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 or 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
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 now 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,
So, this axiom is verified. Now, let’s verify (i). If x is any element of V and c and k are any two
scalars then,
We’ve got the final axiom to go here and that’s a fairly
simple one to verify.
Just remember that 1x
is the notation for scalar multiplication and NOT multiplication of x by the number 1.
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Okay, that was a lot of work and we’re not going to be
showing that much work in the remainder of the examples that are vector
spaces. We’ll leave that up to you to
check most of the axioms now that you’ve seen one done completely out. For those examples that aren’t a vector space
we’ll show the details on at least one of the axioms that fails. For these examples you should check the other
axioms to see if they are valid or fail.
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Example 6 Let
the set V be the points on a line
through the origin in with the standard addition and scalar
multiplication. Then V is a vector space.
First, let’s think about just what V is. The set V is all the points that are on some
line through the origin in . So, we know that the line must have the
equation,
for some a and
some b, at least one not zero. Also note that a and b are fixed
constants and aren’t allowed to change.
In other words we are always on the same line. Now, a point will be on the line, and hence in V, provided it satisfies the equation
above.
We’ll show that V
is closed under addition and scalar multiplication and leave it to you to
verify the remaining axioms. Let’s
first show that V is closed under
addition. To do this we’ll need the
sum of two random points from V,
say and ,
and we’ll need to show that is also in V. This amounts to showing
that this point satisfies the equation of the line and that’s fairly simple
to do, just plug the coordinates into the equation and verify we get zero.
So, after some rearranging and using the fact that both u and v were both in V (and
so satisfied the equation of the line) we see that the sum also satisfied the
line and so is in V. We’ve now shown that V is closed under addition.
To show that V
is closed under scalar multiplication we’ll need to show that for any u from V and any scalar, c,
the is also in V. This is done pretty
much as we did closed under addition.
So, cu is on the line and hence in V.
V is therefore closed under
scalar multiplication.
Again we’ll leave it to you to verify the remaining
axioms. Note however, that because
we’re working with the standard addition that the zero vector and negative
are the standard zero vector and negative that we’re used to dealing with,
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Note that we can extend this example to a line through the
origin in and still have a vector space. Showing that this set is a vector space can
be a little difficult if you don’t know the equation of a line in however, as many of you probably don’t, and so
we won’t show the work here.
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Example 7 Let
the set V be the points on a line
that does NOT go through the origin in with the standard addition and scalar
multiplication. Then V is not a vector space.
In this case the equation of the line will be,
for fixed constants a,
b, and c where at least one of a
and b is non-zero and c is not zero. This set is not closed under addition or
scalar multiplication. Here is the
work showing that it’s not closed under addition. Let and be any two points from V (and so they satisfy the equation above). Then,
So the sum, ,
does not satisfy the equation and hence is not in V and so V is not
closed under addition.
We’ll leave it to you to verify that this particular V is not closed under scalar
multiplication.
Also, note that since we are working on a set of points
from with the standard addition then the zero
vector must be ,
but because this doesn’t satisfy the equation it is not in V and so axiom (e) is also not satisfied.
In order for V to be a
vector space it must contain the zero vector 0!
You should go through the remaining axioms and see if
there are any others that fail.
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Before moving on we should note that prior to this example
all the sets that have not been vector spaces we’ve not been operating under
the standard addition and/or scalar multiplication. In this example we’ve now seen that for some
sets under the standard addition and scalar multiplication will not be vector
spaces either.
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Example 8 Let
the set V be the points on a plane
through the origin in with the standard addition and scalar
multiplication. Then V is a vector space.
The equation of a plane through the origin in is,
where a, b, and c are fixed constants and at least one is not zero.
Given the equation you can (hopefully) see that this will
work in pretty much the same manner as the Example 6 and so we won’t show any
work here.
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Okay, we’ve seen quite a few examples to this point, but
they’ve all involved sets that were some or all of and so we now need to see a couple of examples
of vector spaces whose elements (and hence the “vectors” of the set) are not
points in .
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Example 9 Let
n and m be fixed numbers and let represent the set of all matrices.
Also let addition and scalar multiplication on be the standard matrix addition and standard
matrix scalar multiplication. Then is a vector space.
If we let c be
any scalar and let the “vectors” u
and v represent any two matrices (i.e. they are both objects in ) then we know from our work in the
first chapter that the sum, ,
and the scalar multiple, cu, are also matrices and hence are in . So is closed under addition and scalar
multiplication.
Next, if we define the zero vector, 0, to be the zero matrix and if the “vector” u is some ,
A, we can define the negative, ,
to be the matrix -A then the properties of matrix arithmetic
will show that the remainder of the axioms are valid.
Therefore, is a vector space.
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Note that this example now gives us a whole host of new
vector spaces. For instance, the set of matrices, ,
is a vector space and the set of all matrices, ,
is a vector space, etc.
Also, the “vectors” in this vector space are really
matrices!
Here’s another important example that may appear to be even
stranger yet.
We should make a couple of quick comments about this vector
space before we move on. First, recall
that the represents the interval (i.e.
we include the endpoints). We could also
look at the set which is the set of all real valued functions
that are defined on (,
no endpoints) or the set of all real valued functions defined
on and we’ll still have a vector space.
Also, depending upon the interval we choose to work with we
may have a different set of functions in the set. For instance, the function would be in but not in because of division by zero.
In this case the “vectors” are now functions so again we
need to be careful with the term vector.
It can mean a lot of different things depending upon what type of vector
space we’re working with.
Both of the vector spaces from Examples 9 and 10 are fairly
important vector spaces and as we’ll look at them again in the next section
where we’ll see some examples of some related vector spaces.
There is one final example that we need to look at in this
section.
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Example 11 Let
V consist of a single object,
denoted by 0, and define
Then V is a
vector space and is called the zero
vector space.
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The last thing that we need to do in this section before
moving on is to get a nice set of facts that fall pretty much directly from the
axioms and will be true for all vector spaces.
The proofs of these are really quite simple, but they only
appear that way after you’ve seen them.
Coming up with them on your own can be difficult sometimes. We’ll give the proof for two of them and you
should try and prove the other two on your own.
Proof :
(a) Now, this can
seem tricky, but each of these steps will come straight from a property of real
numbers or one of the axioms above.
We’ll start with 0u and use
the fact that we can always write and then we’ll use axiom (h).
This may have seemed like a silly and/or strange step, but
it was required. We couldn’t just add a
0u onto one side because this would,
in essence, be using the fact that and that’s what we’re trying to prove!
So, while we don’t know just what 0u is as a vector, it is in the vector space and so we know from
axiom (f) that it has a negative
which we’ll denote by -0u. Add the negative to both sides and then use
axiom (f) again to say that
Finally, use axiom (e)
on the right side to get,
and we’ve proven (a).
(c) In this case
if we can show that then from axiom (f) we’ll know that is the negative of u, or in other words that . This isn’t too hard to show. We’ll start with and use axiom (j) to rewrite the first u
as follows,
Next, use axiom (h)
on the right side and then a nice property of real numbers.
Finally, use part (a)
of this theorem on the right side and we get,
