1550078481-Ordinary_Differential_Equations__Roberts_

Linear Systems of First-Order Differential Equations 345

SOLUTION

Forming the matrix Y whose jth column is Yj and computing det Y , we

find

(

1 0 2)

det Y = det 0 1 0 = 1 + 0 + 0 - 0 - 0 + 2 = 3 -=I-0.

-1 1 1

Therefore, the set of vectors {y 1 , Y2, y3} is linearly independent.

DEFINITIONS Linearly Dependent and Linearly Independent

Sets of Vector Functions

Let {y1 ( x), Y2 ( x),.. ., y m ( x)} be a set of vector functions which are all

the same size-say, n x 1. Thus, each component of Yj(x) is a function of

x. Let

Yj(x) = (~~~-~~~).

Ynj.(x)

Suppose each vector function y j ( x) is defined on the interval [a, b]- that

is, suppose Yij(x) is defined on [a, b] for all i = 1, 2, ... , n and all j =

1, 2, ... , m. The concept of linear dependence and linear independence for

vector functions is then defined as follows:

If there exists a set of scalar constants c1, c2, ... , Cm not all zero such that

C1Y1(x) + C2Y2(x) + · · · + CmYm(x) = 0 for all x E [a, b],

then the set of vector functions {YI(x),y2(x),.. .,Ym(x)} is linearly de-

pendent on the interval [a, b].

The set of vector functions {YI(x), Y2(x), .. ., Ym(x)} is linearly inde-

pendent on the interval [a, b], if

C1Y1(x) + C2Y2(x) + · · · + CmYm(x) = 0 for all x E [a, b]

implies C1 = c2 = · · · = Cm = 0.

Determining linear independence or linear dependence of a set of vector
functions on an interval is more complicated than determining linear indepen-
dence or linear dependence for a set of constant vectors. If the set of vector

functions {y 1 ( x ), y 2 ( x), ... , y m ( x)} is linearly dependent on an interval, then

it is linearly dependent at each point in the interval. However, if the set is