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As noted in the introduction to this chapter vectors do not
have to represent directed line segments in space. When we first start looking at many of the
concepts of a vector space we usually start with the directed line segment idea
and their natural extension to vectors in 
because it is something that most people can
visualize and get their hands on. So,
the first thing that we need to do in this chapter is to define just what a
vector space is and just what vectors really are.
However, before we actually do that we should point out that
because most people can visualize directed line segments most of our examples
in these notes will revolve around vectors in . We will try to always include an example or
two with vectors that aren’t in just to make sure that we don’t forget
that vectors are more general objects, but the reality is that most of the
examples will be in .
So, with all that out of the way let’s go ahead and get the
definition of a vector and a vector space out of the way.
We should make a couple of comments about these axioms at
this point. First, do not get too locked
into the “standard” ways of defining addition and scalar multiplication. For the most part we will be doing addition
and scalar multiplication in a fairly standard way, but there will be the
occasional example where we won’t. In
order for something to be a vector space it simply must have an addition and
scalar multiplication that meets the above axioms and it doesn’t matter how strange
the addition of scalar multiplication might be.
Next, the first two axioms may seem a little strange at
first glance. It might seem like these
two will be trivially true for any definition of addition or scalar
multiplication, however, we will see at least one example in this section of a
set that is not closed under a particular scalar multiplication.
Finally, with the exception of the first two these axioms
should all seem familiar to you. All of
these axioms were in one of the theorems from the discussion on vectors and/or
Euclidean n-space in the previous
chapter. However, in this case they
aren’t properties, they are axioms. What
that means is that they aren’t to be proven.
Axioms are simply the rules under which we’re going to operate when we
work with vector spaces. Given a
definition of addition and scalar multiplication we’ll simply need to verify
that the above axioms are satisfied by our definitions.
We should also make a quick comment about the scalars that
we’ll be using here. To this point, and
in all the examples we’ll be looking at in the future, the scalars are real
numbers. However, they don’t have to be
real numbers. They could be complex
numbers. When we restrict the scalars to
real numbers we generally call the vector space a real vector space and when we allow the scalars to be complex
numbers we generally call the vector space a complex vector space. We
will be working exclusively with real vector spaces and from this point on when
we see vector space it is to be understood that we mean a real vector space.
We should now look at some examples of vector spaces and at
least a couple of examples of sets that aren’t vector spaces. Some of these will be fairly standard vector
spaces while others may seem a little strange at first but are fairly important
to other areas of mathematics.
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Example 1 If
n is any positive integer then the
set with the standard addition and scalar
multiplication as defined in the Euclidean n-space section is a vector space.
Technically we should show that the axioms are all met
here, however that was done in Theorem
1 from the Euclidean n-space
section and so we won’t do that for this example.
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Note that from this point on when we refer to the standard
vector addition and standard vector scalar multiplication we are referring to
that we defined in the Euclidean n-space
section.
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Example 2 The
set with the standard vector addition and scalar
multiplication defined as,
is NOT a vector space.
Showing that something is not a vector space can be tricky
because it’s completely possible that only one of the axioms fails. In this case because we’re dealing with the
standard addition all the axioms involving the addition of objects from V
(a, c, d, e, and f) will be valid.
Also, in this case of all the axioms involving the scalar
multiplication (b, g, h, i, and j), only (h) is not valid. We’ll
show this in a bit, but the point needs to be made here that only one of the
axioms will fail in this case and that is enough for this set under this
definition of addition and multiplication to not be a vector space.
First we should at least show that the set meets axiom (b) and this is easy enough to show,
in that we can see that the result of the scalar multiplication is again a
point in and so the set is closed under scalar
multiplication. Again, do not get used
to this happening. We will see at
least one example later in this section of a set that is not closed under
scalar multiplication as we’ll define it there.
Now, to show that (h)
is not valid we’ll need to compute both sides of the equality and show that
they aren’t equal.
So, we can see that because the first components are not the
same. This means that axiom (h) is not valid for this definition
of scalar multiplication.
