When the n × n matrix A possesses n distinct real eigenvalues λ₁, λ₂, … , λ_n, then a set of n linearly independent eigenvectors K₁, K₂, … , K_n can always be found and

is a fundamental set of solutions of (2) on the interval (–∞, ∞).

EXAMPLE 1 Distinct Eigenvalues

Phase Portrait

For the sake of review, you should keep firmly in mind that a solution of a system of linear first-order differential equations, when written in terms of matrices, is simply an alternative to the method that we employed in Section 3.12—namely, listing the individual functions and the relationship between the constants. If we add the vectors on the right side of (5) and then equate the entries with the corresponding entries in the vector of the left, we obtain the more familiar statement

x = c₁e^–t + 3c₂e^4t, y = –c₁e^–t + 2c₂e^4t.

As pointed out in Section 10.1, we can interpret these equations as parametric equations of a curve or trajectory in the xy-plane or phase plane. The three graphs shown in FIGURE 10.2.1, x(t) in the tx-plane, y(t) in the ty-plane, and the trajectory in the phase plane, correspond to the choice of constants c₁ = c₂ = 1 in the solution. A collection of trajectories in the phase plane as shown in FIGURE 10.2.2 is said to be a phase portrait of the given linear system. What appears to be two red lines in Figure 10.2.2 are actually four half-lines defined parametrically in the first, second, third, and fourth quadrants by the solutions X₂, –X₁, –X₂, and X₁, respectively. For example, the Cartesian equations y = x, x > 0, and y = –x, x > 0, of the half-lines in the first and fourth quadrants were obtained by eliminating the parameter t in the solutions x = 3e^4t, y = 2e^4t, and x = e^–t, y = –e^–t, respectively. Moreover, each eigenvector can be visualized as a two-dimensional vector lying along one of these half-lines. The eigenvector K₂ = lies along y = x in the first quadrant and K₁ = lies along y = –x in the fourth quadrant; each vector starts at the origin with K₂ terminating at the point (2, 3) and K₁ terminating at (1, –1).

The origin is not only a constant solution, x = 0, y = 0, of every 2 × 2 homogeneous linear system X′ = AX but is as well an important point in the qualitative study of such systems. If we think in physical terms, the arrowheads on a trajectory in Figure 10.2.2 indicate the direction that a particle with coordinates (x(t), y(t)) on a trajectory at time t would move as time increases. Observe that the arrowheads, with only those on the half-lines in the second and fourth quadrants being the exception, indicate that a particle would move away from the origin as time t increases. If we imagine time ranging from –∞ to ∞, then inspection of the solution x = c₁e^–t + 3c₂e^4t, y = –c₁e^–t + 2c₂e^4t, c₁ ≠ 0, c₂ ≠ 0, shows that a trajectory, or moving particle, “starts” asymptotic to one of the half-lines defined by X₁ or –X₁ (since e^4t is negligible for t → –∞) and “finishes” asymptotic to one of the half-lines defined by X₂ and –X₂ (since e^–t is negligible for t → ∞).

We note in passing that Figure 10.2.2 represents a phase portrait that is typical of all 2 × 2 homogeneous linear systems X′ = AX with real eigenvalues of opposite signs. See Problem 19 in Exercises 10.2. Moreover, phase portraits in the two cases when distinct real eigenvalues have the same algebraic sign would be typical portraits of all such 2 × 2 linear systems; the only difference is that the arrowheads would indicate that a particle would move away from the origin on any trajectory as t → ∞ when both λ₁ and λ₂ are positive and would move toward the origin on any trajectory when both λ₁ and λ₂ are negative. Consequently, it is common to call the origin a repeller in the case λ₁ > 0, λ₂ > 0, and an attractor in the case λ₁ < 0, λ₂ < 0. See Problem 20 in Exercises 10.2. The origin in Figure 10.2.2 is neither a repeller nor an attractor. Investigation of the remaining case when λ = 0 is an eigenvalue of a 2 × 2 homogeneous linear system is left as an exercise. See Problem 52 in Exercises 10.2.

EXAMPLE 2 Distinct Eigenvalues

Solve

(6)

SOLUTION

Using the cofactors of the third row, we find

det (A − λI) = = −(λ + 3)(λ + 4)(λ − 5) = 0,

and so the eigenvalues are λ₁ = −3, λ₂ = −4, λ₃ = 5.

