span the full solution subspace, requiring generalized eigenvectors.

2. Chains of Generalized Eigenvectors

A chain of generalized eigenvectors $\{\vec{v}_1, \vec{v}_2, \dots, \vec{v}_k\}$ associated with eigenvalue $\lambda$ satisfies:

$$\begin{aligned} (\mathbf{A} - \lambda \mathbf{I}) \vec{v}_1 &= \vec{0} \quad (\text{genuine eigenvector}) \\ (\mathbf{A} - \lambda \mathbf{I}) \vec{v}_2 &= \vec{v}_1 \implies (\mathbf{A} - \lambda \mathbf{I})^2 \vec{v}_2 = \vec{0} \\ (\mathbf{A} - \lambda \mathbf{I}) \vec{v}_3 &= \vec{v}_2 \implies (\mathbf{A} - \lambda \mathbf{I})^3 \vec{v}_3 = \vec{0} \\ &\;\;\vdots \\ (\mathbf{A} - \lambda \mathbf{I}) \vec{v}_k &= \vec{v}_{k-1} \implies (\mathbf{A} - \lambda \mathbf{I})^k \vec{v}_k = \vec{0} \end{aligned}$$

3. Form of Independent Solutions

For a chain of length 2 ($\vec{v}_1, \vec{v}_2$), the two linearly independent solutions are:

$$\begin{aligned} \vec{x}_1(t) &= e^{\lambda t} \vec{v}_1 \\ \vec{x}_2(t) &= e^{\lambda t} (t \vec{v}_1 + \vec{v}_2) \end{aligned}$$

For a chain of length 3 ($\vec{v}_1, \vec{v}_2, \vec{v}_3$):

$$\vec{x}_3(t) = e^{\lambda t} \left( \frac{t^2}{2} \vec{v}_1 + t \vec{v}_2 + \vec{v}_3 \right)$$

This matches the Taylor expansion of $e^{\mathbf{A}t} \vec{v}_k = e^{\lambda t} e^{(\mathbf{A} - \lambda \mathbf{I})t} \vec{v}_k$.

TIERED UNIVERSITY HONORS PROBLEMS

Step-by-Step Solved Examination Problems

Comprehensive analytical derivations, multi-tier solutions (Foundational, Intermediate Exam, and Honors/Proof Challenge) with complete line-by-line verification.

Tier 1 • Foundational Example 2.1: Matrix Exponential for Defective Repeated Eigenvalue Matrix

Compute the matrix exponential $e^{\mathbf{A}t}$ for the defective matrix:

\mathbf{A} = \begin{pmatrix} 2 & 1 \\ 0 & 2 \end{pmatrix}

Step 1: Nilpotent Decomposition
Notice that $\mathbf{A}$ can be split into a scalar multiple of identity and a nilpotent upper-triangular matrix:

\mathbf{A} = 2\mathbf{I} + \mathbf{N}, \quad \text{where } \mathbf{N} = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}

Observe that $\mathbf{N}^2 = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix} \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix} = \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix}$. Thus $\mathbf{N}$ is nilpotent of degree 2.

Step 2: Commuting Matrices Property
Since $(2\mathbf{I}t)(\mathbf{N}t) = (\mathbf{N}t)(2\mathbf{I}t)$, we can multiply their exponentials:

e^{\mathbf{A}t} = e^{2\mathbf{I}t + \mathbf{N}t} = e^{2\mathbf{I}t} e^{\mathbf{N}t}

Step 3: Exponentiation of Terms

e^{2\mathbf{I}t} = e^{2t} \mathbf{I} = \begin{pmatrix} e^{2t} & 0 \\ 0 & e^{2t} \end{pmatrix}

e^{\mathbf{N}t} = \mathbf{I} + t\mathbf{N} + \frac{t^2}{2!} \mathbf{N}^2 + \dots = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} + t \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix} = \begin{pmatrix} 1 & t \\ 0 & 1 \end{pmatrix}

Step 4: Matrix Multiplication

e^{\mathbf{A}t} = e^{2t} \begin{pmatrix} 1 & t \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} e^{2t} & t e^{2t} \\ 0 & e^{2t} \end{pmatrix}

Final Answer & Physical Insight

$e^{\mathbf{A}t} = \begin{pmatrix} e^{2t} & t e^{2t} \\ 0 & e^{2t} \end{pmatrix}$

Tier 2 • Intermediate Exam Example 2.2: Putzer's Algorithm for 3x3 Nilpotent Jordan Block

Use Putzer's algorithm to compute the matrix exponential $e^{\mathbf{A}t}$ for the $3 \times 3$ matrix:

\mathbf{A} = \begin{pmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{pmatrix}

Step 1: Eigenvalues
The characteristic polynomial is $\det(\mathbf{A} - \lambda \mathbf{I}) = -\lambda^3 = 0 \implies \lambda_1 = \lambda_2 = \lambda_3 = 0$.

