Linear Algebra, 2020a

112 Chapter Two. Vector Spaces
are no such c 2 and c 3 . But knowing that the first vector is not dependent on the
other two is not enough. This person would have to go on to try ⃗v 2 = c 1 ⃗v 1 +c 3 ⃗v 3 ,
in order to find the dependence c 1 = 0, c 3 = 1/2. Lemma 1.5 gets the same
conclusion with only one computation.
1.11 Example The empty subset of a vector space is linearly independent. There
is no nontrivial linear relationship among its members as it has no members.
1.12 Example In any vector space, any subset containing the zero vector is linearly
dependent. One example is, in the space P 2 of quadratic polynomials, the subset
{1 + x, x + x 2 ,0}. It is linearly dependent because 0 · ⃗v 1 + 0 · ⃗v 2 + 1 · ⃗0 = ⃗0 is a
nontrivial relationship, since not all of the coefficients are zero.
There is a subtle point that we shall see a number of times and that bears
on the prior example. It is about the trivial sum, the sum of the empty set.
One way to see how to define the trivial sum is to consider the progression
⃗v 1 + ⃗v 2 + ⃗v 3 , followed by ⃗v 1 + ⃗v 2 , followed by ⃗v 1 . The difference between the
sum of three vectors and the sum of two is ⃗v 3 . Then the difference between the
sum of two and the sum of one is ⃗v 2 . In next passing to the trivial sum, the
sum of zero-many vectors, we can expect to subtract ⃗v 1 . So we define the sum
of zero-many vectors to be the zero vector.
The relation with the prior example is that if the zero vector is in a set then
that set has an element that is a combination of a subset of other vectors from
the set, specifically, the zero vector is a combination of the empty subset. Even
the set S = {⃗0} is linearly dependent, because ⃗0 is the sum of the empty set and
the empty set is a subset of S.
1.13 Remark The definition of linear independence, Definition 1.4, refers to a
‘set’ of vectors. Sets are the most familiar kind of collection and in practice
everyone refers to these collections as sets. But to be complete we will note that
sets are not quite the right kind of collection for this purpose.
Recall that a set is a collection with two properties: (i) order does not matter,
so that the set {1, 2} equals the set {2, 1}, and (ii) duplicates collapse, so that
the set {1, 1, 2} equals the set {1, 2}.
Now consider this matrix reduction.
⎛
⎜
1 1 1
⎞ ⎛
⎟
⎝2 2 2⎠ (1/2)ρ 2 ⎜
1 1 1
⎞
⎟
−→ ⎝1 1 1⎠
1 2 3
1 2 3
On the left the set of matrix rows {(1 1 1), (2 2 2), (1 2 3)} is linearly dependent.
On the right the set of rows is {(1 1 1), (1 1 1), (1 2 3)}. Because
duplicates collapse, that equals the set {(1 1 1), (1 2 3)}, which is linearly
independent. That’s a problem because Gauss’s Method should preserve linear
dependence.