Linear Algebra

106 Chapter 2. Vector Spaces
Looking for a linear relationship
⎛
c1 ⎝ 1
⎞ ⎛
0⎠
+ c2 ⎝
0
0
⎞ ⎛
2⎠
+ c3 ⎝
0
1
⎞ ⎛
2⎠
+ c4 ⎝
0
0
⎞ ⎛
−1⎠
+ c5 ⎝
1
3
⎞ ⎛
3⎠
= ⎝
0
0
⎞
0⎠
0
gives a three equations/five unknowns linear system whose solution set can be
paramatrized in this way.
⎛ ⎞ ⎛ ⎞ ⎛ ⎞
c1 −1 −3
⎜c2⎟
⎜
⎜ ⎟ ⎜−1⎟
⎜
⎟ ⎜−3/2
⎟ �
{ ⎜c3⎟
⎜ ⎟ = c3
⎜ 1 ⎟ + c5
⎜ 0 ⎟ �
⎟ c3,c5 ∈ R}
⎝c4⎠
⎝ 0 ⎠ ⎝ 0 ⎠
0
1
c5
Setting, say, c3 =0andc5 = 1 shows that the fifth vector is a linear combination
of the first two. Thus, Lemma 1.1 gives that this set
⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞
1 0 1 0
S1 = { ⎝0⎠
, ⎝2⎠
, ⎝2⎠
, ⎝−1⎠}
0 0 0 1
has the same span as S0. Similarly, the third vector of the new set S1 is a linear
combination of the first two and we get
⎛ ⎞ ⎛ ⎞ ⎛ ⎞
1 0 0
S2 = { ⎝0⎠
, ⎝2⎠
, ⎝−1⎠}
0 0 1
with the same span as S1 and S0, but with one difference. This last set is
linearly independent (this is easily checked), and so removal of any of its vectors
will shrink the span.
We finish this subsection by recasting that example as a theorem that any
finite spanning set has a subset with the same span that is linearly independent.
To prove that result we will first need some facts about how linear independence
and dependence, which are properties of sets, interact with the subset relation
between sets.
1.13 Lemma Any subset of a linearly independent set is also linearly independent.
Any superset of a linearly dependent set is also linearly dependent.
Proof. This is clear. QED
Restated, independence is preserved by subset and dependence is preserved
by superset. Those are two of the four possible cases of interaction that we
can consider. The third case, whether linear dependence is preserved by the
subset operation, is covered by Example 1.12, which gives a linearly dependent
set S0 with a subset S1 that is linearly dependent and another subset S2 that
is linearly independent.
That leaves one case, whether linear independence is preserved by superset.
The next example shows what can happen.