Many of the properties of
are shared by
For instance, the scalar multiplicative
identity is the scalar 1 and the additive identity in
is
The
standard
basis
for
is simply
which is the standard basis for
Because this basis contains
n
vectors, it follows that the
dimension of
is
n.
Other bases exist; in fact, any linearly independent set of
n
vectors in
can be used, as demonstrated in Example 2.
E X A M P L E 2
Verifying a Basis
Show that
is a basis for
Solution
Because
has a dimension of 3, the set
will be a basis if it is linearly inde-
pendent. To check for linear independence, set a linear combination of the vectors in
S
equal
to
0
as follows.
This implies that
So,
and you can conclude that
is linearly independent.
E X A M P L E 3
Representing a Vector in C
n
by a Basis
Use the basis
S
in Example 2 to represent the vector
v
2,
i
, 2
i
.
v
1
,
v
2
,
v
3
c
1
c
2
c
3
0,
c
3
i
0.
c
2
i
0
c
1
c
2
i
0
c
1
c
2
i
,
c
2
i
,
c
3
i
0, 0, 0
c
1
i
,
0, 0
c
2
i
,
c
2
i
, 0
0, 0,
c
3
i
0, 0, 0
c
1
v
1
c
2
v
2
c
3
v
3
0, 0, 0
v
1
,
v
2
,
v
3
C
3
C
3
.
S
i
, 0, 0 ,
i
,
i
, 0 , 0, 0,
i
C
n
C
n
R
n
.
e
n
0, 0, 0, . . . , 1
.
.
.
e
2
0, 1, 0, . . . , 0
e
1
1, 0, 0, . . . , 0
C
n
0
0, 0, 0, . . . , 0 .
C
n
C
n
.
R
n
494
CHAPTER 8
COMPLEX VECTOR SPACES
v
1
v
2
v
3