Constant Coefficient Systems

Substituting the trial solution $𝐲 = e^{λ t} 𝐯$ into \mathbf{y}' = \mathbf{A}\mathbf{y} turns the differential equation into the eigenvalue equation $𝐀 𝐯 = λ 𝐯$ . When $𝐀$ has $n$ independent eigenvectors, the $n$ resulting solutions form a fundamental set and the problem is finished.

Quick Reference

Item	Statement
Matrix form	$\frac{d 𝐲}{d t} = 𝐀 𝐲$
Trial solution	$𝐲 (t) = e^{λ t} 𝐯$
Resulting condition	$𝐀 𝐯 = λ 𝐯$ , so $λ$ is an eigenvalue and $𝐯$ an eigenvector
Characteristic equation	$p (λ) = det (𝐀 - λ 𝐈) = 0$
Non-defective matrix	$𝐀$ has $n$ linearly independent eigenvectors
General solution	$𝐲 = c_{1} e^{λ_{1} t} 𝐯^{(1)} + \dots + c_{n} e^{λ_{n} t} 𝐯^{(n)}$
Guaranteed non-defective if	$𝐀$ is symmetric, skew-symmetric, or has $n$ distinct eigenvalues
Complex eigenvalue $α + i β$	Real solutions are $Re (e^{λ t} 𝐯)$ and $Im (e^{λ t} 𝐯)$
Conjugate pairs	If $λ$ is an eigenvalue of a real $𝐀$ , so is $\overset{―}{λ}$ , with eigenvector $\overset{―}{𝐯}$

Matrix Form

Consider the system of differential equations

\left\{ \begin{aligned} \frac{dy_1}{dt} &= a_{11}y_1 + a_{12}y_2 + \cdots + a_{1n}y_n \\ \frac{dy_2}{dt} &= a_{21}y_1 + a_{22}y_2 + \cdots + a_{2n}y_n \\ &\ \ \vdots \\ \frac{dy_n}{dt} &= a_{n1}y_1 + a_{n2}y_2 + \cdots + a_{nn}y_n \end{aligned} \right.

where the $a_{i j}$ are constants. We want to find $y_{1} (t), y_{2} (t), \dots, y_{n} (t)$ satisfying these equations simultaneously.

The system can be written as

\frac{d 𝐲}{d t} = 𝐀 𝐲

where

𝐀 = [\begin{matrix} a_{11} & \dots & a_{1 n} \\ a_{21} & \dots & a_{2 n} \\ ⋮ & ⋱ & ⋮ \\ a_{n 1} & \dots & a_{n n} \end{matrix}], 𝐲 = [\begin{matrix} y_{1} \\ y_{2} \\ ⋮ \\ y_{n} \end{matrix}], \frac{d 𝐲}{d t} = [\begin{matrix} \frac{d y_{1}}{d t} \\ \frac{d y_{2}}{d t} \\ ⋮ \\ \frac{d y_{n}}{d t} \end{matrix}] .

Differentiating a vector means differentiating each component, so $\frac{d 𝐲}{d t}$ is defined entrywise.

Why Eigenvectors Appear

We try a solution in which every component is the same exponential times a constant:

y_{1} (t) = v_{1} e^{λ t}, \dots, y_{n} (t) = v_{n} e^{λ t}

where $λ$ and $v_{1}, \dots, v_{n}$ are constants to be determined. In vector form,

\underset{𝐲 (t)}{\underset{⏟}{[\begin{matrix} y_{1} (t) \\ ⋮ \\ y_{n} (t) \end{matrix}]}} = e^{λ t} \underset{𝐯}{\underset{⏟}{[\begin{matrix} v_{1} \\ ⋮ \\ v_{n} \end{matrix}]}}

