Let’s go back to the previous section for a second and
examine Example 1 and Example 6. In Example 1 we saw that 
was a vector space with the standard addition
and scalar multiplication for any positive integer n. So, in particular is a vector space with the standard addition
and scalar multiplication. In Example 6
we saw that the set of points on a line through the origin in with the standard addition and vector space
multiplication was also a vector space.
So, just what is so important about these two examples? Well first notice that they both are using
the same addition and scalar multiplication.
In and of itself that isn’t important, but it will be important for the
end result of what we want to discuss here.
Next, the set of points in the vector space of Example 6 are also in the
set of points in the vector space of Example 1.
While it’s not important to the discussion here note that the opposite
isn’t true, given a line we can find points in that aren’t on the line.
What we’ve seen here is that, at least for some vector spaces,
it is possible to take certain subsets of the original vector space and as long
as we retain the definition of addition and scalar multiplication we will get a
new vector space. Of course, it is
possible for some subsets to not be a new vector space. To see an example of this see Example 7 from the previous section. In that example we’ve got a subset of with the standard addition and scalar
multiplication and yet it’s not a vector space.
We want to investigate this idea in more detail and we’ll
start off with the following definition.
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Definition 1 Suppose
that V is a vector space and W is a subset of V. If, under the addition
and scalar multiplication that is defined on V, W is also a vector
space then we call W a subspace of V.
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Now, technically if we wanted to show that a subset W of a vector space V was a subspace we’d need to show that all 10 of the axioms from
the definition of a vector space are
valid, however, in reality that doesn’t need to be done.
Many of the axioms (c,
d, g, h, i, and j) deal with how addition and scalar multiplication work, but W is inheriting the definition of
addition and scalar multiplication from V. Therefore, since elements of W are also elements of V the six axioms listed above are
guaranteed to be valid on W.
The only ones that we really need to worry about are the
remaining four, all of which require something to be in the subset W.
The first two (a, and b) are the closure axioms that require
that the sum of any two elements from W
is back in W and that the scalar
multiple of any element from W will
be back in W. Note that the sum and scalar multiple will
be in V we just don’t know if it will
be in W. We also need to verify that the zero vector
(axiom e) is in W and that each element of W
has a negative that is also in W
(axiom f).
As the following theorem shows however, the only two axioms
that we really need to worry about are the two closure axioms. Once we have those two axioms valid, we will
get the zero vector and negative vector for free.
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Theorem 1 Suppose
that W is a non-empty (i.e. at least one element in it)
subset of the vector space V then W will be a subspace if the following
two conditions are true.
(a) If
u and v are in W then is also in W (i.e. W is closed under addition).
(b) If
u is in W and c is any scalar
then cu is also in W (i.e. W is closed under scalar multiplication).
Where the definition of addition and scalar multiplication
on W are the same as on V.
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Proof : To prove
this theorem all we need to do is show that if we assume the two closure axioms
are valid the other 8 axioms will be given to us for free. As we discussed above the axioms c, d,
g, h, i, and j are true simply based on the fact
that W is a subset of V and it uses the same addition and
scalar multiplication and so we get these for free.
We only need to verify that assuming the two closure
condition we get axioms e and f as well. From the second condition above we see that
we are assuming that W is closed
under scalar multiplication and so both 0u
and must be in W,
but from Theorem 1 from the previous
section we know that,
But this means that the zero vector and the negative of u must be in W and so we’re done.
Be careful with this proof.
On the surface it may look like we never used the first condition of
closure under addition and we didn’t use that to show that axioms e and f were valid. However, in
order for W to be a vector space it
must be closed under addition and so without that first condition we can’t know
whether or not W is in fact a vector
space. Therefore, even though we didn’t
explicitly use it in the proof it was required in order to guarantee that we
have a vector space.
Next we should acknowledge the following fact.
Fact
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Every vector space, V,
has at least two subspaces. Namely, V itself and (the zero space).
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Because V can be
thought of as a subset of itself we can also think of it as a subspace of
itself. Also, the zero space which is
the vector space consisting only of the zero vector, is a subset of V and is a vector space in its own right and so will be a subspace
of V.
At this point we should probably take a look at some
examples. In all of these examples we
assume that the standard addition and scalar multiplication are being used in
each case unless otherwise stated.
