Homogeneous Systems of Linear Constant-Coefficient Differential Equations

Introduction

A system of linear constant-coefficient differential equations can be written in matrix form as

where $X\in\mathbb{R}^n$ is an unknown vector-valued function, $A\in\mathbb{R}^{n\times n}$ is a constant coefficient matrix, and $F\in\mathbb{R}^n$ is the forcing term. When $F=0$, we have a homogeneous system. In the homogeneous case, $X\equiv 0$ is always a solution and the set of all solutions forms an $n$-dimensional vector space.

As in the single dimensional case, linearity leads to superposition, and specifying an initial condition $X(t_0)=X_0$ determines a unique solution.

A key feature of the system $X'=AX$ is that all solution behavior is encoded in the matrix $A$. The eigenvalues of $A$ determine the behavior of the system, and the eigenvectors (or generalized eigenvectors) determine the directions in which these behaviors occur.

Motivation: One-Dimensional Constant-Coefficient Linear ODEs

Recall that for a one-dimensional constant-coefficient linear ODE, we typically seek an exponential trial solution of the form $e^{rt}$. For example, for the second-order homogeneous equation

trying $x=e^{rt}$ leads to the characteristic equation

which produces three standard cases:

1. **Distinct real roots $r_1\neq r_2$:} $x(t)=C_1e^{r_1 t}+C_2e^{r_2 t}$.
2. **Repeated real root $r_1=r_2$:} $x(t)=(C_1+C_2 t)e^{rt}$.
3. **Complex roots $r=\alpha\pm i\beta$:} $x(t)=e^{\alpha t}\big(C_1\cos(\beta t)+C_2\sin(\beta t)\big)$.

Generalization to $n$ Dimensions: Eigenvalues and Eigenvectors

For the homogeneous system

we use the analogous exponential ansatz

where $v$ is a nonzero constant vector. Differentiating gives

Substituting into $X'=AX$ and cancelling the always nonzero factor $e^{\lambda t}$ yields

so nontrivial solutions exist only when

You may recall that this is precisely the characteristic equation of the matrix $A$. Solving the equation for $\lambda$ retrieves the eigenvalues of $A$.

For each eigenvalue—eigenvector pair $(\lambda,v)$, we obtain a solution $X(t)=e^{\lambda t}v$. The eigenvector $v$ specifies the direction in space along which the exponential behavior $e^{\lambda t}$ occurs.

As with scalar equations, there are three important cases for systems. In each case, the solution is built from eigenvalues and (generalized) eigenvectors, in direct analogy with characteristic roots and repeated roots in the one-dimensional setting.

Distinct Real Eigenvalues

Assume $A\in\mathbb{R}^{n\times n}$ has $n$ distinct real eigenvalues $\lambda_1,\dots,\lambda_n$ with corresponding eigenvectors $v_1,\dots,v_n$. Then the $n$ vector functions

are linearly independent solutions of $X'=AX$. The general solution is the linear combination

where the constants $c_1,\dots,c_n$ are determined by an initial condition $X(0)=X_0$.

Repeated Eigenvalues (Generalized Eigenvectors)

Suppose $\lambda$ is an eigenvalue of $A$ with multiplicity $m>1$. There are two cases for the eigenvectors:

1. **$m$ eigenvectors:} we get $m$ linearly independent eigenvectors.
2. **$n<m$ eigenvectors:} we get $n$ linearly independent eigenvectors, where $n<m$.

Recall that we will need $m$ linearly independent eigenvectors to create a basis. We first count how many eigenvectors we have, and then use generalized eigenvectors to make up the difference.

To find eigenvectors for $\lambda$, recall that we solve the homogeneous linear system

Row-reducing $A-\lambda I$ to RREF can let us count the number of eigenvectors relatively quickly: if the RREF has $p$ pivots, then there are $n-p$ free variables, so we get $n-p$ linearly independent eigenvectors for $\lambda$.

If there are $n-p < m$ eigenvectors, the eigenvectors will not give us a complete basis of solutions. To complete the basis for our solution space, we must find generalized eigenvectors.

Let $v$ be an eigenvector for $\lambda$, so $(A-\lambda I)v=0$. A generalized eigenvector $w$ is a vector that solves the linear system

Once we have $v$ and $w$, we obtain two independent solutions

Then the general solution contributed by this repeated eigenvalue is

If we still need more eigenvectors, we can continue the process by solving $(A-\lambda I)u=w$, and so on. Each new generalized eigenvector produces another solution with a higher power of $t$ multiplying $e^{\lambda t}$.

In general, if $\lambda$ has algebraic multiplicity $m$ and we can build generalized eigenvectors $v_1,v_2,\dots,v_m$ with the recurrence relation

then we obtain $m$ independent solutions

We can think of this as an analogy to the exponential response formula.

Complex Eigenvalues

For real matrices $A$, nonreal eigenvalues occur in conjugate pairs $\lambda=\alpha\pm i\beta$ with $\beta\neq 0$. If $v=p+iq$ is an eigenvector for $\alpha+i\beta$ (with $p,q\in\mathbb{R}^n$), then the complex solution

has real and imaginary parts that are real solutions. Using Euler’s formula, we obtain the real solution pair

This same construction generalizes directly to $n$ dimensions: each complex conjugate eigenvalue pair produces two linearly independent real solutions obtained from the real and imaginary parts of the corresponding complex solution.

If $\alpha\pm i\beta$ is a single complex conjugate pair, then the general (real) solution contributed by this pair is

with real constants $c_1,c_2$.

Examples

Distinct Real Eigenvalues

Consider the initial value problem

Since $A$ is upper triangular, its eigenvalues are the diagonal entries $\lambda_1=3$ and $\lambda_2=2$.

For $\lambda_1=3$, we solve $(A-3I)v=0$:

For $\lambda_2=2$, we solve $(A-2I)v=0$:

Thus the general solution is

Applying the initial condition and solving for $c_1$ and $c_2$

gives $c_2=-1$ and $c_1=2$, so

Repeated Eigenvalues with Generalized Eigenvector

Consider the homogeneous system

The characteristic polynomial is $(2-\lambda)^2$, so $A$ has a repeated eigenvalue $\lambda=2$.

Solving $(A-2I)v=0$ gives

Note that as there is 1 pivot, we have $n-p \to 2 - 1 = 1$ inearly independent eigenvectors.

To obtain a second independent solution, we find a generalized eigenvector $w$ by solving the linear system

Here

so $w$ must satisfy

From the first row we get $w_2=1$, and $w_1$ is free, so one convenient choice is

Then two independent solutions are

and therefore the general solution is

Repeated Eigenvalues with Two Eigenvectors

Consider

Here $\lambda=2$ has multiplicity $2$, and every nonzero vector is an eigenvector. In particular, two linearly independent eigenvectors are

Thus two independent solutions are $e^{2t}v_1$ and $e^{2t}v_2$, and the general solution is

Complex Eigenvalues

Consider

The eigenvalues are $\lambda=\pm i$ (a complex conjugate pair). Using the formulas from the complex-eigenvalue case, we obtain two real solutions

and therefore the general solution is