Differentiating both sides,

\underbrace{\begin{bmatrix} y_1'(t) \\ \vdots \\ y_n'(t) \end{bmatrix}}_{d\mathbf{y}/dt} = \lambda e^{\lambda t}\mathbf{v}

and substituting into \mathbf{y}' = \mathbf{A}\mathbf{y},

\begin{aligned} \lambda e^{\lambda t}\mathbf{v} &= \mathbf{A}\left(e^{\lambda t}\mathbf{v}\right) \\ &= e^{\lambda t}\mathbf{A}\mathbf{v} \end{aligned}

where the scalar $e^{λ t}$ was moved past the matrix, which is legitimate since scalars commute with matrices. Because $e^{λ t} \neq 0$ , we may cancel it from both sides:

𝐀 𝐯 = λ 𝐯 .

This means $𝐯$ is an eigenvector and $λ$ an eigenvalue of $𝐀$ . So $e^{λ t} 𝐯$ is a nontrivial solution of the system if and only if $λ$ is an eigenvalue and $𝐯$ is a corresponding eigenvector.

For a system of $n$ linear differential equations, two cases must be distinguished:

$𝐀$ has $n$ linearly independent eigenvectors. Such a matrix is called non-defective.
$𝐀$ does not have $n$ linearly independent eigenvectors. Such a matrix is called defective, and is treated in Section 5.2.

Non-defective Coefficient Matrix

Assume $𝐀$ has $n$ linearly independent eigenvectors $𝐯^{(1)}, \dots, 𝐯^{(n)}$ corresponding to eigenvalues $λ_{1}, \dots, λ_{n}$ .

This assumption holds in most applications, and in particular whenever $𝐀$

is symmetric, that is, $a_{i j} = a_{j i}$ ;
is skew-symmetric, that is, $a_{i j} = - a_{j i}$ , in which case all diagonal entries must be zero;
has $n$ distinct eigenvalues.

In this case, $n$ corresponding solutions are

𝐲^{(1)} = e^{λ_{1} t} 𝐯^{(1)}, \dots, 𝐲^{(n)} = e^{λ_{n} t} 𝐯^{(n)} .

Computing the Wronskian of these solutions shows that they are linearly independent, so the general solution of \mathbf{y}' = \mathbf{A}\mathbf{y} is

Find the general solution of

\left\{ \begin{aligned} \frac{dy_1}{dt} &= 3y_1 - 2y_2 \\[1ex] \frac{dy_2}{dt} &= -y_1 + 2y_2 \end{aligned} \right.

Solution

Rewrite the system as

\frac{d 𝐲}{d t} = 𝐀 𝐲, 𝐀 = [\begin{matrix} 3 & - 2 \\ - 1 & 2 \end{matrix}] .

Step 1: eigenvalues.

\begin{aligned} p(\lambda) &= \det\left(\mathbf{A} - \lambda\mathbf{I}\right) \\ &= \begin{vmatrix} 3-\lambda & -2 \\ -1 & 2-\lambda \end{vmatrix} \\ &= (3-\lambda)(2-\lambda) - 2 \\ &= \lambda^{2} - 5\lambda + 4 = (\lambda-1)(\lambda-4) = 0 \end{aligned}

Therefore the eigenvalues are $λ_{1} = 1$ and $λ_{2} = 4$ .

Step 2: eigenvectors. Because $𝐀$ has two distinct eigenvalues, we know it has two linearly independent eigenvectors. For $λ_{1} = 1$ , solve $(𝐀 - λ_{1} 𝐈) 𝐯^{(1)} = 0$ :

\begin{bmatrix} 2 & -2 \\ -1 & 1 \end{bmatrix} \begin{bmatrix} v_1^{(1)} \\ v_2^{(1)} \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix} \quad \Longrightarrow \quad \left\{ \begin{aligned} 2v_1^{(1)} - 2v_2^{(1)} &= 0 \\ -v_1^{(1)} + v_2^{(1)} &= 0 \end{aligned} \right.