For λ₁ = −3, Gauss–Jordan elimination gives

Therefore k₁ = k₃ and k₂ = 0. The choice k₃ = 1 gives an eigenvector and corresponding solution vector

K₁ = , X₁ = e^−3t.(7)

Similarly, for λ₂ = −4,

implies k₁ = 10k₃ and k₂ = −k₃. Choosing k₃ = 1, we get a second eigenvector and solution vector

(8)

Finally, when λ₃ = 5, the augmented matrices

yield (9)

The general solution of (6) is a linear combination of the solution vectors in (7), (8), and (9):

≡

Use of Computers

Software packages such as MATLAB, Mathematica, and Maple can be real time savers in finding eigenvalues and eigenvectors of a matrix. For example, to find the eigenvalues and eigenvectors of the matrix of coefficients in (6) using Mathematica, we first input the definition of the matrix by rows:

m = { {−4, 1, 1}, { 1, 5, −1}, { 0, 1, −3} }.

The commands Eigenvalues[m] and Eigenvectors[m] given in sequence yield

{5, −4, −3} and { { 1, 8, 1}, { 10, −1, 1}, { 1, 0, 1} },

respectively. In Mathematica, eigenvalues and eigenvectors can also be obtained at the same time by using Eigensystem[m].

Of course, not all of the n eigenvalues λ₁, λ₂, … , λ_n of an n × n matrix A need be distinct; that is, some of the eigenvalues may be repeated. For example, the characteristic equation of the coefficient matrix in the system

X′ = X(10)

is readily shown to be (λ + 3)² = 0, and therefore λ₁ = λ₂ = −3 is a root of multiplicity two. For this value we find the single eigenvector

(11)

is one solution of (10). But since we are obviously interested in forming the general solution of the system, we need to pursue the question of finding a second solution.

In general, if m is a positive integer and (λ − λ₁)^m is a factor of the characteristic equation while (λ − λ₁)^m+1 is not a factor, then λ₁ is said to be an eigenvalue of multiplicity m. The next three examples illustrate the following cases:

1. For some n × n matrices A it may be possible to find m linearly independent eigenvectors K₁, K₂, … , K_m corresponding to an eigenvalue λ₁ of multiplicity m ≤ n. In this case, the general solution of the system contains the linear combination

   

2. If there is only one eigenvector corresponding to the eigenvalue λ₁ of multiplicity m, then m linearly independent solutions of the form

   

   where K_ij are column vectors, can always be found.

Eigenvalue of Multiplicity Two

We begin by considering eigenvalues of multiplicity two. In the first example we illustrate a matrix for which we can find two distinct eigenvectors corresponding to a repeated eigenvalue.

EXAMPLE 3 Repeated Eigenvalues

Solve X′ = X.

SOLUTION

Expanding the determinant in the characteristic equation

det(A − λI) = = 0

yields −(λ + 1)² (λ − 5) = 0. We see that λ₁ = λ₂ = −1 and λ₃ = 5.

For λ₁ = −1, Gauss–Jordan elimination immediately gives

The first row of the last matrix means k₁ − k₂ + k₃ = 0 or k₁ = k₂ − k₃. The choices k₂ = 1, k₃ = 0 and k₂ = 1, k₃ = 1 yield, in turn, k₁ = 1 and k₁ = 0. Thus two eigenvectors corresponding to λ₁ = −1 are

Since neither eigenvector is a constant multiple of the other, we have found, corresponding to the same eigenvalue, two linearly independent solutions

Lastly, for λ₃ = 5, the reduction

implies k₁ = k₃ and k₂ = −k₃. Picking k₃ = 1 gives k₁ = 1, k₂ = −1, and thus a third eigenvector is

We conclude that the general solution of the system is

≡

The matrix of coefficients A in Example 3 is a special kind of matrix known as a symmetric matrix. An n × n matrix A is said to be symmetric if its transpose A^T (where the rows and columns are interchanged) is the same as A; that is, if A^T = A. It can be proved that if the matrix A in the system X′ = AX is symmetric and has real entries, then we can always find n linearly independent eigenvectors K₁, K₂, … , K_n, and the general solution of such a system is as given in Theorem 10.2.1. As illustrated in Example 3, this result holds even when some of the eigenvalues are repeated.

