From A First Course in Linear Algebra
Version 2.12
© 2004.
Licensed under the GNU Free Documentation License.
http://linear.ups.edu/
A subspace is a vector space that is contained within another vector space. So
every subspace is a vector space in its own right, but it is also defined relative to
some other (larger) vector space. We will discover shortly that we are already
familiar with a wide variety of subspaces from previous sections. Here’s the
definition.
Definition S
Subspace
Suppose that
and are two
vector spaces that have identical definitions of vector addition and scalar multiplication,
and that is
a subset of ,
. Then
is a
subspace of .
Lets look at an example of a vector space inside another vector space.
Example SC3
A subspace of
We know that
is a vector space (Example VSCV). Consider the subset,
It is clear that , since the objects in are column vectors of size 3. But is a vector space? Does it satisfy the ten properties of Definition VS when we use the same operations? That is the main question. Suppose and are vectors from . Then we know that these vectors cannot be totally arbitrary, they must have gained membership in by virtue of meeting the membership test. For example, we know that must satisfy while must satisfy . Our first property (Property AC) asks the question, is ? When our set of vectors was , this was an easy question to answer. Now it is not so obvious. Notice first that
and we can test this vector for membership in as follows,
and by this computation we see that . One property down, nine to go.
If is a scalar and , is it always true that ? This is what we need to establish Property SC. Again, the answer is not as obvious as it was when our set of vectors was all of . Let’s see.
and we can test this vector for membership in with
and we see that indeed . Always.
If has a zero vector, it will be unique (Theorem ZVU). The zero vector for should also perform the required duties when added to elements of . So the likely candidate for a zero vector in is the same zero vector that we know has. You can check that is a zero vector in too (Property Z).
With a zero vector, we can now ask about additive inverses (Property AI). As you might suspect, the natural candidate for an additive inverse in is the same as the additive inverse from . However, we must insure that these additive inverses actually are elements of . Given , is ?
and we can test this vector for membership in with
and we now believe that .
Is the vector addition in commutative (Property C)? Is ? Of course! Nothing about restricting the scope of our set of vectors will prevent the operation from still being commutative. Indeed, the remaining five properties are unaffected by the transition to a smaller set of vectors, and so remain true. That was convenient.
So satisfies all ten properties, is therefore a vector space, and thus earns the title of being a subspace of .
In Example SC3 we proceeded through all ten of the vector space properties before believing that a subset was a subspace. But six of the properties were easy to prove, and we can lean on some of the properties of the vector space (the superset) to make the other four easier. Here is a theorem that will make it easier to test if a subset is a vector space. A shortcut if there ever was one.
Theorem TSS
Testing Subsets for Subspaces
Suppose that is a
vector space and
is a subset of ,
. Endow
with the same
operations as .
Then
is a subspace if and only if three conditions are met
Proof () We have the hypothesis that is a subspace, so by Definition VS we know that contains a zero vector. This is enough to show that . Also, since is a vector space it satisfies the additive and scalar multiplication closure properties, and so exactly meets the second and third conditions. If that was easy, the the other direction might require a bit more work.
() We have three properties for our hypothesis, and from this we should conclude that has the ten defining properties of a vector space. The second and third conditions of our hypothesis are exactly Property AC and Property SC. Our hypothesis that is a vector space implies that Property C, Property AA, Property SMA, Property DVA, Property DSA and Property O all hold. They continue to be true for vectors from since passing to a subset, and keeping the operation the same, leaves their statements unchanged. Eight down, two to go.
Suppose . Then by the third part of our hypothesis (scalar closure), we know that . By Theorem AISM , so together these statements show us that . is the additive inverse of in , but will continue in this role when viewed as element of the subset . So every element of has an additive inverse that is an element of and Property AI is completed. Just one property left.
While we have implicitly discussed the zero vector in the previous paragraph, we need to be certain that the zero vector (of ) really lives in . Since is non-empty, we can choose some vector . Then by the argument in the previous paragraph, we know . Now by Property AI for and then by the second part of our hypothesis (additive closure) we see that
So contain the zero vector from . Since this vector performs the required duties of a zero vector in , it will continue in that role as an element of . This gives us, Property Z, the final property of the ten required. (Sarah Fellez contributed to this proof.)