The two equations reduce to the single equation $v_{1}^{(1)} - v_{2}^{(1)} = 0$ . The rows becoming dependent is exactly what must happen, since $λ$ was chosen to make the determinant vanish. Thus $v_{1}^{(1)} = v_{2}^{(1)}$ and we may take

𝐯^{(1)} = [\begin{matrix} 1 \\ 1 \end{matrix}] .

Similarly, for $λ_{2} = 4$ ,

[\begin{matrix} - 1 & - 2 \\ - 1 & - 2 \end{matrix}] [\begin{matrix} v_{1}^{(2)} \\ v_{2}^{(2)} \end{matrix}] = [\begin{matrix} 0 \\ 0 \end{matrix}] ⟹ v_{1}^{(2)} + 2 v_{2}^{(2)} = 0

so we may take

𝐯^{(2)} = [\begin{matrix} 2 \\ - 1 \end{matrix}] .

Step 3: assemble. Two linearly independent solutions are

𝐲^{(1)} = e^{t} [\begin{matrix} 1 \\ 1 \end{matrix}] and 𝐲^{(2)} = e^{4 t} [\begin{matrix} 2 \\ - 1 \end{matrix}],

and the general solution is

[\begin{matrix} y_{1} \\ y_{2} \end{matrix}] = c_{1} e^{t} [\begin{matrix} 1 \\ 1 \end{matrix}] + c_{2} e^{4 t} [\begin{matrix} 2 \\ - 1 \end{matrix}]

or, componentwise,

y_{1} (t) = c_{1} e^{t} + 2 c_{2} e^{4 t} and y_{2} (t) = c_{1} e^{t} - c_{2} e^{4 t} .

Complex-Valued Eigenvalues

Consider a system of $n$ linear homogeneous equations with constant coefficients

\mathbf{y}' = \mathbf{A}\mathbf{y} . \tag{i}

When the elements of $𝐀$ are real, if $α + i β$ is an eigenvalue of $𝐀$ , then its complex conjugate $α - i β$ is also an eigenvalue. Two linearly independent real solutions of (i) corresponding to these two eigenvalues are

𝐲^{(1)} = Re (e^{(α + i β) t} 𝐯^{(1)}), 𝐲^{(2)} = Im (e^{(α + i β) t} 𝐯^{(1)})

where $𝐯^{(1)}$ is an eigenvector corresponding to $α + i β$ , and $Re$ and $Im$ denote the real and imaginary parts of the vector.

Why the conjugate is automatic, and why the real and imaginary parts work

If the components of $𝐀$ are real, its characteristic equation $p (λ) = det (𝐀 - λ 𝐈) = 0$ is a polynomial with real coefficients. Therefore any complex roots occur in conjugate pairs: if $λ_{1} = α + i β$ is an eigenvalue, so is $\overset{―}{λ_{1}} = α - i β$ .

Suppose $𝐯^{(1)}$ is an eigenvector corresponding to $λ_{1}$ , so

(𝐀 - λ_{1} 𝐈) 𝐯^{(1)} = 0 .

Taking the complex conjugate of both sides,

\begin{aligned} \overline{\left(\mathbf{A} - \lambda_1\mathbf{I}\right)\mathbf{v}^{(1)}} &= \overline{\mathbf{0}} \\ \left(\mathbf{A} - \overline{\lambda_1}\mathbf{I}\right)\overline{\mathbf{v}^{(1)}} &= \mathbf{0} \end{aligned}

using $\overset{―}{𝐀} = 𝐀$ since $𝐀$ is real, along with $\overset{―}{𝐈} = 𝐈$ and $\overset{―}{0} = 0$ . This shows $\overset{―}{𝐯^{(1)}}$ is an eigenvector corresponding to $\overset{―}{λ_{1}}$ , so the second eigenvector requires no computation.