Second Solution

Now suppose that λ₁ is an eigenvalue of multiplicity two and that there is only one eigenvector associated with this value. A second solution can be found of the form

(12)

where

To see this we substitute (12) into the system X′ = AX and simplify:

Since this last equation is to hold for all values of t, we must have

(A − λ₁I)K = 0(13)

and (A − λ₁I)P = K.(14)

Equation (13) simply states that K must be an eigenvector of A associated with λ₁. By solving (13), we find one solution X₁ = K. To find the second solution X₂ we need only solve the additional system (14) for the vector P.

EXAMPLE 4 Repeated Eigenvalues

Find the general solution of the system given in (10).

SOLUTION

From (11) we know that λ₁ = −3 and that one solution is X₁ = e^−3t. Identifying K = and P = , we find from (14) that we must now solve

Since this system is obviously equivalent to one equation, we have an infinite number of choices for p₁ and p₂. For example, by choosing p₁ = 1 we find p₂ = . However, for simplicity, we shall choose p₁ = so that p₂ = 0. Hence P = . Thus from (12),

The general solution of (10) is then

≡

By assigning various values to c₁ and c₂ in the solution in Example 4, we can plot trajectories of the system in (10). A phase portrait of (10) is given in FIGURE 10.2.3. The solutions X₁ and −X₁ determine two half-lines y = x, x > 0, and y = x, x < 0, respectively, that are shown in red in Figure 10.2.3. Because the single eigenvalue is negative and e^–3t → 0 as t → ∞ on every trajectory, we have (x(t), y(t)) → (0, 0) as t → ∞. This is why the arrowheads in Figure 10.2.3 indicate that a particle on any trajectory would move toward the origin as time increases and why the origin is an attractor in this case. Moreover, a moving particle on a trajectory x = 3c₁e^–3t + c₂(te^–3t + e^–3t), y = c₁e^−3t + c₂te^−3t, c₂ ≠ 0, approaches (0, 0) tangentially to one of the half-lines as t → ∞. In contrast, when the repeated eigenvalue is positive the situation is reversed and the origin is a repeller. See Problem 23 in Exercises 10.2. Analogous to Figure 10.2.2, Figure 10.2.3 is typical of all 2 × 2 homogeneous linear systems X′ = AX that have repeated negative eigenvalues. See Problem 34 in Exercises 10.2.

Eigenvalue of Multiplicity Three

When the coefficient matrix A has only one eigenvector associated with an eigenvalue λ₁ of multiplicity three, we can find a solution of the form (12) and a third solution of the form

(15)

where

By substituting (15) into the system X′ = AX, we find that the column vectors K, P, and Q must satisfy

(A − λ₁I)K = 0(16)

(A − λ₁I)P = K(17)

and (A − λ₁I)Q = P.(18)

Of course, the solutions of (16) and (17) can be used in forming the solutions X₁ and X₂.

EXAMPLE 5 Repeated Eigenvalues

Solve X′ = X.

SOLUTION

The characteristic equation (λ − 2)³ = 0 shows that λ₁ = 2 is an eigenvalue of multiplicity three. By solving (A − 2I)K = 0 we find the single eigenvector

We next solve the systems (A − 2I)P = K and (A − 2I)Q = P in succession and find that

Using (12) and (15), we see that the general solution of the system is

If λ₁ = α + iβ and λ₂ = α − iβ, β > 0, i² = −1, are complex eigenvalues of the coefficient matrix A, we can then certainly expect their corresponding eigenvectors to also have complex entries.*

For example, the characteristic equation of the system

(19)

is

From the quadratic formula we find λ₁ = 5 + 2i, λ₂ = 5 − 2i.

Now for λ₁ = 5 + 2i we must solve

Since k₂ = (1 − 2i)k₁,^† the choice k₁ = 1 gives the following eigenvector and a solution vector:

In like manner, for λ₂ = 5 − 2i we find

We can verify by means of the Wronskian that these solution vectors are linearly independent, and so the general solution of (19) is

(20)

Note that the entries in K₂ corresponding to λ₂ are the conjugates of the entries in K₁ corresponding to λ₁. The conjugate of λ₁ is, of course, λ₂. We write this as λ₂ = and K₂ = . We have illustrated the following general result.

It is desirable and relatively easy to rewrite a solution such as (20) in terms of real functions. To this end we first use Euler’s formula to write

Then (20) becomes

If we let and then the last line can be written

X = C₁X₁ + C₂X₂,(21)

where

and

It is now important to realize that the two vectors X₁ and X₂ in (21) are themselves linearly independent real solutions of the original system. Consequently, we are justified in ignoring the relationship between C₁, C₂ and c₁, c₂, and we can regard C₁ and C₂ as completely arbitrary and real. In other words, the linear combination (21) is an alternative general solution of (19).