So just three conditions, plus being a subset of a known vector space, gets us all ten properties. Fabulous! This theorem can be paraphrased by saying that a subspace is “a non-empty subset (of a vector space) that is closed under vector addition and scalar multiplication.”
You might want to go back and rework Example SC3 in light of this result, perhaps seeing where we can now economize or where the work done in the example mirrored the proof and where it did not. We will press on and apply this theorem in a slightly more abstract setting.
Example SP4
A subspace of
is the vector space of polynomials with degree at most
(Example VSP).
Define a subset
as
so is the collection of those polynomials (with degree 4 or less) whose graphs cross the -axis at . Whenever we encounter a new set it is a good idea to gain a better understanding of the set by finding a few elements in the set, and a few outside it. For example , while .
Is nonempty? Yes, .
Additive closure? Suppose and . Is ? and are not totally arbitrary, we know that and . Then we can check for membership in ,
so we see that qualifies for membership in .
Scalar multiplication closure? Suppose that and . Then we know that . Testing for membership,
so .
We have shown that meets the three conditions of Theorem TSS and so qualifies as a subspace of . Notice that by Definition S we now know that is also a vector space. So all the properties of a vector space (Definition VS) and the theorems of Section VS apply in full.
Much of the power of Theorem TSS is that we can easily establish new vector spaces if we can locate them as subsets of other vector spaces, such as the ones presented in Subsection VS.EVS.
It can be as instructive to consider some subsets that are not subspaces. Since Theorem TSS is an equivalence (see Technique E) we can be assured that a subset is not a subspace if it violates one of the three conditions, and in any example of interest this will not be the “non-empty” condition. However, since a subspace has to be a vector space in its own right, we can also search for a violation of any one of the ten defining properties in Definition VS or any inherent property of a vector space, such as those given by the basic theorems of Subsection VS.VSP. Notice also that a violation need only be for a specific vector or pair of vectors.
Example NSC2Z
A non-subspace in ,
zero vector
Consider the subset
below as a candidate for being a subspace of
The zero vector of , will need to be the zero vector in also. However, since . So has no zero vector and fails Property Z of Definition VS. This subspace also fails to be closed under addition and scalar multiplication. Can you find examples of this?
Example NSC2A
A non-subspace in ,
additive closure
Consider the subset
below as a candidate for being a subspace of
You can check that , so the approach of the last example will not get us anywhere. However, notice that and . Yet
So fails the additive closure requirement of either Property AC or Theorem TSS, and is therefore not a subspace.
Example NSC2S
A non-subspace in ,
scalar multiplication closure
Consider the subset
below as a candidate for being a subspace of
is the set of integers, so we are only allowing “whole numbers” as the constituents of our vectors. Now, , and additive closure also holds (can you prove these claims?). So we will have to try something different. Note that and , but
So fails the scalar multiplication closure requirement of either Property SC or Theorem TSS, and is therefore not a subspace.
There are two examples of subspaces that are trivial. Suppose that is any vector space. Then is a subset of itself and is a vector space. By Definition S, qualifies as a subspace of itself. The set containing just the zero vector is also a subspace as can be seen by applying Theorem TSS or by simple modifications of the techniques hinted at in Example VSS. Since these subspaces are so obvious (and therefore not too interesting) we will refer to them as being trivial.
Definition TS
Trivial Subspaces
Given the vector space ,
the subspaces and
are each called a
trivial subspace.
We can also use Theorem TSS to prove more general statements about subspaces, as illustrated in the next theorem.
Theorem NSMS
Null Space of a Matrix is a Subspace
Suppose that is
an matrix. Then
the null space of ,
, is a
subspace of .
Proof We will examine the three requirements of Theorem TSS. Recall that .
First, , which can be inferred as a consequence of Theorem HSC. So .
Second, check additive closure by supposing that and . So we know a little something about and : and , and that is all we know. Question: Is ? Let’s check.
So, yes, qualifies for membership in .
Third, check scalar multiplication closure by supposing that and . So we know a little something about : , and that is all we know. Question: Is ? Let’s check.
So, yes, qualifies for membership in .
Having met the three conditions in Theorem TSS we can now say that the null space of a matrix is a subspace (and hence a vector space in its own right!).
Here is an example where we can exercise Theorem NSMS.