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Example 1 Determine
if the given set is a subspace of the given vector space.
(a) Let
W be the set of all points, ,
from in which . Is this a subspace of ? [Solution]
(b) Let
W be the set of all points from of the form . Is this a subspace of ? [Solution]
(c) Let
W be the set of all points from of the form . Is this a subspace of ? [Solution]
Solution
In each of these cases we need to show either that the set
is closed under addition and scalar multiplication or it is not closed for at
least one of those.
(a) Let W be the set of all points, , from in
which .
Is this a subspace of ?
This set is closed under addition because,
and since we also have and so the resultant point is back in W.
However, this set is not closed under scalar
multiplication. Let c be any negative scalar and further assume that then,
Then because and we must have and so the resultant point is not in W because the first component is
neither zero nor positive.
Therefore, W is
not a subspace of V.
[Return to Problems]
(b) Let W be the set of all points from of
the form .
Is this a subspace of ?
This one is fairly simple to check a point will be in W if the first component is zero. So, let and be any two points in W and let c be any
scalar then,
So, both and cx are in W and so W is closed
under addition and scalar multiplication and so W is a subspace.
[Return to Problems]
(c) Let W be the set of all points from of
the form .
Is this a subspace of ?
This one is here just to keep us from making any
assumptions based on the previous part.
This set is closed under neither addition nor scalar
multiplication. In order for points to
be in W in this case the first
component must be a 1. However, if and be any two points in W and let c be any
scalar other than 1 we get,
Neither of which is in W
and so W is not a subspace.
[Return to Problems]
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Do not read too much into the result from part (c) of this example. In general the set of upper triangular matrices (without restrictions, unlike part (c) from above) is a subspace of and the set of lower triangular matrices is also a subspace of . You should verify this for the practice.
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Example 3 Determine
if the given set is a subspace of the given vector space.
(a) Let
be the set of all continuous functions on
the interval . Is this a subspace of ,
the set of all real valued functions on the interval . [Solution]
(b) Let
be the set of all polynomials of degree n or less. Is this a subspace of ? [Solution]
(c) Let
W be the set of all polynomials of
degree exactly n. Is this a
subspace of ? [Solution]
(d) Let
W be the set of all functions such
that . Is this a subspace of where we have ? [Solution]
Solution
(a) Let be
the set of all continuous functions on the interval .
Is this a subspace of , the set of all real valued functions on
the interval .
Okay, if you’ve not had Calculus you may not know what a
continuous function is. A quick and
dirty definition of continuity (not mathematically correct, but useful if you
haven’t had Calculus) is that a function is continuous on if there are no holes or breaks in the
graph. Put in another way. You can sketch the graph of the function
from a to b without ever picking up your pencil of pen.
A fact from Calculus (which if you haven’t had please just
believe this) is that the sum of two continuous functions is continuous and
multiplying a continuous function by a constants will give a new continuous
function. So, what this fact tells us
is that the set of continuous functions is closed under standard function
addition and scalar multiplication and that is what we’re working with here.
So, is a subspace of .
[Return to Problems]
(b) Let be
the set of all polynomials of degree n
or less. Is this a subspace of ?
First recall that a polynomial is said to have degree n if its largest exponent is n.
Okay, let and and let c
be any scalar. Then,
In both cases the degree of the new polynomial is not
greater than n. Of course in the case of scalar
multiplication it will remain degree n,
but with the sum, it is possible that some of the coefficients cancel out to
zero and hence reduce the degree of the polynomial.
The point is that is closed under addition and scalar
multiplication and so will be a subspace of .
[Return to Problems]
(c) I Let W be the set of all polynomials of
degree exactly n. Is this a
subspace of ?
In this case W is
not closed under addition. To see this let’s take a look at the case to keep things simple (the same
argument will work for other values of n)
and consider the following two polynomials,
where a is not
zero, we know this is true because each polynomial must have degree 2. The other constants may or may not be
zero. Both are polynomials of exactly
degree 2 (since a is not zero) and
if we add them we get,
So, the sum had degree 1 and so is not in W.
Therefore for W
is not closed under addition.