We’ll not verify that the remaining scalar multiplication
axioms are valid for this definition of scalar multiplication. We’ll leave those to you. All you need to do is compute both sides of
the equal sign and show that you get the same thing on each side.
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Example 3 The
set with the standard vector addition and scalar
multiplication defined as,
is NOT a vector space.
Again, there is a single axiom that fails in this
case. We’ll leave it to you to verify
that the others hold. In this case it
is the last axiom, (j), that fails
as the following work shows.
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Example 4 The
set with the standard scalar multiplication and
addition defined as,
Is NOT a vector space.
To see that this is not a vector space let’s take a look
at the axiom (c).
So, because only the first component of the second point
listed gets multiplied by 2 we can see that and so this is not a vector space.
You should go through the other axioms and determine if
they are valid or not for the practice.
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So, we’ve now seen three examples of sets of the form that are NOT vector spaces so hopefully it is
clear that there are sets out there that aren’t vector spaces. In each case we had to change the definition
of scalar multiplication or addition to make the set fail to be a vector
space. However, don’t read too much into
that. It is possible for a set under the
standard scalar multiplication and addition to fail to be a vector space as
we’ll see in a bit. Likewise, it’s
possible for a set of this form to have a non-standard scalar multiplication
and/or addition and still be a vector space.
In fact, let’s take a look at the following example. This is probably going to be the only example
that we’re going to go through and do in excruciating detail in this
section. We’re doing this for two
reasons. First, you really should see
all the detail that needs to go into actually showing that a set along with a
definition of addition and scalar multiplication is a vector space. Second, our definitions are NOT going to be
standard here and it would be easy to get confused with the details if you had
to go through them on your own.
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Example 5 Suppose
that the set V is the set of
positive real numbers (i.e. ) with addition and scalar
multiplication defined as follows,
This set under this addition and scalar multiplication is
a vector space.
First notice that we’re taking V to be only a portion of . If we took it to be all of we would not have a vector space. Next, do not get excited about the
definitions of “addition” and “scalar multiplication” here. Even though they are not they are not
addition and scalar multiplication as we think of them we are still going to
call them the addition and scalar multiplication operations for this vector
space.
Okay, let’s go through each of the axioms and verify that
they are valid.
First let’s take a look at the closure axioms, (a) and (b). Since by x and y are positive numbers their product xy is a positive real number and so the V is closed under addition.
Since x is positive then for
any c is a positive real number and so V is closed under scalar
multiplication.
Next we’ll verify (c). We’ll do this one with some detail pointing
out how we do each step. First assume
that x and y are any two elements of V
(i.e. they are two positive real
numbers).
We’ll now verify (d). Again, we’ll make it clear how we’re going
about each step with this one. Assume
that x, y, and z are any three
elements of V.
Next we need to find the zero vector, 0, and we need to be careful here. We use 0
to denote the zero vector but it does NOT have to be the number zero. In fact in this case it can’t be zero if
for no other reason than the fact that the number zero isn’t in the set V !
We need to find an element that is in V so that under our definition of addition we have,
It looks like we should define the “zero vector” in this
case as : 0=1. In other words the zero vector for this set
will be the number 1! Let’s see how
that works and remember that our “addition” here is really multiplication and
remember to substitute the number 1 in for 0. If x is any element of V,
Sure enough that does what we want it to do.
We next need to define the negative, ,
for each element x that is in V.
As with the zero vector to not confuse with “minus x”, this is just the notation we use to denote the negative of x.
In our case we need an element of V
(so it can’t be minus x since that
isn’t in V) such that
and remember that 0=1
in our case!
Given an x in V we know that x is strictly positive and so is defined (since x isn’t zero) and is positive (since x is positive) and therefore is in V. Also, under our definition of addition and
the zero vector we have,
Therefore, for the set V
the negative of x is .
So, at this point we’ve taken care of the closure and
addition axioms we not just need to deal with the axioms relating to scalar
multiplication.
We’ll start with (g). We’ll do this one in some detail so you can
see what we’re doing at each step. If x and y are any two elements of V
and c is any scalar then,
So, it looks like we’ve verified (g).
Let’s now verify (h). If x
is any element of V and c and k are any two scalars then,
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