A First Course in Linear Algebra, 2017a

470 Vector Spaces
The augmented matrix and resulting reduced row-echelon form are given by
⎡
⎤ ⎡ ⎤
1 2 0
1 0 0
⎣ 2 −1 0 ⎦ →···→⎣
0 1 0 ⎦
−1 3 0
0 0 0
Hence the solution is a = b = 0 and the set is linearly independent.
♠
The next example shows us what it means for a set to be dependent.
Example 9.19: Dependent Set
Determine if the set S given below is independent.
⎧⎡
⎤ ⎡
⎨ −1
S = ⎣ 0 ⎦, ⎣
⎩
1
1
1
1
⎤
⎡
⎦, ⎣
1
3
5
⎤⎫
⎬
⎦
⎭
Solution. To determine if S is linearly independent, we look for solutions to
⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤
−1 1 1 0
a⎣
0 ⎦ + b⎣
1 ⎦ + c⎣
3 ⎦ = ⎣ 0 ⎦
1 1 5 0
Notice that this equation has nontrivial solutions, for example a = 2, b = 3andc = −1. Therefore S is
dependent.
♠
The following is an important result regarding dependent sets.
Lemma 9.20: Dependent Sets
Let V be a vector space and suppose W = {⃗v 1 ,⃗v 2 ,···,⃗v k } is a subset of V. ThenW is dependent if
and only if⃗v i can be written as a linear combination of {⃗v 1 ,⃗v 2 ,···,⃗v i−1 ,⃗v i+1 ,···,⃗v k } for some i ≤ k.
Revisit Example 9.19 with this in mind. Notice that we can write one of the three vectors as a combination
of the others.
⎡ ⎤
1
⎡ ⎤
−1
⎡ ⎤
1
⎣ 3 ⎦ = 2⎣
5
0
1
⎦ + 3⎣
1 ⎦
1
By Lemma 9.20 this set is dependent.
If we know that one particular set is linearly independent, we can use this information to determine if
a related set is linearly independent. Consider the following example.
Example 9.21: Related Independent Sets
Let V be a vector space and suppose S ⊆ V is a set of linearly independent vectors given by
S = {⃗u,⃗v,⃗w}. Let R ⊆ V be given by R = { 2⃗u −⃗w,⃗w +⃗v,3⃗v + 1 2 ⃗u} . Show that R is also linearly
independent.