3.4 Eigenvalue Method

In this section we will learn how to solve linear homogeneous constant coefficient systems of ODEs by the eigenvalue method. Suppose we have such a system

${\vec{x}}^{'} = P \vec{x},$

where $P$ is a constant square matrix. We wish to adapt the method for the single constant coefficient equation by trying the function $e^{λ t}$ . However, $\vec{x}$ is a vector. So we try $\vec{x} = \vec{v} e^{λ t}$ , where $\vec{v}$ is an arbitrary constant vector. We plug this $\vec{x}$ into the equation to get

$\underset{{\vec{x}}^{'}}{\underset{⏟}{λ \vec{v} e^{λ t}}} = \underset{P \vec{x}}{\underset{⏟}{P \vec{v} e^{λ t}}} .$

We divide by $e^{λ t}$ and notice that we are looking for a scalar $λ$ and a vector $\vec{x}$ that satisfy the equation

$λ \vec{v} = P \vec{v} .$

To solve this equation we need a little bit more linear algebra, which we now review.

Eigenvalues and Eigenvectors of a Matrix

Let $A$ be a constant square matrix. Suppose there is a scalar $λ$ and a nonzero vector $\vec{v}$ such that

$A \vec{v} = λ \vec{v} .$

We then call $λ$ an eigenvalue of $A$ and $\vec{x}$ is said to be a corresponding eigenvector.

Example 1

The matrix $2 1; 0 1$ has an eigenvalue of $λ equals 2$ with a corresponding eigenvector $1; 0$

because

$2 1; 0 1$

Let us see how to compute the eigenvalues for any matrix. We rewrite the equation for an eigenvalue as

$(A - λ I) \vec{v} = \vec{0} .$

We notice that this equation has a nonzero solution $\vec{v}$ only if $A - λ I$ is not invertible. Were it invertible, we could write $(A - λ I)^{- 1} (A - λ I) \vec{v} = (A - λ I)^{- 1} \vec{0},$ which implies $v → equals 0 →.$ Therefore, $A$ has the eigenvalue $λ$ if and only if $λ$ solves the equation

$d e t (A - λ I) = 0 .$

Consequently, we will be able to find an eigenvalue of $A$ without finding a corresponding eigenvector. An eigenvector will have to be found later, once $λ$ is known.

Example 2

Find all eigenvalues of $2 1 1; 1 2 0; 0 0 2$ .

Show solution

We write

$\begin{matrix} d e t ([\begin{matrix} 2 & 1 & 1 \\ 1 & 2 & 0 \\ 0 & 0 & 2 \end{matrix}] - λ [\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}]) = d e t ([\begin{matrix} 2 - λ & 1 & 1 \\ 1 & 2 - λ & 0 \\ 0 & 0 & 2 - λ \end{matrix}]) \\ = (2 - λ) ((2 - λ)^{2} - 1) = - (λ - 1) (λ - 2) (λ - 3) . \end{matrix}$

So the eigenvalues are $λ = 1, λ = 2,$ and $λ equals 3$ .

Note that for an $n \times n$ matrix, the polynomial we get by computing $d e t (A - λ I)$ will be of degree $n$ , and hence we will in general have $n$ eigenvalues. Some may be repeated, some may be complex.

To find an eigenvector corresponding to an eigenvalue $λ$ , we write

$(A - λ I) \vec{v} = \vec{0},$

and solve for a nontrivial (nonzero) vector $\vec{v}$ . If $λ$ is an eigenvalue, there will be at least one free variable, and so for each distinct eigenvalue $λ$ , we can always find an eigenvector

Example 3

Find an eigenvector of $2 1 1; 1 2 0; 0 0 2$ corresponding to the eigenvalue $λ equals 3$ .

Show solution

We write

$2 1 1; 1 2 0; 0 0 2$

It is easy to solve this system of linear equations. We write down the augmented matrix

$\begin{matrix} - 1000 \\ 1 - 100 \\ 00 - 10 \end{matrix}$

and perform row operations (exercise: which ones?) until we get:

$1 minus 1 0 0; 0 0 1 0; 0 0 0 0$

The entries of $\vec{v}$ have to satisfy the equations $v_{1} - v_{2} = 0, v_{3} = 0$ and $v sub 2$ is a free variable. We can pick $v sub 2$ to be arbitrary (but nonzero), let $v sub 1 equals v sub 2,$ and of course $v sub 3 equals 0.$ For example, if we pick $v sub 2 equals 1,$ then $1; 1; 0$ Let us verify that $\vec{v}$ really is an eigenvector corresponding to $λ equals 3$ :

$2 1 1; 1 2 0; 0 0 2$

Yay! It worked.

