8.4 COMPLEX VECTOR SPACES AND INNER PRODUCTS
All the vector spaces you have studied thus far in the text are
real vector spaces
because
the scalars are real numbers. A
complex vector space
is one in which the scalars are com-
plex numbers. So, if
are vectors in a complex vector space, then a linear
combination is of the form
where the scalars
are complex numbers. The complex version of
is the
complex vector space
consisting of ordered
n
-tuples of complex numbers. So, a vector
in
has the form
It is also convenient to represent vectors in
by column matrices of the form
As with
the operations of addition and scalar multiplication in
are performed com-
ponent by component.
EXAMPLE
1
Vector Operations in C
n
Let
and
be vectors in the complex vector space
Determine each vector.
(a)
(b)
(c)
Solution
(a) In column matrix form, the sum
is
(b) Because
and
you have
(c)
(c)
(c)
5
s
12
2
i
,
2
11
1
i
d
5
s
3
1
6
i
, 9
2
3
i
d
2
s
2
9
1
7
i
, 20
2
4
i
d
3
v
2
s
5
2
i
d
u
5
3
s
1
1
2
i
, 3
2
i
d
2
s
5
2
i
ds
2
2
1
i
, 4
d
s
2
1
i
d
v
5
s
2
1
i
ds
1
1
2
i
, 3
2
i
d
5
s
5
i
, 7
1
i
d
.
s
2
1
i
ds
3
2
i
d
5
7
1
i
,
s
2
1
i
ds
1
1
2
i
d
5
5
i
v
1
u
5
3
1
1
2
i
3
2
i
4
1
3
2
2
1
i
4
4
5
3
2
1
1
3
i
7
2
i
4
.
v
1
u
3
v
2
s
5
2
i
d
u
s
2
1
i
d
v
v
1
u
C
2
.
u
5
s
2
2
1
i
, 4
d
v
5
s
1
1
2
i
, 3
2
i
d
C
n
R
n
,
v
5
3
a
1
1
b
1
i
a
2
1
b
2
i
.
.
.
a
n
1
b
n
i
4
.
C
n
v
5
s
a
1
1
b
1
i
,
a
2
1
b
2
i
, . . . ,
a
n
1
b
n
i
d
.
C
n
C
n
R
n
c
1
,
c
2
, . . . ,
c
m
c
1
v
1
1
c
2
v
2
1
? ?
?
1
c
m
v
m
v
1
,
v
2
, . . . ,
v
m
SECTION 8.4
COMPLEX VECTOR SPACES AND INNER PRODUCTS
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*Sign up*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.
EXAMPLE 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.
EXAMPLE 3
Representing a Vector in C
n
by a Basis
Use the basis
S
in Example 2 to represent the vector
v
5
s
2,
i
, 2
2
i
d
.
H
v
1
,
v
2
,
v
3
J
c
1
5
c
2
5
c
3
5
0,
c
3
i
5
0.
c
2
i
5
0
s
c
1
1
c
2
d
i
5
0
ss
c
1
1
c
2
d
i
,
c
2
i
,
c
3
i
d
5
s
0, 0, 0
d
s
c
1
i
,
0, 0
d
1
s
c
2
i
,
c
2
i
, 0
d
1
s
0, 0,
c
3
i
d
5
s
0, 0, 0
d
c
1
v
1
1
c
2
v
2
1
c
3
v
3
5
s
0, 0, 0
d
H
v
1
,
v
2
,
v
3
J
C
3
C
3
.
S

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