Note: 2 lectures, §5.2 in [EP], part of §7.3, §7.5, and §7.6 in [BD]
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
~x0 = P~x,
where P is a constant square matrix. We wish to adapt the method for the single constant coefficient equation by trying the function eλt. However, ~x is a vector. So we try ~x = ~veλt, where ~v is an arbitrary constant vector. We plug this ~xinto the equation to get
λ~veλt= P~veλt.
We divide by eλtand notice that we are looking for a scalar λ and a vector ~v that satisfy the equation λ~v = P~v.
To solve this equation we need a little bit more linear algebra, which we now review.
3.4.1 Eigenvalues and eigenvectors of a matrix
Let A be a constant square matrix. Suppose there is a scalar λ and a nonzero vector ~v such that A~v= λ~v.
We then call λ an eigenvalue of A and ~v is said to be a corresponding eigenvector.
Example 3.4.1: The matrix2 1
0 1 has an eigenvalue of λ= 2 with a corresponding eigenvector 10 because
"2 1 0 1
# "1 0
#
= "2 0
#
= 2"1 0
# .
Let us see how to compute the eigenvalues for any matrix. We rewrite the equation for an eigenvalue as
(A − λI)~v= ~0.
We notice that this equation has a nonzero solution ~v only if A − λI is not invertible. Were it invertible, we could write (A − λI)−1(A − λI)~v = (A − λI)−1~0, which implies ~v = ~0. Therefore, A has the eigenvalue λ if and only if λ solves the equation
det(A − λI)= 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 3.4.2: Find all eigenvalues of
Note that for an n × n matrix, the polynomial we get by computing det(A − λI) 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 − λI)~v= ~0,
and solve for a nontrivial (nonzero) vector ~v. If λ is an eigenvalue, this will always be possible.
Example 3.4.3: Find an eigenvector of
2 1 1
1 2 0 0 0 2
corresponding to the eigenvalue λ = 3.
We write
It is easy to solve this system of linear equations. We write down the augmented matrix
and perform row operations (exercise: which ones?) until we get
. Let us verify that we really have an eigenvector corresponding to λ = 3:
Exercise 3.4.1 (easy): Are eigenvectors unique? Can you find a different eigenvector for λ = 3 in the example above? How are the two eigenvectors related?
Exercise 3.4.2: Note that when the matrix is 2 × 2 you do not need to write down the augmented matrix and do row operations when computing eigenvectors (if you have computed the eigenvalues correctly). Can you see why? Try it for the matrix2 1
1 2.
3.4.2 The eigenvalue method with distinct real eigenvalues
OK. We have the system of equations
~x0 = P~x.
We find the eigenvalues λ1, λ2, . . . , λnof the matrix P, and corresponding eigenvectors ~v1, ~v2, . . . , ~vn. Now we notice that the functions ~v1eλ1t, ~v2eλ2t, . . . , ~vneλnt are solutions of the system of equations and hence ~x= c1~v1eλ1t+ c2~v2eλ2t + · · · + cn~vneλnt is a solution.
Theorem 3.4.1. Take ~x0 = P~x. If P is an n × n constant matrix that has n distinct real eigenvalues λ1,λ2, . . . ,λn, then there exist n linearly independent corresponding eigenvectors ~v1, ~v2, . . . , ~vn, and the general solution to ~x0 = P~x can be written as
~x = c1~v1eλ1t + c2~v2eλ2t+ · · · + cn~vneλnt.
The corresponding fundamental matrix solution is X(t)= [~v1eλ1t ~v2eλ2t · · · ~vneλnt]. That is, X(t) is the matrix whose jth column is ~vjeλjt.
Example 3.4.4: Suppose we take the system
~x0 =
We have found the eigenvalues 1, 2, 3 earlier. We have found the eigenvector
1
1 0
for the eigenvalue 3. Similarly we find the eigenvector
1
−1 0
for the eigenvalue 1, and
0
−11
for the eigenvalue 2 (exercise: check). Hence our general solution is
~x = c1
In terms of a fundamental matrix solution
~x = X(t)~c =
Exercise 3.4.3: Check that this ~x really solves the system.
Note: If we write a homogeneous linear constant coefficient nthorder equation as a first order system (as we did in § 3.1), then the eigenvalue equation
det(P − λI) = 0
is essentially the same as the characteristic equation we got in § 2.2 and § 2.3.
3.4.3 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
~x0 =" 1 1
−1 1
#
~x.
Let us compute the eigenvalues of the matrix P= −1 11 1.
det(P − λI)= det "1 − λ 1
−1 1 − λ
#!
= (1 − λ)2+ 1 = λ2− 2λ+ 2 = 0.
Thus λ= 1 ± i. The corresponding eigenvectors will also be complex. First take λ = 1 − i, (P − (1 − i)I)~v= ~0,
" i 1
−1 i
#
~v = ~0.