Two solutions of \mathbf{y}' = \mathbf{A}\mathbf{y} corresponding to these eigenvalues are

{\tilde{𝐲}}^{(1)} = e^{λ_{1} t} 𝐯^{(1)}, {\tilde{𝐲}}^{(2)} = e^{\overset{―}{λ_{1}} t} \overset{―}{𝐯^{(1)}},

which are complex conjugates of each other. Since the system is linear with real coefficients, any combination of solutions is a solution, so we obtain two real-valued ones by taking

𝐲^{(1)} = \frac{1}{2} ({\tilde{𝐲}}^{(1)} + {\tilde{𝐲}}^{(2)}) = Re ({\tilde{𝐲}}^{(1)})

and

𝐲^{(2)} = \frac{1}{2 i} ({\tilde{𝐲}}^{(1)} - {\tilde{𝐲}}^{(2)}) = Im ({\tilde{𝐲}}^{(1)}) .

These are the familiar identities $Re (z) = \frac{1}{2} (z + \overset{―}{z})$ and $Im (z) = \frac{1}{2 i} (z - \overset{―}{z})$ , applied componentwise.

Practical consequence. Compute the eigenvector for only one member of a conjugate pair, expand $e^{(α + i β) t} 𝐯$ using Euler's formula, and read off the real and imaginary parts. The conjugate eigenvalue contributes nothing new.

Solve the initial-value problem

\mathbf{y}' = \begin{bmatrix} 1 & -1 \\ 1 & 1 \end{bmatrix}\mathbf{y}, \qquad \mathbf{y}(0) = \begin{bmatrix} 1 \\ 2 \end{bmatrix} .

Solution

Step 1: eigenvalues.

\begin{aligned} \det\left(\mathbf{A} - \lambda\mathbf{I}\right) &= \begin{vmatrix} 1-\lambda & -1 \\ 1 & 1-\lambda \end{vmatrix} \\ &= (1-\lambda)^{2} + 1 \\ &= \lambda^{2} - 2\lambda + 2 = 0 . \end{aligned}

The eigenvalues are $λ_{1} = 1 + i$ and $λ_{2} = 1 - i$ .

Step 2: one eigenvector. From $(𝐀 - λ_{1} 𝐈) 𝐯^{(1)} = 0$ ,

[\begin{matrix} - i & - 1 \\ 1 & - i \end{matrix}] [\begin{matrix} v_{1}^{(1)} \\ v_{2}^{(1)} \end{matrix}] = [\begin{matrix} 0 \\ 0 \end{matrix}] .

The first row gives $- i v_{1}^{(1)} - v_{2}^{(1)} = 0$ , so $v_{2}^{(1)} = - i v_{1}^{(1)}$ . Choosing $v_{1}^{(1)} = 1$ ,

𝐯^{(1)} = [\begin{matrix} 1 \\ - i \end{matrix}] .

Without further calculation, the eigenvector corresponding to $λ_{2} = 1 - i$ is the conjugate,

\overset{―}{[\begin{matrix} 1 \\ - i \end{matrix}]} = [\begin{matrix} 1 \\ i \end{matrix}] .

Step 3: expand and take real and imaginary parts. Two complex solutions are

{\tilde{𝐲}}^{(1)} = e^{(1 + i) t} [\begin{matrix} 1 \\ - i \end{matrix}], {\tilde{𝐲}}^{(2)} = e^{(1 - i) t} [\begin{matrix} 1 \\ i \end{matrix}] .

Expanding the first with $e^{(1 + i) t} = e^{t} (\cos t + i \sin t)$ , and noting that $- i (\cos t + i \sin t) = \sin t - i \cos t$ ,

{\tilde{𝐲}}^{(1)} = e^{t} [\begin{matrix} \cos t + i \sin t \\ \sin t - i \cos t \end{matrix}] .

Therefore a set of real-valued solutions is

𝐲^{(1)} = Re ({\tilde{𝐲}}^{(1)}) = [\begin{matrix} e^{t} \cos t \\ e^{t} \sin t \end{matrix}], 𝐲^{(2)} = Im ({\tilde{𝐲}}^{(1)}) = [\begin{matrix} e^{t} \sin t \\ - e^{t} \cos t \end{matrix}] .