Example RSNS
Recasting a subspace as a null space
Consider the subset of
defined as
It is possible to show that is a subspace of by checking the three conditions of Theorem TSS directly, but it will get tedious rather quickly. Instead, give a fresh look and notice that it is a set of solutions to a homogeneous system of equations. Define the matrix
and then recognize that . By Theorem NSMS we can immediately see that is a subspace. Boom!
The span of a set of column vectors got a heavy workout in Chapter V and Chapter M. The definition of the span depended only on being able to formulate linear combinations. In any of our more general vector spaces we always have a definition of vector addition and of scalar multiplication. So we can build linear combinations and manufacture spans. This subsection contains two definitions that are just mild variants of definitions we have seen earlier for column vectors. If you haven’t already, compare them with Definition LCCV and Definition SSCV.
Definition LC
Linear Combination
Suppose that is a
vector space. Given
vectors
and
scalars ,
their linear combination is the vector
Example LCM
A linear combination of matrices
In the vector space
of
matrices, we have the vectors
and we can form linear combinations such as
When we realize that we can form linear combinations in any vector space, then it is natural to revisit our definition of the span of a set, since it is the set of all possible linear combinations of a set of vectors.
Definition SS
Span of a Set
Suppose that is a vector
space. Given a set of vectors ,
their span, ,
is the set of all possible linear combinations of
.
Symbolically,
Theorem SSS
Span of a Set is a Subspace
Suppose is a vector space.
Given a set of vectors ,
their span, , is
a subspace.
Proof We will verify the three conditions of Theorem TSS. First,
So we have written as a linear combination of the vectors in and by Definition SS and therefore .
Second, suppose and . Can we conclude that ? What do we know about and by virtue of their membership in ? There must be scalars from , and so that
Then
Since each is again a scalar from we have expressed the vector sum as a linear combination of the vectors from , and therefore by Definition SS we can say that .
Third, suppose and . Can we conclude that ? What do we know about by virtue of its membership in ? There must be scalars from , so that
Then
Since each is again a scalar from we have expressed the scalar multiple as a linear combination of the vectors from , and therefore by Definition SS we can say that .
With the three conditions of Theorem TSS met, we can say that is a subspace (and so is also vector space, Definition VS). (See Exercise SS.T20, Exercise SS.T21, Exercise SS.T22.)
Example SSP
Span of a set of polynomials
In Example SP4 we proved that
is a subspace of , the vector space of polynomials of degree at most 4. Since is a vector space itself, let’s construct a span within . First let
and verify that is a subset of by checking that each of these two polynomials has as a root. Now, if we define , then Theorem SSS tells us that is a subspace of . So quite quickly we have built a chain of subspaces, inside , and inside .
Rather than dwell on how quickly we can build subspaces, let’s try to gain a better understanding of just how the span construction creates subspaces, in the context of this example. We can quickly build representative elements of ,
and
and each of these polynomials must be in since it is closed under addition and scalar multiplication. But you might check for yourself that both of these polynomials have as a root.
I can tell you that is not in , but would you believe me? A first check shows that does have as a root, but that only shows that . What does have to do to gain membership in ? It must be a linear combination of the vectors in , and . So let’s suppose that is such a linear combination,
Notice that operations above are done in accordance with the definition of the vector space of polynomials (Example VSP). Now, if we equate coefficients, which is the definition of equality for polynomials, then we obtain the system of five linear equations in two variables
Build an augmented matrix from the system and row-reduce,
With a leading 1 in the final column of the row-reduced augmented matrix, Theorem RCLS tells us the system of equations is inconsistent. Therefore, there are no scalars, and , to establish as a linear combination of the elements in . So .
Let’s again examine membership in a span.
Example SM32
A subspace of
The set of all
matrices forms a vector space when we use the operations of matrix addition
(Definition MA) and scalar matrix multiplication (Definition MSM), as was show
in Example VSM. Consider the subset
and define a new subset of vectors in using the span (Definition SS), . So by Theorem SSS we know that is a subspace of . While is an infinite set, and this is a precise description, it would still be worthwhile to investigate whether or not contains certain elements.