3.4.2Eigenvalue Method with Distinct Real Eigenvalues

We have the system of equations

${\vec{x}}^{'} = P \vec{x} .$

We find the eigenvalues $λ sub 1, λ sub 2, …, λ sub n$ of the matrix $P$ , and corresponding eigenvectors $x → sub 1, x → sub 2, …, x → sub n.$ Now we notice that the functions ${\vec{v}}_{1} e^{λ_{1} t}, {\vec{v}}_{2} e^{λ_{2} t}, \dots, {\vec{v}}_{n} e^{λ_{n} t}$ are solutions of the system of equations and hence $\vec{x} = c_{1} {\vec{v}}_{1} e^{λ_{1} t} + c_{2} {\vec{v}}_{2} e^{λ_{2} t} + \dots + c_{n} {\vec{v}}_{n} e^{λ_{n} t}$ is a solution.

Note

Take ${\vec{x}}^{'} = P \vec{x}$ . If $P$ is an $n \times n$ constant matrix that has $n$ distinct real eigenvalues $λ sub 1, λ sub 2, …, λ sub n,$ then there exist $n$ linearly independent corresponding eigenvectors $v → sub 1, v → sub 2, …, v → sub n,$ and the general solution to ${\vec{x}}^{'} = P \vec{x}$ can be written as

$\vec{x} = c_{1} {\vec{v}}_{1} e^{λ_{1} t} + c_{2} {\vec{v}}_{2} e^{λ_{2} t} + \dots + c_{n} {\vec{v}}_{n} e^{λ_{n} t} .$

The corresponding fundamental matrix solution is

$X (t) = [{\vec{v}}_{1} e^{λ_{1} t} {\vec{v}}_{2} e^{λ_{2} t} \dots {\vec{v}}_{n} e^{λ_{n} t}] .$

That is,

X (t)

is the matrix whose

j^{t h}

column is

{\vec{v}}_{j} e^{λ_{j} t}

Example 4

Consider the system

$2 1 1; 1 2 0; 0 0 2$

Find the general solution.

Show solution

Earlier, we found the eigenvalues are $1, 2, 3.$ We found the eigenvector $1; 1; 0$ for the eigenvalue 3. Similarly we find the eigenvector $1; minus 1; 0$ for the eigenvalue 1, and $0; 1; minus 1$ for the eigenvalue 2 (exercise: check). Hence our general solution is

$1; minus 1; 0$

In terms of a fundamental matrix solution

$\begin{matrix} e^{t} 0 e^{3 t} \\ - e^{t} e^{2 t} e^{3 t} \\ 0 - e^{2 t} 0 \end{matrix}$

Complex Eigenvalues

A matrix might very well have complex eigenvalues even if all the entries are real. For example, suppose that we have the system

$1 1; minus 1 1$

Let us compute the eigenvalues of the matrix $1 1; minus 1 1$

$\begin{matrix} 1 - λ 1 \\ - 11 - λ \end{matrix}$

Thus $λ equals 1 ± i.$ The corresponding eigenvectors are also complex. First take $λ = 1 - i,$

$\begin{matrix} (P - (1 - i) I) \vec{v} = \vec{0}, \\ [\begin{matrix} i & 1 \\ - 1 & i \end{matrix}] \vec{v} = \vec{0} . \end{matrix}$

The equations $i v_{1} + v_{2} = 0$ and $- v_{1} + i v_{2} = 0$ are multiples of each other. So we only need to consider one of them. After picking $v sub 2 equals 1$ , for example, we have an eigenvector $i; 1$ . In similar fashion we find that $minus i; 1$ is an eigenvector corresponding to the eigenvalue $1 plus i$ .

We could write the solution as

$i; 1$

We would then need to look for complex values $c sub 1$ and $c sub 2$ to solve any initial conditions. It is perhaps not completely clear that we get a real solution. We could use Euler’s formula and do the whole song and dance we did before, but we will not. We will do something a bit smarter first.

We claim that we did not have to look for a second eigenvector (nor for the second eigenvalue). All complex eigenvalues come in pairs (because the matrix $P$ is real).