Step 4: general solution.

𝐲 = [\begin{matrix} y_{1} (t) \\ y_{2} (t) \end{matrix}] = c_{1} [\begin{matrix} e^{t} \cos t \\ e^{t} \sin t \end{matrix}] + c_{2} [\begin{matrix} e^{t} \sin t \\ - e^{t} \cos t \end{matrix}] .

Step 5: apply the initial condition. At $t = 0$ , $\cos 0 = 1$ and $\sin 0 = 0$ , so

[\begin{matrix} 1 \\ 2 \end{matrix}] = c_{1} [\begin{matrix} 1 \\ 0 \end{matrix}] + c_{2} [\begin{matrix} 0 \\ - 1 \end{matrix}] ⟹ c_{1} = 1, c_{2} = - 2 .

Therefore the solution of the initial-value problem is

[\begin{matrix} y_{1} (t) \\ y_{2} (t) \end{matrix}] = [\begin{matrix} e^{t} \cos t \\ e^{t} \sin t \end{matrix}] - 2 [\begin{matrix} e^{t} \sin t \\ - e^{t} \cos t \end{matrix}]

y_{1} (t) = e^{t} (\cos t - 2 \sin t) and y_{2} (t) = e^{t} (\sin t + 2 \cos t) .

The trajectory is a spiral: the factor $e^{t}$ grows while the trigonometric factors rotate, so the solution spirals outward from the origin.

Exercises

Solve \mathbf{y}' = \begin{bmatrix} 1 & 2 \\ 2 & 1 \end{bmatrix}\mathbf{y}.

Solution

det [\begin{matrix} 1 - λ & 2 \\ 2 & 1 - λ \end{matrix}] = (1 - λ)^{2} - 4 = λ^{2} - 2 λ - 3 = (λ - 3) (λ + 1) .

For $λ = 3$ : $[\begin{matrix} - 2 & 2 \\ 2 & - 2 \end{matrix}] 𝐯 = 0$ gives $𝐯 = (1, 1)^{T}$ .
For $λ = - 1$ : $[\begin{matrix} 2 & 2 \\ 2 & 2 \end{matrix}] 𝐯 = 0$ gives $𝐯 = (1, - 1)^{T}$ .

𝐲 = c_{1} e^{3 t} [\begin{matrix} 1 \\ 1 \end{matrix}] + c_{2} e^{- t} [\begin{matrix} 1 \\ - 1 \end{matrix}] .

Note that $𝐀$ is symmetric here, which guaranteed in advance that it would be non-defective.

Solve \mathbf{y}' = \begin{bmatrix} 0 & -4 \\ 1 & 0 \end{bmatrix}\mathbf{y} and describe the trajectories.

Solution

det [\begin{matrix} - λ & - 4 \\ 1 & - λ \end{matrix}] = λ^{2} + 4 = 0 ⟹ λ = \pm 2 i .

For $λ = 2 i$ : $[\begin{matrix} - 2 i & - 4 \\ 1 & - 2 i \end{matrix}] 𝐯 = 0$ . The second row gives $v_{1} = 2 i v_{2}$ , so with $v_{2} = 1$ , $𝐯 = (2 i, 1)^{T}$ .

Then

e^{2 i t} [\begin{matrix} 2 i \\ 1 \end{matrix}] = (\cos 2 t + i \sin 2 t) [\begin{matrix} 2 i \\ 1 \end{matrix}] = [\begin{matrix} - 2 \sin 2 t + 2 i \cos 2 t \\ \cos 2 t + i \sin 2 t \end{matrix}] .

Therefore

𝐲 = c_{1} [\begin{matrix} - 2 \sin 2 t \\ \cos 2 t \end{matrix}] + c_{2} [\begin{matrix} 2 \cos 2 t \\ \sin 2 t \end{matrix}] .