First, is
in ? To answer this, we want to determine if can be written as a linear combination of the five matrices in . Can we find scalars, so that
Using our definition of matrix equality (Definition ME) we can translate this statement into six equations in the five unknowns,
This is a linear system of equations, which we can represent with an augmented matrix and row-reduce in search of solutions. The matrix that is row-equivalent to the augmented matrix is
So we recognize that the system is consistent since there is no leading 1 in the final column (Theorem RCLS), and compute free variables (Theorem FVCS). While there are infinitely many solutions, we are only in pursuit of a single solution, so let’s choose the free variable for simplicity’s sake. Then we easily see that , , , . So the scalars , , , , will provide a linear combination of the elements of that equals , as we can verify by checking,
So with one particular linear combination in hand, we are convinced that deserves to be a member of . Second, is
in ? To answer this, we want to determine if can be written as a linear combination of the five matrices in . Can we find scalars, so that
Using our definition of matrix equality (Definition ME) we can translate this statement into six equations in the five unknowns,
This is a linear system of equations, which we can represent with an augmented matrix and row-reduce in search of solutions. The matrix that is row-equivalent to the augmented matrix is
With a leading 1 in the last column Theorem RCLS tells us that the system is inconsistent. Therefore, there are no values for the scalars that will place in , and so we conclude that .
Notice how Example SSP and Example SM32 contained questions about membership in a span, but these questions quickly became questions about solutions to a system of linear equations. This will be a common theme going forward.
Several of the subsets of vectors spaces that we worked with in Chapter M are also subspaces — they are closed under vector addition and scalar multiplication in .
Theorem CSMS
Column Space of a Matrix is a Subspace
Suppose that
is an matrix.
Then is a
subspace of .
Proof Definition CSM shows us that is a subset of , and that it is defined as the span of a set of vectors from (the columns of the matrix). Since is a span, Theorem SSS says it is a subspace.
That was easy! Notice that we could have used this same approach to prove that the null space is a subspace, since Theorem SSNS provided a description of the null space of a matrix as the span of a set of vectors. However, I much prefer the current proof of Theorem NSMS. Speaking of easy, here is a very easy theorem that exposes another of our constructions as creating subspaces.
Theorem RSMS
Row Space of a Matrix is a Subspace
Suppose that
is an matrix.
Then is a
subspace of .
Proof Definition RSM says , so the row space of a matrix is a column space, and every column space is a subspace by Theorem CSMS. That’s enough.
One more.
Theorem LNSMS
Left Null Space of a Matrix is a Subspace
Suppose that
is an matrix.
Then is a
subspace of .
Proof Definition LNS says , so the left null space is a null space, and every null space is a subspace by Theorem NSMS. Done.
So the span of a set of vectors, and the null space, column space, row space and left null space of a matrix are all subspaces, and hence are all vector spaces, meaning they have all the properties detailed in Definition VS and in the basic theorems presented in Section VS. We have worked with these objects as just sets in Chapter V and Chapter M, but now we understand that they have much more structure. In particular, being closed under vector addition and scalar multiplication means a subspace is also closed under linear combinations.
Is the set a subspace of ? Explain your answer.
C15 Working within the vector space , determine if is in the subspace ,
Contributed by Chris Black Solution [922]
C16 Working within the vector space , determine if is in the subspace ,
Contributed by Chris Black Solution [922]
C17 Working within the vector space , determine if is in the subspace ,
Contributed by Chris Black Solution [923]
C20 Working within the vector space of polynomials of degree 3 or less, determine if is in the subspace below.
Contributed by Robert Beezer Solution [924]
C21 Consider the subspace
of the vector space of
matrices, .
Is an
element of ?
Contributed by Robert Beezer Solution [926]
C25 Show that the set
from Example NSC2Z fails Property AC and Property SC.
Contributed by Robert Beezer
C26 Show that the set
from Example NSC2S has Property AC.
Contributed by Robert Beezer
M20 In , the vector space of column vectors of size 3, prove that the set is a subspace.
Contributed by Robert Beezer Solution [927]
T20 A square matrix
of size is upper
triangular if
whenever .
Let
be the set of all upper triangular matrices of size
. Prove
that
is a subspace of the vector space of all square matrices of size
,
.
Contributed by Robert Beezer Solution [930]
T30 Let
be the set of all polynomials, of any degree. The set
is a vector
space. Let be
the subset of
consisting of all polynomials with only terms of even degree. Prove or disprove: the
set is a
subspace of .
Contributed by Chris Black Solution [932]
T31 Let
be the set of all polynomials, of any degree. The set
is a vector
space. Let be
the subset of
consisting of all polynomials with only terms of odd degree. Prove or disprove: the set
is a subspace
of .
