Suppose
X
1
, X
2
,
· · ·
X
m
is a finite set of vectors in
E
n
. The
span
of this
set,
sp
{
X
1
, X
2
,
· · ·
X
m
}
, is the set of all vectors
X
which are linear combina
tions of
X
1
, X
2
,
· · ·
X
m
. An easy computation shows that
sp
{
X
1
, X
2
,
· · ·
X
m
}
is a subspace of
E
n
: the subspace
spanned by
X
1
, X
2
,
· · ·
X
m
.
Example
Again let
S
=
{
X
∈
E
n

x
1
+
x
2
+
· · ·
+
x
n
= 0
}
.
We claim this space is the span of the
n

1 vectors
D
j
described in the
preceding example.
First of all, it is clear that any linear combination of
the
D
j
must have the indicated property since the
D
j
themselves have this
property. On the other hand, given any vector
X
such that
x
1
+
x
2
+
· · ·
+
x
n
=
0 it is easy to see that
X
=
x
1
D
2
+ (
x
1
+
x
2
)
D
3
+
· · ·
+ (
x
1
+
x
2
+
· · ·
+
x
n

1
)
D
n
.
(Working through the arithmetic the right hand side is seen to be the vector
(
x
1
, x
2
,
· · ·
, x
n

1
,
(

x
1

x
2
· · · 
x
n

1
)) and the last component is just
x
n
since
x
1
+
x
2
+
· · ·
+
x
n
= 0.
A set of vectors
X
1
, X
2
,
· · ·
, X
m
in
E
n
is
linearly independent
if none of
these vectors can be written as a linear combination of the others; that is, it
is not possible, for any
j
, to write
X
j
=
X
k
6
=
j
d
k
X
k
for any scalar coefficients
d
k
, k
6
=
j
. Another way to say this is that it is not
possible to find coefficients
c
k
, k
= 1
,
2
,
· · ·
, m
, not all zero, such that
c
1
X
1
+
c
2
X
2
+
· · ·
+
c
m
X
m
= 0
2