We looked at only to make it somewhat easier to write
down the two example polynomials. We
could just have easily done the work for general n and we’d get the same result and so W is not a subspace.
[Return to Problems]
(d) Let W be the set of all functions such
that .
Is this a subspace of where we have ?
First notice that if we don’t have then this problem makes no sense, so we will
assume that .
In this case suppose that we have two elements from W, and . This means that and . In order for W to be a subspace we’ll need to show that the sum and a scalar
multiple will also be in W. In other words, if we evaluate the sum or
the scalar multiple at 6 we’ll get a result of 10. However, this won’t happen. Let’s take a look at the sum. The sum is,
and so the sum will not be in W. Likewise, if c is any scalar that isn’t 1 we’ll
have,
and so the scalar is not in W either.
Therefore W is
not closed under addition or scalar multiplication and so is not a subspace.
[Return to Problems]
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Before we move on let’s make a couple of observations about
some of the sets we looked at in this example.
First, we should just point out that the set of all
continuous functions on the interval ,
,
is a fairly important vector space in its own right to many areas of
mathematical study.
Next, we saw that the set of all polynomials of degree less
than or equal to n, ,
was a subspace of . However, if you’ve had Calculus you’ll know
that polynomials are continuous and so can also be thought of as a subspace of as well.
In other words, subspaces can have subspaces themselves.
Finally, here is something for you to think about. In the last part we saw that the set of all
functions for which was not a subspace of with . Let’s take a more general look at this. For some fixed number k let W be the set of all
real valued functions for which . Are there any values of k for which W will be a
subspace of with ? Go back and think about how we did the work
for that part and that should show you that there is one value of k (and only one) for which W will be a subspace. Can you figure out what that number has to
be?
We now need to look at a fairly important subspace of that we’ll be seeing in future sections.
Let’s see some examples of null spaces that are easy to
find.
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Example 4 Determine
the null space of each of the following matrices.
(a) [Solution]
(b) [Solution]
(c) [Solution]
Solution
(a)
To find the null space of A we’ll need to solve the following system of equations.
We’ve given this in both matrix form and equation
form. In equation form it is easy to
see that the only solution is . In terms of vectors from the solution consists of the single vector and hence the null space of A is .
[Return to Problems]
(b)
Here is the system that we need to solve for this part.
Now, we can see that these two equations are in fact the
same equation and so we know there will be infinitely many solutions and that
they will have the form,
If you need a refresher on solutions to system take a look
at the first section of the first chapter.
So, the since the null space of B consists of all the solutions to .
Therefore, the null space of B
will consist of all the vectors from that are in the form,
We’ll see a better way to write this answer in the next
section.
In terms of equations, rather than vectors in ,
let’s note that the null space of B
will be all of the points that are on the equation through the origin give by
.
[Return to Problems]
(c)
In this case we’re going to be looking for solutions to
However, if you think about it, every vector x in will be a solution to this system since we
are multiplying x by the zero
matrix.
Hence the null space of 0 is all of .
[Return to Problems]
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To see some examples of a more complicated null space check
out Example 7 from the section on Basis
and Example 2 in the
Fundamental Subspace section. Both of
these examples have more going on in them, but the first step is to write down
the null space of a matrix so you can check out the first step of the examples
and then ignore the remainder of the examples.
Now, let’s go back and take a look at all the null spaces
that we saw in the previous example. The
null space for the first matrix was . For the second matrix the null space was the
line through the origin given by . The null space for the zero matrix was all of
. Thinking back to the early parts of this
section we can see that all of these are in fact subspaces of .
In fact, this will always be the case as the following
theorem shows.
Proof : We know
that the null space of A consists of
all the solution to the system . First, we should point out that the zero
vector, 0, in will be a solution to this system and so we
know that the null space is not empty.
This is a good thing since a vector space (subspace or not) must contain
at least one element.
Now that we know that the null space is not empty let x and y be two elements from the null space and let c be any scalar. We just
need to show that the sum and scalar multiple of these are also in the null
space and we’ll be done.
Let’s start with the sum.
The sum, is a solution to and so is in the null space. The null space is therefore closed under
addition.
Next, let’s take a look at the scalar multiple.
The scalar multiple is also in the null space and so the
null space is closed under scalar multiplication.
Therefore the null space is a subspace of .