In this section we will cover a topic that we’ll see off and
on over the course of this chapter.
Let’s start off by going back to part (b) of Example 4 from
the previous section. In that example we
saw that the null space of the given matrix consisted of all the vectors of the
form
We would like a more compact way of stating this result and
by the end of this section we’ll have that.
Let’s first revisit an idea that we saw quite some time
ago. In the section on Matrix Arithmetic
we looked at linear combinations of matrices and columns of matrices. We can also talk about linear combinations of
vectors.
So, we can see that the null space we were looking at above
is in fact all the linear combinations of the vector (7,1). It may seem strange to talk about linear
combinations of a single vector since that is really scalar multiplication, but
we can think of it as that if we need to.
The null space above was not the first time that we’ve seen
linear combinations of vectors however.
When we were looking at Euclidean n-space
we introduced these things called the standard basis vectors. The standard basis vectors for were defined as,
We saw that we could take any vector from and write it as,
Or, in other words, we could write u as a linear combination of the standard basis vectors, . We will be revisiting this idea again in a
couple of sections, but the point here is simply that we’ve seen linear
combinations of vectors prior to us actually discussing them here.
Let’s take a look at an example or two.
So, this example was kept fairly simple, but if we add in
more components and/or more vectors to the set the problem will work in
essentially the same manner.
Now that we’ve seen how linear combinations work and how to
tell if a vector is a linear combination of a set of other vectors we need to
move into the real topic of this section.
In the opening of this section we recalled a null space that we’d looked
at in the previous section. We can now
see that the null space from that example is nothing more than all the linear
combinations of the vector (7,1) (and again, it is kind of strange to be
talking about linear combinations of a single vector).
As pointed out at the time we’re after a more compact
notation for denoting this. It is now
time to give that notation.
So, with this notation we can now see that the null space
that we examined at the start of this section is now nothing more than,
Before we move on to some examples we should get a nice
theorem out of the way.
Proof :
(a) So, we need
to show that W is closed under
addition and scalar multiplication. Let u and w be any two vectors from W. Now, since W is the set of all linear combinations of that means that both u and w must be a linear
combination of these vectors. So, there
are scalars and so that,
Now, let’s take a look at the sum.
So the sum, ,
is a linear combination of the vectors and hence must be in W and so W is closed
under addition.
Now, let k be any
scalar and let’s take a look at,
As we can see the scalar multiple, ku, is a linear
combination of the vectors and hence must be in W and so W is closed
under scalar multiplication.
Therefore, W must
be a vector space.
(b) In these
cases when we say that W is the
smallest vector space that contains the set of vectors we’re really saying that if is also a vector space that contains then it will also contain a complete copy of W as well.
So, let’s start this off by noticing that W does in fact contain each of the ’s since,
Now, let be a vector space that contains and consider any vector u from W. If we can show that u must also be in then we’ll have shown that contains a copy of W since it will contain all the vectors in W. Now, u is in W and so must be
a linear combination of ,
Each of the terms in this sum, ,
is a scalar multiple of a vector that is in and since is a vector space it must be closed under
scalar multiplication and so each is in . But this means that u is the sum of a bunch of vectors that are in which is closed under addition and so that
means that u must in fact be in .
We’ve now shown that contains every vector from W and so must contain W itself.
Now, let’s take a look at some examples of spans.
Now, let’s see if we can determine a set of vectors that
will span some of the common vector spaces that we’ve seen. What we’ll need in each of these examples is
a set of vectors with which we can write a general vector from the space as a
linear combination of the vectors in the set.
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Example 4 Determine
a set of vectors that will exactly span each of the following vector spaces.
(a) [Solution]
(b) [Solution]
(c) [Solution]
Solution
Okay, before we start this let’s think about just what we
need to show here. We’ll need to find
a set of vectors so that the span of that set will be exactly the space
given. In other words, we need to show
that the span of our proposed set of vectors is in fact the same set as the
vector space.
So just what do we need to do to mathematically show that
two sets are equal? Let’s suppose that
we want to show that A and B are equal sets. To so this we’ll need to show that each a in A will be in B and in
doing so we’ll have shown that B
will at the least contain all of A. Likewise, we’ll need to show that each b in B will be in A and in
doing that we’ll have shown that A
will contain all of B. However, the only way that A can contain all of B and B can contain all of A
is for A and B to be the same set.
