Page 209 - 35Linear Algebra

10.3 From Dependent Independent 209

10.3 From Dependent Independent

Now suppose vectors v 1 , . . . , v n are linearly dependent,
1
2
n
c v 1 + c v 2 + · · · + c v n = 0
1
with c 6= 0. Then:
span{v 1 , . . . , v n } = span{v 2 , . . . , v n }
because any x ∈ span{v 1 , . . . , v n } is given by

1 n
x = a v 1 + · · · + a v n
2 n
c c
2
n
= a 1 − v 2 − · · · − v n + a v 2 + · · · + a v n
c 1 c 1
2 n
c c
2
n
= a − a 1 v 2 + · · · + a − a 1 v n .
c 1 c 1
Then x is in span{v 2 , . . . , v n }.
When we write a vector space as the span of a list of vectors, we would like
that list to be as short as possible (this idea is explored further in chapter 11).
This can be achieved by iterating the above procedure.
Example 120 In the above example, we found that v 4 = v 1 + v 2 . In this case,
any expression for a vector as a linear combination involving v 4 can be turned into a
combination without v 4 by making the substitution v 4 = v 1 + v 2 .
Then:
2
2
2
2
S = span{1 + t, 1 + t , t + t , 2 + t + t , 1 + t + t }
2
2
2
= span{1 + t, 1 + t , t + t , 1 + t + t }.
1
1
2
2
2
Now we notice that 1 + t + t = 1 (1 + t) + (1 + t ) + (t + t ). So the vector
2
2
2
2
1 + t + t = v 5 is also extraneous, since it can be expressed as a linear combination of
the remaining three vectors, v 1 , v 2 , v 3 . Therefore
2
2
S = span{1 + t, 1 + t , t + t }.
In fact, you can check that there are no (non-zero) solutions to the linear system
2
3
1
2
2
c (1 + t) + c (1 + t ) + c (t + t ) = 0.
2
2
Therefore the remaining vectors {1 + t, 1 + t , t + t } are linearly independent, and
span the vector space S. Then these vectors are a minimal spanning set, in the sense
that no more vectors can be removed since the vectors are linearly independent. Such
a set is called a basis for S.
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