1550078481-Ordinary_Differential_Equations__Roberts_

360 Ordinary Differential Equations

single eigenvector we calculated by hand in example 2 was x = ( -~).

Notice that x 1 and x 2 are both scalar multiples of x. This example

is intended to show that we can so metimes determine directly from

computer output when eigenvalues of multiplicity m > 1 have fewer

than m linearly indep endent associated eigenvectors.

THE H.i.TRIX liJHOSE EIGENVALUES AND EIGENVECTORS ARE TO BE CALCULATED IS

3.0000EtOO l.OOOOEtOO

-1.0000EtOO l.OOOOEtOO

AN EIGENVALUE IS 2.00000EtOO t O.OOOOOEtOO I

THE ASSOCIATED EIGENVECTOR IS

7.07107E-Ol t O.OOOOOEtOO I

-7.07107E-Ol t O.OOOOOEtOO I

AN EIGENVALUE IS 2.00000EtOO t O.OOOOOEtOO I

THE ASSOCIATED EIGENVECTOR IS

3.18453Etl5. t O.OOOOOEtOO I

-3.18453Etl5 t O.OOOOOEtOO I

Figure 8.2 Eigenvalues and Eigenvectors for Example 5.2.

3. In example 3 we manually calculated t he eigenvalues of the given matrix

to be .A 1 = -1, >. 2 = 2 and >. 3 = 2. And we found the following set of

three linearly independent associated eigenvectors

Using EIGEN, we found, as shown in Figure 8.3, that the eigenvalues of

the given matrix are -1, 2, 2 and that the associated eigenvectors are

respectively

(

.745356)

Z1 = .745356 ,

-.745356 (

.894427)

Z2 = -.447214 ,

.447214 (

.311803)

Z3 = .344098.

.655902

Notice that z 1 = -.745356x 1 , but z2 is not a multiple of x2 or x 3

and z 3 is not a multiple of x 2 or x 3. The easiest way for us to show

that z 2 and z 3 are linearly independent eigenvectors associated wit h the

eigenvalue >. = 2 of multiplicity 2 is to show that the set { z 1 , z 2 , zs} is

linearly independent. We do so by comput ing det (z 1 z 2 z 3 ) and finding

its value to b e -3(.745356)(.447214) = -1 # 0.