So, for our example we’ll need to determine a possible set
of spanning vectors show that every
vector from our vector space is in the span of our set of vectors. Next we’ll need to show that each vector in
our span will also be in the vector space.
(a)
We’ve pretty much done this one already. Earlier in the section we showed that any vector from can be written as a linear combination of
the standard basis vectors, and so at the least the span of the standard
basis vectors will contain all of . However, since any linear combination of
the standard basis vectors is going to be a vector in we can see that must also contain the span of the standard
basis vectors.
Therefore, the span of the standard basis vectors must be .
[Return to Problems]
(b)
We can use result of Example 3(a) above as a guide
here. In that example we saw a set of
matrices that would span all the diagonal matrices in and so we can do a natural extension to get
a set that will span all of . It looks like the following set should do
it.
Clearly any
linear combination of these four matrices will be a matrix and hence in and so the span of these matrices must be
contained in .
Likewise, given
any matrix from ,
we can write it
as the following linear combination of these “vectors”.
and so must be contained in the span of these
vectors and so these vectors will span .
[Return to Problems]
(c)
We can use Example 3(b) to help with this one. First recall that is the set of all polynomials of degree n or less. Using Example 3(b) as a guide it looks like
the following set of “vectors” will work for us.
Note that used
subscripts that matched the degree of the term and so started at instead of the usual .
It should be
clear (hopefully) that a linear combination of these is a polynomial of degree
n or less and so will be in . Therefore the span of these vectors will be
contained in .
Likewise, we
can write a general polynomial of degree n
or less,
as the following
linear combination
Therefore is contained in the span of these vectors
and this means that the span of these vectors is exactly .
[Return to Problems]
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There is one last idea about spans that we need to discuss
and its best illustrated with an example.
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Example 5 Determine
if the following sets of vectors will span .
(a) ,
,
and . [Solution]
(b) ,
,
and . [Solution]
Solution
(a) ,
,
and .
Okay let’s think about how we’ve got to approach
this. Clearly the span of these
vectors will be in since they are vectors from . The real question is whether or not will be contained in the span of these
vectors, . In the previous example our set of vectors
contained vectors that we could easily show this. However, in this case its not so
clear. So to answer that question here
we’ll do the following.
Choose a general vector from ,
,
and determine if we can find scalars ,
,
and so that u
is a linear combination of the given vectors.
Or,
If we set
components equal we arrive at the following system of equations,
In matrix form
this is,
What we need to
do is to determine if this system will be consistent (i.e. have at least one solution) for every possible choice of . Nicely enough this is very easy to do if
you recall Theorem 9
from the section on Determinant Properties.
This theorem tells us that this system will be consistent for every
choice of provided the coefficient matrix is
invertible and we can check that be doing a quick determinant
computation. So, if we denote the
coefficient matrix as A we’ll leave
it to you to verify that .
Therefore the
coefficient matrix is invertible and so this system will have a solution for
every choice of . This in turn tells us that is contained in and so we’ve now shown that
[Return to Problems]
(b) ,
,
and .
We’ll do this
one a little quicker. As with the
first part, let’s choose a general vector form and form up the system that we need to
solve. We’ll leave it to you to verify
that the matrix form of this system is,
This system
will have a solution for every choice of if the coefficient matrix, A, is invertible. However, in this case we have (you should verify this) and so the
coefficient matrix is not invertible.
This in turn
tells us that there is at least one choice of for which this system will not have a
solution and so cannot be written as a linear combination of
these three vectors. Note that there
are in fact infinitely many choices of that will not yield solutions!
Now, we know that
is contained in ,
but we’ve just shown that there is at least one vector from that is not contained in and so the span of these three vectors will
not be all of .
[Return to Problems]
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This example has shown us two things. First, it has shown us that we can’t just
write down any set of three vectors and expect to get those three vectors to
span . This is an idea we’re going to be looking at
in much greater detail in the next couple of sections.
Secondly, we’ve now seen at least two different sets of
vectors that will span . There are the three vectors from Example 5(a)
as well as the standard basis vectors for . This tells us that the set of vectors that
will span a vector space is not unique.
In other words, we can have more than one set of vectors span the same
vector space.