First a small side note. The real part of a complex number $z$ can be computed as $\frac{z + \bar{z}}{2}$ , where the bar above $z$ means $\overset{―}{a + i b} = a - i b$ . This operation is called the complex conjugate. If $a$ is a real number, then $\bar{a} = a$ . Similarly we can bar whole vectors or matrices by taking the complex conjugate of every entry. If a matrix $P$ is real, then $\bar{P} = P$ . We note that $\overset{―}{P \vec{x}} = \bar{P} \bar{\vec{x}} = P \bar{\vec{x}} .$ . Also the complex conjugate of $0$ is still $0$ , therefore,

$\vec{0} = \overset{―}{\vec{0}} = \overset{―}{(P - λ I) \vec{v}} = (P - \bar{λ} I) \bar{\vec{v}} .$

So if $\vec{v}$ is an eigenvector corresponding to the eigenvalue $λ = a + i b$ , then $\bar{\vec{v}}$ is an eigenvector corresponding to the eigenvalue $\bar{λ} = a - i b$ .

Suppose that $a + i b$ is a complex eigenvalue of $P$ , and $\vec{v}$ is a corresponding eigenvector. Then

${\vec{x}}_{1} = \vec{v} e^{(a + i b) t}$

is a solution (complex valued) of ${\vec{x}}^{'} = P \vec{x}$ . Euler’s formula shows that $\overset{―}{e^{a + i b}} = e^{a - i b}$ , and so

${\vec{x}}_{2} = \overset{―}{\vec{x_{1}}} = \bar{\vec{v}} e^{(a + i b) t}$

is also a solution. As $x sub 1 →$ and $x sub 2 →$ are solutions, the function

${\vec{x}}_{3} = R e {\vec{x}}_{1} = R e \vec{v} e^{(a + i b) t} = \frac{{\vec{x}}_{1} + \overset{―}{\vec{x_{1}}}}{2} = \frac{{\vec{x}}_{1} + \vec{x_{2}}}{2} = \frac{1}{2} \vec{x_{1}} + \frac{1}{2} \vec{x_{2}}$

is also a solution. And $x → sub 3$ is real-valued! Similarly as $I m z = \frac{z - \bar{z}}{2 i}$ is the imaginary part, we find that

${\vec{x}}_{4} = I m {\vec{x}}_{1} = \frac{{\vec{x}}_{1} - \overset{―}{\vec{x_{1}}}}{2 i} = \frac{{\vec{x}}_{1} - \vec{x_{2}}}{2 i}$

is also a real-valued solution. It turns out that $x → sub 3$ and $x → sub 4$ are linearly independent. We will use Euler’s formula to separate out the real and imaginary part.

Returning to our problem,

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

Then

$\begin{matrix} e^{t} \sin t \\ e^{t} \cos t \end{matrix}$

are the two real-valued linearly independent solutions we seek.

The general solution is

$\begin{matrix} e^{t} \sin t \\ e^{t} \cos t \end{matrix}$

This solution is real-valued for real $c sub 1$ and $c sub 2$ . At this point, we would solve for any initial conditions we may have to find $c sub 1$ and $c sub 2$ .

Let us summarize the discussion as a theorem.

Note

Let $P$ be a real-valued constant matrix. If $P$ has a complex eigenvalue $a + i b$ and a corresponding eigenvector $\vec{v}$ , then $P$ also has a complex eigenvalue $a - i b$ with a corresponding eigenvector $\bar{\vec{v}}$ . Furthermore, ${\vec{x}}^{'} = P \vec{x}$ has two linearly independent real-valued solutions

${\vec{x}}_{1} = R e \vec{v} e^{(a + i b) t}, a n d {\vec{x}}_{2} = I m \vec{v} e^{(a + i b) t} .$

For each pair of complex eigenvalues $a + i b$ and $a - i b$ , we get two real-valued linearly independent solutions. We then go on to the next eigenvalue, which is either a real eigenvalue or another complex eigenvalue pair. If we have $n$ distinct eigenvalues (real or complex), then we end up with $n$ linearly independent solutions. If we had only two equations $open parenthesis n equals 2 close parenthesis$ as in the example above, then once we found two solutions we are finished, and our general solution is

$\vec{x} = c_{1} {\vec{x}}_{1} + c_{2} {\vec{x}}_{2} = c_{1} (Re \vec{v} e^{(a + i b) t}) + c_{2} (Im \vec{v} e^{(a + i b) t}) .$

We can now find a real-valued general solution to any homogeneous system where the matrix has distinct eigenvalues. When we have repeated eigenvalues, matters get a bit more complicated and we will look at that situation in Section 3.7.

Adapted from Differential Equations for Engineers by Jiří Lebl (Oklahoma State University), hosted on LibreTexts (math.libretexts.org) and licensed under CC BY-SA 4.0. Changes were made. License: CC-BY-SA-4.0.