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Transcript
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“main”
2007/2/16
page 271
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4.5
Linear Dependence and Linear Independence
271
2. Any set of vectors in V containing the zero vector is linearly dependent.
Remark We emphasize that the first result in Proposition 4.5.7 holds only for the
case of two vectors. It cannot be applied to sets containing more than two vectors.
Example 4.5.8
If v1 = (1, 2, −9) and v2 = (−2, −4, 18), then {v1 , v2 } is linearly dependent in R3 ,
since v2 = −2v1 . Geometrically, v1 and v2 lie on the same line.
Example 4.5.9
If
A1 =
2 1
,
3 4
A2 =
0 0
,
0 0
A3 =
2 5
,
−3 2
then {A1 , A2 , A3 } is linearly dependent in M2 (R), since it contains the zero vector from
M2 (R).
For more complicated situations, we must resort to Definitions 4.5.3 and 4.5.4,
although conceptually it is always helpful to keep in mind that the essence of the problem
we are solving is to determine whether a vector in a given set can be expressed as a linear
combination of the remaining vectors in the set. We now give some examples to illustrate
the use of Definitions 4.5.3 and 4.5.4.
Example 4.5.10
If v1 = (1, 2, −1) v2 = (2, −1, 1), and v3 = (8, 1, 1), show that {v1 , v2 , v3 } is linearly
dependent in R3 , and determine the linear dependency relationship.
Solution: We must first establish that there are values of the scalars c1 , c2 , c3 , not all
zero, such that
c1 v1 + c2 v2 + c3 v3 = 0.
(4.5.1)
Substituting for the given vectors yields
c1 (1, 2, −1) + c2 (2, −1, 1) + c3 (8, 1, 1) = (0, 0, 0).
That is,
(c1 + 2c2 + 8c3 , 2c1 − c2 + c3 , −c1 + c2 + c3 ) = (0, 0, 0).
Equating corresponding components on either side of this equation yields
c1 + 2c2 + 8c3 = 0,
2c1 − c2 + c3 = 0,
−c1 + c2 + c3 = 0.
The reduced row-echelon form of the augmented matrix of this system is


1 0 2 0
0 1 3 0.
0 0 0 0
Consequently, the system has an infinite number of solutions for c1 , c2 , c3 , so the vectors
are linearly dependent.
In order to determine a specific linear dependency relationship, we proceed to find
c1 , c2 , and c3 . Setting c3 = t, we have c2 = −3t and c1 = −2t. Taking t = 1 and
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