Page 188 - Advanced engineering mathematics

168 CHAPTER 6 Vectors and Vector Spaces

EXAMPLE 6.14
Let
F 1 =< 1,0,1,0 >,F 2 =< 0,1,1,0 > and F 3 =< 2,3,5,0 >.
These vectors are linearly dependent in R because
4
F 3 = 2F 1 + 3F 2 .
4
The subspace S of R spanned by F 1 and F 2 is the same as the subspace spanned by all three of
these vectors. Indeed, any linear combination of all three vectors is a linear combination of the
first two:
c 1 F 1 + c 2 F 2 + c 3 F 3
= c 1 F 1 + c 2 F 2 + c 3 (2F 1 + 3F 2 )
= (c 1 + 2c 3 )F 1 + (c 2 + 3c 3 )F 2 .
F 1 and F 2 contain all of the information needed to specify S.

There is an important characterization of linear independence and dependence that is used
frequently.

THEOREM 6.1 Linear Dependence and Independence

Let F 1 ,F 2 ,··· ,F k be vectors in R . Then
n
1. F 1 ,F 2 ,··· ,F k are linearly dependent if and only if there are real numbers α 1 ,α 2 ,··· ,α k ,
not all zero, such that
α 1 F 1 + α 2 F 2 + ··· + α k F k = O.
2. F 1 ,F 2 ,··· ,F k are linearly independent if and only if an equation
α 1 F 1 + α 2 F 2 + ··· + α k F k = O,
can hold only if each coefficient is zero:
α 1 = α 2 = ··· = α k = 0.
Proof To prove (1), suppose first that F 1 ,F 2 ,··· ,F k are linearly dependent. Then at least one
of these vectors is a linear combination of the others. As a convenience, suppose
F 1 = α 2 F 2 + ··· + α k F k .
Then
F 1 − α 2 F 2 − ··· − α k F k = O.
This is a linear combination of F 1 ,F 2 ,··· ,F k adding up to the zero vector, and having at least
one nonzero coefficient (the coefficient of F 1 is 1).
Conversely, suppose there are real numbers α 1 ,··· ,α k , not all zero, such that
α 1 F 1 + α 2 F 2 + ··· + α k F k = O.
By assumption at least one of the coefficient is not zero. Suppose, for convenience, that α k = 0.
Then
α 1 α k−1
F k =− F 1 − ··· − F k−1 ,
α k α k
so F k is a linear combination of F 1 ,···, F k−1 and F 1 ,F 2 ,··· ,F k are linearly dependent.
Part (2) of the theorem is proved similarly.

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October 14, 2010 14:21 THM/NEIL Page-168 27410_06_ch06_p145-186