Linear Algebra

Section II. Linear Independence 103
1.2 Example In R3 , where
⎛
�v1 = ⎝ 1
⎞
0⎠
,
⎛
�v2 = ⎝
0
0
⎞
1⎠
,
⎛
�v3 = ⎝
0
2
⎞
1⎠
0
the spans [{�v1,�v2}] and [{�v1,�v2,�v3}] are equal since �v3 is in the span [{�v1,�v2}].
The lemma says that if we have a spanning set then we can remove a �v to
get a new set S with the same span if and only if �v is a linear combination of
vectors from S. Thus, under the second sense described above, a spanning set
is minimal if and only if it contains no vectors that are linear combinations of
the others in that set. We have a term for this important property.
1.3 Definition A subset of a vector space is linearly independent if none of its
elements is a linear combination of the others. Otherwise it is linearly dependent.
Here is a small but useful observation: although this way of writing one
vector as a combination of the others
�s0 = c1�s1 + c2�s2 + ···+ cn�sn
visually sets �s0 off from the other vectors, algebraically there is nothing special
in that equation about �s0. For any �si with a coefficient ci that is nonzero, we
can rewrite the relationship to set off �si.
�si =(1/ci)�s0 +(−c1/ci)�s1 + ···+(−cn/ci)�sn
When we don’t want to single out any vector by writing it alone on one side of
the equation we will instead say that �s0,�s1,...,�sn areinalinear relationship and
write the relationship with all of the vectors on the same side. The next result
rephrases the linear independence definition in this style. It gives what is usually
the easiest way to compute whether a finite set is dependent or independent.
1.4 Lemma A subset S of a vector space is linearly independent if and only if
for any distinct �s1,...,�sn ∈ S the only linear relationship among those vectors
is the trivial one: c1 =0,..., cn =0.
c1�s1 + ···+ cn�sn = �0 c1,...,cn ∈ R
Proof. This is a direct consequence of the observation above.
If the set S is linearly independent then no vector �si can be written as a linear
combination of the other vectors from S so there is no linear relationship where
some of the �s ’s have nonzero coefficients. If S is not linearly independent then
some �si is a linear combination �si = c1�s1+···+ci−1�si−1+ci+1�si+1+···+cn�sn of
other vectors from S, and subtracting �si from both sides of that equation gives
a linear relationship involving a nonzero coefficient, namely the −1 infrontof
�si. QED