Step 2: Putzer's polynomial matrices $\mathbf{P}_k$

\mathbf{P}_0 = \mathbf{I} = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}

\mathbf{P}_1 = \mathbf{A} - \lambda_1 \mathbf{I} = \mathbf{A} = \begin{pmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{pmatrix}

\mathbf{P}_2 = (\mathbf{A} - \lambda_2 \mathbf{I})\mathbf{P}_1 = \mathbf{A}^2 = \begin{pmatrix} 0 & 0 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix}

Step 3: Scalar differential equations for $r_k(t)$
$r_1'(t) = 0, r_1(0) = 1 \implies r_1(t) = 1$.
$r_2'(t) = r_1(t) = 1, r_2(0) = 0 \implies r_2(t) = t$.
$r_3'(t) = r_2(t) = t, r_3(0) = 0 \implies r_3(t) = \frac{t^2}{2}$.

Step 4: Matrix exponential assembly

e^{\mathbf{A}t} = r_1(t)\mathbf{P}_0 + r_2(t)\mathbf{P}_1 + r_3(t)\mathbf{P}_2 = \mathbf{I} + t\mathbf{A} + \frac{t^2}{2}\mathbf{A}^2 = \begin{pmatrix} 1 & t & t^2/2 \\ 0 & 1 & t \\ 0 & 0 & 1 \end{pmatrix}

Final Answer & Physical Insight

$e^{\mathbf{A}t} = \begin{pmatrix} 1 & t & t^2/2 \\ 0 & 1 & t \\ 0 & 0 & 1 \end{pmatrix}$

Tier 3 • Honors Challenge Example 2.3: Jordan Canonical Form Derivation for Highly Defective 3x3 System

Solve the defective system $\vec{x}' = \mathbf{A}\vec{x}$ where $\mathbf{A} = \begin{pmatrix} 3 & 1 & 0 \\ 0 & 3 & 1 \\ 0 & 0 & 3 \end{pmatrix}$ using generalized eigenvector chains.

Step 1: Spectral Analysis
The characteristic polynomial is $(\lambda - 3)^3 = 0 \implies \lambda = 3$ with algebraic multiplicity $m_a = 3$. The null space $(\mathbf{A} - 3\mathbf{I})\vec{v} = \vec{0}$ has rank 2, so the geometric multiplicity is $m_g = 3 - 2 = 1$. The matrix is defective with a single Jordan block of size 3.

Step 2: Construct the Jordan Chain
We seek a generalized eigenvector $\vec{v}_3$ of rank 3:

(\mathbf{A} - 3\mathbf{I})^3 = \mathbf{0}, \quad (\mathbf{A} - 3\mathbf{I})^2 \vec{v}_3 \ne \vec{0}

\mathbf{A} - 3\mathbf{I} = \begin{pmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{pmatrix}, \quad (\mathbf{A} - 3\mathbf{I})^2 = \begin{pmatrix} 0 & 0 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix}

Choose $\vec{v}_3 = \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix}$. Then:

\vec{v}_2 = (\mathbf{A} - 3\mathbf{I}) \vec{v}_3 = \begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix}

\vec{v}_1 = (\mathbf{A} - 3\mathbf{I}) \vec{v}_2 = \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix} \quad (\text{genuine eigenvector})

Step 3: Construct Three Linearly Independent Solutions

\begin{aligned} \vec{x}_1(t) &= e^{3t} \vec{v}_1 = e^{3t} \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix} \\ \vec{x}_2(t) &= e^{3t} (t \vec{v}_1 + \vec{v}_2) = e^{3t} \begin{pmatrix} t \\ 1 \\ 0 \end{pmatrix} \\ \vec{x}_3(t) &= e^{3t} \left( \frac{t^2}{2} \vec{v}_1 + t \vec{v}_2 + \vec{v}_3 \right) = e^{3t} \begin{pmatrix} t^2/2 \\ t \\ 1 \end{pmatrix} \end{aligned}

Step 4: General Solution

\vec{x}(t) = e^{3t} \begin{pmatrix} c_1 + c_2 t + c_3 \frac{t^2}{2} \\ c_2 + c_3 t \\ c_3 \end{pmatrix}

Final Answer & Physical Insight

$\vec{x}(t) = e^{3t} \begin{pmatrix} c_1 + c_2 t + c_3 \frac{t^2}{2} \\ c_2 + c_3 t \\ c_3 \end{pmatrix}$