Since $α = 0$ , there is no growth or decay: the trajectories are closed ellipses traversed with period $π$ . This is the phase portrait of an undamped oscillator.

Solve the initial value problem \mathbf{y}' = \begin{bmatrix} 2 & 3 \\ 0 & -1 \end{bmatrix}\mathbf{y}, $𝐲 (0) = [\begin{matrix} 4 \\ 1 \end{matrix}]$ .

Solution

The matrix is triangular, so the eigenvalues are the diagonal entries $λ = 2$ and $λ = - 1$ .

For $λ = 2$ : $[\begin{matrix} 0 & 3 \\ 0 & - 3 \end{matrix}] 𝐯 = 0$ gives $v_{2} = 0$ , so $𝐯 = (1, 0)^{T}$ .
For $λ = - 1$ : $[\begin{matrix} 3 & 3 \\ 0 & 0 \end{matrix}] 𝐯 = 0$ gives $v_{1} = - v_{2}$ , so $𝐯 = (1, - 1)^{T}$ .

𝐲 = c_{1} e^{2 t} [\begin{matrix} 1 \\ 0 \end{matrix}] + c_{2} e^{- t} [\begin{matrix} 1 \\ - 1 \end{matrix}] .

At $t = 0$ : $c_{1} + c_{2} = 4$ and $- c_{2} = 1$ , so $c_{2} = - 1$ and $c_{1} = 5$ . Therefore

y_{1} = 5 e^{2 t} - e^{- t}, y_{2} = e^{- t} .

Frequently Asked Questions

Why does the trial solution

e^{λ t} 𝐯

force

𝐯

to be an eigenvector?

Substituting gives $λ e^{λ t} 𝐯 = e^{λ t} 𝐀 𝐯$ . Cancelling the nonzero scalar leaves $𝐀 𝐯 = λ 𝐯$ , which is the definition of an eigenpair. Geometrically, the trial solution asks for a direction $𝐯$ that the flow leaves unchanged apart from stretching, and that is exactly what an eigenvector is.

How do I know if a matrix is non-defective?

Three sufficient conditions cover most cases: $n$ distinct eigenvalues, symmetry, or skew-symmetry. Beyond those, you have to compute. Defectiveness can only occur when an eigenvalue repeats, so if the characteristic polynomial has no repeated roots, you are safe.

Does the choice of eigenvector matter?

Only up to a nonzero scalar multiple, which is absorbed into the arbitrary constant $c_{k}$ . Taking $𝐯 = (1, 1)^{T}$ or $(5, 5)^{T}$ gives the same family of solutions. Choose whichever avoids fractions.

Why do the eigenvector equations always turn out to be redundant?

Because $λ$ was chosen precisely so that $det (𝐀 - λ 𝐈) = 0$ , which means the rows of $𝐀 - λ 𝐈$ are linearly dependent. If both equations of a $2 \times 2$ system give genuinely different information, you have made an arithmetic error in the eigenvalue.

Must I compute both eigenvectors of a complex pair?

No. If $𝐯$ is an eigenvector for $λ$ , then $\overset{―}{𝐯}$ is one for $\overset{―}{λ}$ , and the two complex solutions are conjugates. Taking real and imaginary parts of just one of them produces both real solutions. Computing the second eigenvector is wasted effort.

How do the eigenvalues describe the picture in the plane?

The sign of the real part governs growth: negative means solutions decay toward the origin, positive means they grow away from it. The imaginary part governs rotation: nonzero $β$ makes trajectories spiral or, when $α = 0$ , trace closed ellipses. Two real eigenvalues of opposite sign make the origin a saddle point, with solutions approaching along one eigendirection and leaving along the other.

Quick Reference

Matrix Form

Why Eigenvectors Appear

Non-defective Coefficient Matrix

Complex-Valued Eigenvalues

Exercises

Frequently Asked Questions

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