Linear Combinations and Span
Let
v1,
v2,…,
v r be vectors in
R n . A
linear combination of these vectors is any expression of the form
where the coefficients
k1,
k2,…,
kr are scalars.
Example 1: The vector
v = (−7, −6) is a linear combination of the vectors
v1 = (−2, 3) and
v2 = (1, 4), since
v = 2
v1 − 3
v2. The zero vector is also a linear combination of
v1 and
v2, since
0 = 0
v1 + 0
v2. In fact, it is easy to see that the zero vector in
R n is always a linear combination of any collection of vectors
v1,
v2,…,
v r from
R n .
The set of
all linear combinations of a collection of vectors
v1,
v2,…,
v r from
R n is called the
span of {
v1,
v2,…,
v r }. This set, denoted span {
v1,
v2,…,
v r }, is always a subspace of
R n , since it is clearly closed under addition and scalar multiplication (because it contains
all linear combinations of
v1,
v2,…,
v r ). If
V = span {
v1,
v2,…,
v r }, then
V is said to be
spanned by
v1,
v2,…,
v r .
Example 2: The span of the set {(2, 5, 3), (1, 1, 1)} is the subspace of
R3 consisting of all linear combinations of the vectors
v1 = (2, 5, 3) and
v2 = (1, 1, 1). This defines a plane in
R3. Since a normal vector to this plane in
n =
v1 x
v2 = (2, 1, −3), the equation of this plane has the form 2
x +
y − 3
z =
d for some constant
d. Since the plane must contain the origin—it's a subspace—
d must be 0. This is the plane in Example 7.
Example 3: The subspace of
R2 spanned by the vectors
i = (1, 0) and
j = (0, 1) is all of
R2, because
every vector in
R2 can be written as a linear combination of
i and
j:
Let
v1,
v2,…,
v r−1 ,
v r be vectors in
R n . If
v r is a linear combination of
v1,
v2,…,
v r−1 , then
That is, if any one of the vectors in a given collection is a linear combination of the others, then it can be discarded without affecting the span. Therefore, to arrive at the most “efficient” spanning set, seek out and eliminate any vectors that depend on (that is, can be written as a linear combination of) the others.
Example 4: Let
v1 = (2, 5, 3),
v2 = (1, 1, 1), and
v3 = (3, 15, 7). Since
v3 = 4
v1 − 5
v2,
That is, because
v3 is a linear combination of
v1 and
v2, it can be eliminated from the collection without affecting the span. Geometrically, the vector (3, 15, 7) lies in the plane spanned by
v1 and
v2 (see Example 7 above), so adding multiples of
v3 to linear combinations of
v1 and
v2 would yield no vectors off this plane. Note that
v1 is a linear combination of
v2 and
v3 (since
v1 = 5/4
v2 + 1/4
v3), and
v2 is a linear combination of
v1 and
v3 (since
v2 = 4/5
v1 − 1/5
v3). Therefore,
any one of these vectors can be discarded without affecting the span:
Example 5: Let
v1 = (2, 5, 3),
v2 = (1, 1, 1), and
v3 = (4, −2, 0). Because there exist no constants
k1 and
k2 such that
v3 =
k1 v1 +
k2 v2,
v3 is not a linear combination of
v1 and
v2. Therefore,
v3 does not lie in the plane spanned by
v1 and
v2, as shown in Figure
1 :
Consequently, the span of
v1,
v2, and
v3 contains vectors not in the span of
v1 and
v2 alone. In fact,
