Linear Algebra, 2020a

Section II. Linear Independence 111
gives
c 1 + c 2 = 0
c 1 − c 2 = 0
−ρ 1 +ρ 2
−→
c 1 + c 2 = 0
2c 2 = 0
since polynomials are equal only if their coefficients are equal. Thus, the only
linear relationship between these two members of P 2 is the trivial one.
1.8 Remark The lemma specifies that ⃗s i ≠ ⃗s j when i ≠ j because of course if some
vector ⃗s appears twice then we can get a nontrivial c 1 ⃗s 1 + ···+ c n ⃗s n = ⃗0, by
taking the associated coefficients to be 1 and −1. Besides, if some vector appears
more than once in an expression then we can always combine the coefficients.
Note that the lemma allows the opposite of appearing more than once, that
some vectors from S don’t appear at all. For instance, if S is infinite then because
linear relationships involve only finitely many vectors, any such relationship
leaves out many of S’s vectors. However, note also that if S is finite then where
convenient we can take a combination c 1 ⃗s 1 + ···+ c n ⃗s n to contain each of S’s
vectors once and only once. If a vector is missing then we can add it by using a
coefficient of 0.
1.9 Example The rows of this matrix
⎛
⎜
2 3 1 0
⎞
⎟
A = ⎝0 −1 0 −2⎠
0 0 0 1
form a linearly independent set. This is easy to check for this case but also recall
that Lemma One.III.2.5 shows that the rows of any echelon form matrix form a
linearly independent set.
1.10 Example In R 3 , where
⎛
⎜
3
⎞ ⎛
⎟ ⎜
2
⎞ ⎛ ⎞
4
⎟ ⎜ ⎟
⃗v 1 = ⎝4⎠ ⃗v 2 = ⎝9⎠ ⃗v 3 = ⎝18⎠
5
2
4
the set S = {⃗v 1 ,⃗v 2 ,⃗v 3 } is linearly dependent because this is a relationship
0 · ⃗v 1 + 2 · ⃗v 2 − 1 · ⃗v 3 = ⃗0
where not all of the scalars are zero (the fact that some of the scalars are zero
doesn’t matter).
That example illustrates why, although Definition 1.4 is a clearer statement
of what independence means, Lemma 1.5 is better for computations. Working
straight from the definition, someone trying to compute whether S is linearly
independent would start by setting ⃗v 1 = c 2 ⃗v 2 + c 3 ⃗v 3 and concluding that there