Page 207 - 35Linear Algebra
P. 207
10.2 Showing Linear Independence 207
Example 117 Consider the vector space P 2 (t) of polynomials of degree less than or
equal to 2. Set:
v 1 = 1 + t
v 2 = 1 + t 2
v 3 = t + t 2
v 4 = 2 + t + t 2
2
v 5 = 1 + t + t .
The set {v 1 , . . . , v 5 } is linearly dependent, because v 4 = v 1 + v 2 .
10.2 Showing Linear Independence
We have seen two different ways to show a set of vectors is linearly dependent:
we can either find a linear combination of the vectors which is equal to
zero, or we can express one of the vectors as a linear combination of the
other vectors. On the other hand, to check that a set of vectors is linearly
independent, we must check that every linear combination of our vectors
with non-vanishing coefficients gives something other than the zero vector.
Equivalently, to show that the set v 1 , v 2 , . . . , v n is linearly independent, we
must show that the equation c 1 v 1 + c 2 v 2 + · · · + c n v n = 0 has no solutions
other than c 1 = c 2 = · · · = c n = 0.
3
Example 118 Consider the following vectors in R :
0 2 1
2
4
0
v 1 = , v 2 = , v 3 = .
2 1 3
Are they linearly independent?
We need to see whether the system
2
1
3
c v 1 + c v 2 + c v 3 = 0
2
3
1
has any solutions for c , c , c . We can rewrite this as a homogeneous system:
1
c
c
v 1 v 2 v 3 2 = 0.
c 3
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