The foregoing process can be generalized. Let K₁ be an eigenvector of the coefficient matrix A (with real entries) corresponding to the complex eigenvalue λ₁ = α + iβ. Then the two solution vectors in Theorem 10.2.2 can be written as

By the superposition principle, Theorem 10.1.2, the following vectors are also solutions:

For any complex number z = a + ib, both and are real numbers.

Therefore, the entries in the column vectors and are real numbers. By defining

(22)

we are led to the following theorem.

The matrices B₁ and B₂ in (22) are often denoted by

B₁ = Re(K₁) and B₂ = Im(K₁) (24)

since these vectors are, respectively, the real and imaginary parts of the eigenvector K₁. For example, (21) follows from (23) with

EXAMPLE 6 Complex Eigenvalues

Solve the initial-value problem

(25)

SOLUTION

First we obtain the eigenvalues from

The eigenvalues are λ₁ = 2i and λ₂ = = −2i. For λ₁ the system

gives k₁ = −(2 + 2i)k₂. By choosing k₂ = −1 we get

Now from (24) we form

Since α = 0, it follows from (23) that the general solution of the system is

(26)

Some graphs of the curves or trajectories defined by the solution (26) of the system are illustrated in the phase portrait in FIGURE 10.2.4. Now the initial condition X(0) = , or equivalently x(0) = 2, and y(0) = −1, yields the algebraic system 2c₁ + 2c₂ = 2, −c₁ = −1 whose solution is c₁ = 1, c₂ = 0. Thus the solution to the problem is X = . The specific trajectory defined parametrically by the particular solution x = 2 cos 2t – 2 sin 2t, y = –cos 2t is the red curve in Figure 10.2.4. Note that this curve passes through (2, −1).≡

10.2 Exercises Answers to selected odd-numbered problems begin on page ANS-27.

10.2.1 Distinct Real Eigenvalues

In Problems 1–12, find the general solution of the given system.

1. = x + 2y
   = 4x + 3y
2. = 2x + 2y
   = x + 3y
3. = −4x + 2y
   = − x + 2y
4. = − x + 2y
   = x − 2y
5. X′ = X
6. X′ = X
7. = x + y − z
   = 2y
   = y − z
8. = 2x − 7y
   = 5x + 10y + 4z
   = 5y + 2z
9. X′ = X
10. X′ = X
11. X′ = X
12. X′ = X

In Problems 13 and 14, solve the given initial-value problem.

13. 
14. X′ = X, X(0) =
15. Consider the large mixing tanks shown in FIGURE 10.2.5. Suppose that both tanks A and B initially contain 100 gallons of brine. Liquid is pumped in and out of the tanks as indicated in the figure; the mixture pumped between and out of the tanks is assumed to be well-stirred.
    1. Construct a mathematical model in the form of a linear system of first-order differential equations for the number of pounds x₁(t) and x₂(t) of salt in tanks A and B, respectively, at time t. Write the system in matrix form. [Hint: See Section 2.9 and Problem 52, Chapter 2 in Review.]
    2. Use the eigenvalue method of this section to solve the linear system in part (a) subject to x₁(0) = 20, x₂(0) = 5.
    3. Use a graphing utility or CAS to plot the graphs of x₁(t) and x₂(t) in the same coordinate plane.
    4. Suppose the system of mixing tanks is to be turned off when the number of pounds of salt in tank B equals that in tank A. Use a root-finding application of a CAS or calculator to approximate that time.

       

16. In Problem 26 of Exercises 3.12 you were asked to solve the following linear system

    

    using elimination techniques. Recall, this linear system is a mathematical model for the number of pounds of salt x₁(t), x₂(t), and x₃(t) in the connected mixing tanks A, B, and C shown in Figure 2.9.7 on page 102.

    1. Use the eigenvalue method of this section to solve the system subject to x₁(0) = 15, x₂(0) = 10, x₃(0) = 5.
    2. What are and Interpret this result.

Computer Lab Assignments

In Problems 17 and 18, use a CAS or linear algebra software as an aid in finding the general solution of the given system.

