A First Course in Linear Algebra, 2017a

9.3. Linear Independence 471
Solution. To determine if R is linearly independent, we write
a(2⃗u − ⃗w)+b(⃗w +⃗v)+c(3⃗v + 1 2 ⃗u)= ⃗0
If the set is linearly independent, the only solution will be a = b = c = 0. We proceed as follows.
a(2⃗u −⃗w)+b(⃗w +⃗v)+c(3⃗v + 1 2 ⃗u)
2a⃗u − a⃗w + b⃗w + b⃗v + 3c⃗v + 1 2 c⃗u
= ⃗0
= ⃗0
(2a + 1 2 c)⃗u +(b + 3c)⃗v +(−a + b)⃗w = ⃗0
We know that the set S = {⃗u,⃗v,⃗w} is linearly independent, which implies that the coefficients in the
last line of this equation must all equal 0. In other words:
2a + 1 2 c = 0
b + 3c = 0
−a + b = 0
The augmented matrix and resulting reduced row-echelon form are given by:
⎡
2 0 1 ⎤ ⎡
⎤
2
0
1 0 0 0
⎣ 0 1 3 0 ⎦ →···→⎣
0 1 0 0 ⎦
−1 1 0 0
0 0 1 0
Hence the solution is a = b = c = 0 and the set is linearly independent.
♠
The following theorem was discussed in terms in R n . We consider it here in the general case.
Theorem 9.22: Unique Representation
Let V be a vector space and let U = {⃗v 1 ,···,⃗v k }⊆V be an independent set. If ⃗v ∈ span U, then⃗v
can be written uniquely as a linear combination of the vectors in U.
Consider the span of a linearly independent set of vectors. Suppose we take a vector which is not in this
span and add it to the set. The following lemma claims that the resulting set is still linearly independent.
Lemma 9.23: Adding to a Linearly Independent Set
Suppose⃗v /∈ span{⃗u 1 ,···,⃗u k } and {⃗u 1 ,···,⃗u k } is linearly independent. Then the set
{⃗u 1 ,···,⃗u k ,⃗v}
is also linearly independent.