1550078481-Ordinary_Differential_Equations__Roberts_

374 Ordinary Differential Equations

THE MATRIX WHOSE EIGENVALUES AND EIGENVECTORS ARE TO BE CALCULATED IS

2.0000EtOO O.OOOOEtOO l.OOOOEtOO

O.OOOOEtOO l.OOOOEtOO O.OOOOEtOO

l.OOOOEtOO O.OOOOEtOO 2.0000EtOO

AN EIGENVALUE I S 3.00000EtOO t O.OOOOOEtOO I

THE ASSOCIATED EIGENVECTOR IS

?. O?lO?E-01 t O.OOOOOEtOO I

O.OOOOOEtOO t O.OOOOOEtOO I

?. O?lO?E-01 t O.OOOOOEtOO I

AN EIGENVALUE I S l. OOOOOEtOO t O.OOOOOEtOO I

THE ASSOCIATED EIGENVECTOR IS

-?. O?lO?E-01 t O.OOOOOEtOO I

O.OOOOOEtOO t O.OOOOOEtOO I

?.O?lO?E-01 t O.OOOOOEtOO I

AN EIGENVALUE I S l. OOOOOEtOO t O.OOOOOEtOO I

THE ASSOCIATED EIGENVECTOR IS

O.OOOOOEtOO t O.OOOOOEtOO I

l.OOOOOEtOO t O.OOOOOEtOO I

O.OOOOOEtOO t O.OOOOOEtOO I

Figure 8.6 Eigenvalues and Eigenvectors of the Matrix in Equation (16)

From this figure, we see that the eigenvalues are 3, 1, 1 and the associat ed
eigenvectors are respectively

(

.707107)

V 1 = 0.000000 ,

.707107 (

v -. 707107) (0)

2 = 0.000000 , and v 3 =^1.

.70 7107 0

We see that the eigenvalue 1 has multiplicity two. Thus, we need to verify that
t he associated eigenvectors v 2 and v3 are linearly independent. (In general,
suppose u =f. 0 and v =f. 0 are linearly dependent vectors. Then, by definition
of linearly dependent, there exist constants c 1 and c 2 not both zero such that

c 1 u + c2v = 0. If c 1 = 0 , then we would have c2v = 0 which implies c2 = 0,

since v =f. 0. Hence, c 1 =f. 0, and by a similar argument c2 =f. 0. Solving

c 1 u+c2v = 0 for u, we find u = -c2v/c 1 = kv, where k =f. 0. Thus, the only

way in which two n onzero vectors can be linearly dependent is for one of them

to b e a nonzero scalar multiple of the other.) Since t here is no scalar constant

k such that v 2 = kv3, we conclude v 2 and v 3 are linearly independent and,

therefore, the general solution of the linear homogeneous system (16) is