The equations iv1+v2 = 0 and −v1+iv2 = 0 are multiples of each other. So we only need to consider one of them. After picking v2 = 1, for example, we have an eigenvector ~v = 1i. In similar fashion we find that−i
1 is an eigenvector corresponding to the eigenvalue 1+ i.
We could write the solution as
~x = c1
" i 1
#
e(1−i)t+ c2
"−i 1
#
e(1+i)t ="c1ie(1−i)t − c2ie(1+i)t c1e(1−i)t + c2e(1+i)t
# .
We would then need to look for complex values c1and c2to 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 z+¯z2 , where the bar above z means a+ ib = a − ib. This operation is called the complex conjugate. Note that if a is
a real number, then ¯a = a. Similarly we can bar whole vectors or matrices. If a matrix P is real, then P= P. We note that P~x = P ~x = P~x. Therefore,
(P − λI)~v= (P − ¯λI)~v.
So if ~v is an eigenvector corresponding to the eigenvalue λ = a + ib, then ~v is an eigenvector corresponding to the eigenvalue ¯λ= a − ib.
Suppose that a+ ib is a complex eigenvalue of P, and ~v is a corresponding eigenvector. Then
~x1 = ~ve(a+ib)t
is a solution (complex valued) of ~x0 = P~x. Then note that ea+ib = ea−ib, and so
~x2 = ~x1 = ~ve(a−ib)t is also a solution. The function
~x3 = Re ~x1 = Re~ve(a+ib)t = ~x1+ ~x1
2 = ~x1+ ~x2
2
is also a solution. And ~x3is real-valued! Similarly as Im z= z−¯z2i is the imaginary part, we find that
~x4 = Im ~x1 = ~x1−~x2
2i .
is also a real-valued solution. It turns out that ~x3 and ~x4 are linearly independent. We will use Euler’s formula to separate out the real and imaginary part.
Returning to our problem,
~x1 =" i 1
#
e(1−i)t =" i 1
#
etcos t − ietsin t="ietcos t+ etsin t etcos t − ietsin t
# .
Then
Re ~x1= " etsin t etcos t
# ,
Im ~x1= " etcos t
−etsin t
# ,
are the two real-valued linearly independent solutions we seek.
Exercise 3.4.4: Check that these really are solutions.
The general solution is
~x = c1
" etsin t etcos t
# + c2
" etcos t
−etsin t
#
="c1etsin t+ c2etcos t c1etcos t − c2etsin t
# .
This solution is real-valued for real c1and c2. Now we can solve for any initial conditions that we may have.
Let us summarize as a theorem.
Theorem 3.4.2. Let P be a real-valued constant matrix. If P has a complex eigenvalue a+ ib and a corresponding eigenvector ~v, then P also has a complex eigenvalue a − ib with a corresponding eigenvector ¯~v. Furthermore , ~x0= P~x has two linearly independent real-valued solutions
~x1= Re~ve(a+ib)t, and ~x2= Im~ve(a+ib)t.
So for each pair of complex eigenvalues 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 had n distinct eigenvalues (real or complex), then we will end up with n linearly independent solutions.
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 § 3.7.
3.4.4 Exercises
Exercise 3.4.5 (easy): Let A be a 3 × 3 matrix with an eigenvalue of 3 and a corresponding eigenvector ~v= 1
−13
. Find A~v.
Exercise 3.4.6: a) Find the general solution of x01 = 2x1, x02 = 3x2 using the eigenvalue method (first write the system in the form ~x0 = A~x). b) Solve the system by solving each equation separately and verify you get the same general solution.
Exercise 3.4.7: Find the general solution of x01 = 3x1 + x2, x02 = 2x1+ 4x2 using the eigenvalue method.
Exercise 3.4.8: Find the general solution of x01 = x1− 2x2, x02 = 2x1+ x2 using the eigenvalue method. Do not use complex exponentials in your solution.
Exercise 3.4.9: a) Compute eigenvalues and eigenvectors of A = 9 −2 −6
−8 3 6 10 −2 −6
. b) Find the general solution of ~x0= A~x.
Exercise 3.4.10: Compute eigenvalues and eigenvectors of
−2 −1 −1
3 2 1
−3 −1 0
.
Exercise 3.4.11: Let a, b, c, d, e, f be numbers. Find the eigenvalues ofa b c
0 d e 0 0 f
. Exercise 3.4.101: a) Compute eigenvalues and eigenvectors of A = 1 0 3
−1 0 1 2 0 2
. b) Solve the system
~x0 = A~x.
Exercise 3.4.102: a) Compute eigenvalues and eigenvectors of A = −1 01 1. b) Solve the system
~x0 = A~x.
Exercise 3.4.103: Solve x01 = x2, x02 = x1using the eigenvalue method.
Exercise 3.4.104: Solve x01 = x2, x02 = −x1using the eigenvalue method.