Contributed by Chris Black Solution [934]
C15 Contributed by Chris Black Statement [917]
For to be an
element of
there must be linear combination of the vectors in
that
equals
(Definition SSCV). The existence of such scalars is equivalent to the linear system
being consistent,
where
is the matrix whose columns are the vectors from
(Theorem SLSLC).
So by Theorem RCLS the system is inconsistent, which indicates that is not an element of the subspace .
C16 Contributed by Chris Black Statement [917]
For to be an
element of
there must be linear combination of the vectors in
that
equals
(Definition SSCV). The existence of such scalars is equivalent to the linear system
being consistent,
where
is the matrix whose columns are the vectors from
(Theorem SLSLC).
So by Theorem RCLS the system is consistent, which indicates that is in the subspace .
C17 Contributed by Chris Black Statement [918]
For to be an
element of
there must be linear combination of the vectors in
that
equals
(Definition SSCV). The existence of such scalars is equivalent to the linear system
being consistent,
where
is the matrix whose columns are the vectors from
(Theorem SLSLC).
So by Theorem RCLS the system is consistent, which indicates that is in the subspace .
C20 Contributed by Robert Beezer Statement [918]
The question is if
can be written as a linear combination of the vectors in
. To check
this, we set
equal to a linear combination and massage with the definitions
of vector addition and scalar multiplication that we get with
(Example VSP)
Equating coefficients of equal powers of , we get the system of equations,
The augmented matrix of this system of equations row-reduces to
There is a leading 1 in the last column, so Theorem RCLS implies that the system is inconsistent. So there is no way for to gain membership in , so .
C21 Contributed by Robert Beezer Statement [919]
In order to belong to , we
must be able to express
as a linear combination of the elements in the spanning set of
.
So we begin with such an expression, using the unknowns
for
the scalars in the linear combination.
Massaging the right-hand side, according to the definition of the vector space operations in (Example VSM), we find the matrix equality,
Matrix equality allows us to form a system of four equations in three variables, whose augmented matrix row-reduces as follows,
Since this system of equations is consistent (Theorem RCLS), a solution will provide values for and that allow us to recognize as an element of .
M20 Contributed by Robert Beezer Statement [920]
The membership criteria for
is a single linear equation, which comprises a homogeneous system of equations. As such, we
can recognize
as the solutions to this system, and therefore
is a null space.
Specifically, .
Every null space is a subspace by Theorem NSMS.
A less direct solution appeals to Theorem TSS.
First, we want to be certain is non-empty. The zero vector of , , is a good candidate, since if it fails to be in , we will know that is not a vector space. Check that
so that .
Suppose and are vectors from . Then we know that these vectors cannot be totally arbitrary, they must have gained membership in by virtue of meeting the membership test. For example, we know that must satisfy while must satisfy . Our second criteria asks the question, is ? Notice first that
and we can test this vector for membership in as follows,
and by this computation we see that .
If is a scalar and , is it always true that ? To check our third criteria, we examine
and we can test this vector for membership in with
and we see that indeed . With the three conditions of Theorem TSS fulfilled, we can conclude that is a subspace of .
T20 Contributed by Robert Beezer Statement [920]
Apply Theorem TSS.
First, the zero vector of is the zero matrix, , whose entries are all zero (Definition ZM). This matrix then meets the condition that for and so is an element of .
Suppose . Is ? We examine the entries of “below” the diagonal. That is, in the following, assume that .
which qualifies for membership in .
Suppose and . Is ? We examine the entries of “below” the diagonal. That is, in the following, assume that .
which qualifies for membership in .
Having fulfilled the three conditions of Theorem TSS we see that is a subspace of .
T30 Contributed by Chris Black Statement [920]
Proof: Let be
the subset of
comprised of all polynomials with all terms of even degree. Clearly the set
is non-empty, as
is a polynomial of
even degree. Let and
be arbitrary elements
of . Then there exist
nonnegative integers
and
so that
for some constants and . Without loss of generality, we can assume that . Thus, we have
so has all even terms, and thus . Similarly, let be a scalar. Then
so that also has only terms of even degree, and . Thus, is a subspace of .
T31 Contributed by Chris Black Statement [921]
This conjecture is false. We know that the zero vector in
is the polynomial
, which does not have odd
degree. Thus, the set
does not contain the zero vector, and cannot be a vector space.