17. X′ = X
18. X′ = X
19. 1. Use computer software to obtain the phase portrait of the system in Problem 5. If possible, include the arrowheads as in Figure 10.2.2. Also, include four half-lines in your phase portrait.
    2. Obtain the Cartesian equations of each of the four half-lines in part (a).
    3. Draw the eigenvectors on your phase portrait of the system.
20. Find phase portraits for the systems in Problems 2 and 4. For each system, find any half-line trajectories and include these lines in your phase portrait.

10.2.2 Repeated Eigenvalues

In Problems 21–30, find the general solution of the given system.

21. = 3x − y
    = 9x − 3y
22. = −6x + 5y
    = −5x + 4y
23. X′ = X
24. X′ = X
25. = 3x − y − z
    = x + y − z
    = x − y + z
26. = 3x + 2y + 4z
    = 2x + 2z
    = 4x + 2y + 3z
27. X′ = X
28. X′ = X
29. X′ = X
30. X′ = X

In Problems 31 and 32, solve the given initial-value problem.

31. X′ = X, X(0) =
32. X′ = X, X(0) =
33. Show that the 5 × 5 matrix

    

    has an eigenvalue λ₁ of multiplicity 5. Show that three linearly independent eigenvectors corresponding to λ₁ can be found.

Computer Lab Assignment

34. Find phase portraits for the systems in Problems 22 and 23. For each system, find any half-line trajectories and include these lines in your phase portrait.

10.2.3 Complex Eigenvalues

In Problems 35–46, find the general solution of the given system.

35. = 6x − y
    = 5x + 2y
36. = x + y
    = −2x − y
37. = 5x + y
    = −2x + 3y
38. = 4x + 5y
    = −2x + 6y
39. X′ = X
40. X′ = X
41. = z
    = −z
    = y
42. = 2x + y + 2z
    = 3x + 6z
    = −4x − 3z
43. X′ = X
44. X′ = X
45. X′ = X
46. X′ = X

In Problems 47 and 48, solve the given initial-value problem.

47. X′ = X, X(0) =
48. X′ = X, X(0) =
49. 1. In the closed system shown in FIGURE 10.2.6 the three large tanks A, B, and C initially contain the number of gallons of brine indicated. Construct a mathematical model for the number of pounds of salt x₁(t), x₂(t), and x₃(t) in the tanks A, B, and C at time t, respectively.
    2. Use the eigenvalue method of this section to solve the system in part (a) subject to x₁(0) = 30, x₂(0) = 20, x₃(0) = 5.

       

50. For the system in Problem 49:
    1. Show that Interpret this result.
    2. What are and Interpret this result.

Computer Lab Assignments

51. Find phase portraits for the systems in Problems 38–40.
52. Solve each of the following linear systems.
    1. X′ = X
    2. X′ = X

    Find a phase portrait of each system. What is the geometric significance of the line y = −x in each portrait?

Discussion Problems

53. Consider the 5 × 5 matrix given in Problem 33. Solve the system X′ = AX without the aid of matrix methods, but write the general solution using the matrix notation. Use the general solution as a basis for a discussion on how the system can be solved using the matrix methods of this section. Carry out your ideas.
54. Obtain a Cartesian equation of the curve defined parametrically by the solution of the linear system in Example 6. Identify the curve passing through (2, −1) in Figure 10.2.4. [Hint: Compute x², y², and xy.]
55. Examine your phase portraits in Problem 51. Under what conditions will the phase portrait of a 2 × 2 homogeneous linear system with complex eigenvalues consist of a family of closed curves? Consist of a family of spirals? Under what conditions is the origin (0, 0) a repeller? An attractor?
56. The system of linear second-order differential equations

    (27)

    describes the motion of two coupled spring/mass systems (see Figure 3.12.1). We have already solved a special case of this system in Sections 3.12 and 4.6. In this problem we describe yet another method for solving the system.

    1. Show that the system in (27) can be written as the matrix equation X″ = AX, where

       

    2. If a solution is assumed of the form X = Ke^ωt, show that X″ = AX yields

       (A − λI)K = 0 where λ = ω².

    3. Show that if m₁ = 1, m₂ = 1, k₁ = 3, and k₂ = 2, a solution of the system is

       

    4. Show that the solution in part (c) can be written as

       

       

^* When the characteristic equation has real coefficients, complex eigenvalues always appear in conjugate pairs.

^† Note that the second equation in the system is simply 1 + 2i times the first equation.