S4.1 Real Vector Spaces In this section, we look at a generalization

S4.1 Real Vector Spaces
In this section, we look at a generalization of
and see that
larger class of sets and these sets are called real vector spaces.
is part of a
Definition 1 Let be a nonempty set of objects that is closed under some
defined addition and scalar multiplication i.e.,
based on the definitions given for this addition and scalar
multiplication. Then we say that is a vector space if together with its two
operations satisfy the following eight conditions
:
1.
2.
(addition is commutative).
(
) (
)
(addition is associative).
⃗
⃗ (there is a zero
3. There is an object ⃗
vector/add. identity).
( ) ⃗
4. For each
(
5.
6.
7.
8.
)
(add. Inverse)
(
)
(scalar multiplication distributes over vector
addition).
(
)
(multiplication distributes over scalar addition).
( )
( ) (scalar multiplication is associative).
( one is the multiplicative identity for scalar multiplication).
Example 1 Show that
is a real vector space.
Example 2: let
be the set of all polynomials with real coefficients of degree
{
less than or equal to i.e.,
} . The vector addition in is defined by just adding corresponding coefficients
and scalar multiplication is multiplying all the coefficients of a polynomial by the
scalar, as is done in any algebra class. Show that together with these two
operations is a real vector space.
Closed under vector addition:
Closed under scalar multiplication:
Has a zero vector:
Addition of vectors is commutative:
Each vector has an additive inverse:
The other properties can easily be verified.
Example 3 Let
[
] {
[
]}
[
]
We define vector addition as
Show
] together with these two operations is a real vector space.
that [
Closed under vector addition:
Closed under scalar multiplication:
Has a zero vector:
Addition of vectors is commutative:
Each vector has an additive inverse:
The other properties can easily be
verified.
Example 4: We define
{[
]
}. The
vector addition is just the usual addition of 2 x 2 matrices and scalar multiplication
is just the usual multiplication of a 2x2 matrix by a scalar. Show that
together with these two operations is a real vector space.
Example 5: Show that the subset of
{
operations as .
Theorem 1 Let
be a vector space
statements are always true:
1.
⃗
⃗
2. ⃗
3. ( )
⃗
4.
⃗
given by
} is not a vector space under the same
Then the following
S4.2 Subspaces
Definition 1 If
is a vector space and
and is itself a
vector space under the same operations of addition and scalar multiplication as in
then we say that
.
It is easier to prove that a subset of a vector space is itself a vector space (under
the same operations) because most of the properties are inherited from the
larger vector space.
Theorem 1 (Subspace Theorem) Suppose that
is a vector space and is a
nonempty subset of . Then if is closed under the same scalar multiplication
and vector addition as
then is a subspace of
Proof of the subspace theorem:
This means that to show
10 (easier!).
1)
2)
3)
is a subspace we need only show 3 things instead of
(nonempty- usually we show this by showing ⃗
(closed under addition)
(closed under scalar mult.)
)
Example 1: Suppose that
{
is a given vector. Show that
} is a subspace of
.
Example 2: Show that a line through the origin is a subspace of
.
Example 3 Show that if a plane passes through the origin then it is a subspace of
.
Example 4: Let
. Now we can think of vectors in
as
⃗ } is a subspace of ,
column vectors. Then the set
{
which will call the null space later in this chapter.
Example 5: Show that
Theorem 2 If
Proof:
{
[
] ∫
( )
} is a subspace of [
]
More generally if
are
is also a subspace
{
Definition 2 Suppose that
Then
} is a set containing
( ) {
} i.e., the
( ) is the set of all possible linear combinations of the vectors in
{
}
Examples of span:
1. In
2. In
3. In
({
({
({
⃗⃗
})
.
})
})
.
.
({ })
4. If
5. If
is a plane through the origin in
Theorem 3 Let
addition, if
subset of i.e.,
{
({
})
} is a set containing
. Then
( ) is a subspace of
In
then
( ) is a
( ) is the smallest subspace of
that contains
Example 6: Show that
({(
)(
) (
)}
.
Proof of Theorem 3:
By definition, ⃗
suppose that
(
)
( )
( ), where the
(
)
(
( )
,
are real numbers. Then
) , which is clearly in
. Now
( ), so
( ) is closed under addition.
(
)
Also,
, which is
( ) closed under scalar
in
( ) , since each
is a real number (so
mult.) Thus,
( ) is a subspace of by the Subspace Theorem.
{
Now if
that contains
}, since
contain all linear combinations of {
( ) is a subset of
hence
} then must
is a vector space and

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