# Subspace – Serlo

In this article we consider the *subspace* of a vector space. The subspace is a subset of the vector space, which itself is a vector space.

A subset of the vector space can be identified as a subspace (i.e., it is again a vector space) if and only if the following three properties are satisfied:

- .
- For all we have that .
- For all and for all we have that .

This equivalence is called the subspace criterion. If one of them toes not hold, then we do not have a subspace.

## Motivation[Bearbeiten]

As we have already seen in connection with general algebraic structures like groups or fields, *sub-structures* (like sub-groups or sub-fields) play a major role in mathematics. To repeat: Substructures are (small) subsets of a (large) original structure, which allow for the same computations as the original sets. For example, considering the algebraic structure "group", a subgroup is a subset of a group, which is itself a group. For instance, the set of integer numbers with addition can be seen as a subgroup of the set of rational numbers , which is again a subgroup of the real numbers . In the same way, for the algebraic structure "field", a subfield is a subset of a field, which itself is a field.

In linear algebra we consider a new algebraic structure: the "vector space". As before, we can study the corresponding substructure: a sub-vector space, or simply **subspace is a subset of a vector space, which is again a vector space**.

## Definition of a subspace[Bearbeiten]

**Definition** (subspace)

Let be a -vector space. Consider a subset with the constrained operations and . Then, this subset is called a subspace of if is itself a -vector space.

**Hint**

In this definition, often a little detail is taken for granted: One needs that the constrained operations and have again values only in . That is, for a subspace it is required that for and also and hold. This does not hold for any vector space , see the Example below.

**Hint**

You might recall the notion of a subgroup. We can also think of every vector space as an abelian group . Now if is a subspace of , then forms a subgroup of .

## Subspace criterion [Bearbeiten]

### Derivation of the criterion[Bearbeiten]

How do we find out if a subset of a vector space is a subspace? For to be a vector space, all vector space axioms for must be fulfilled. Let's first use an example to see how this works.

#### Checking the vector space axioms for an example[Bearbeiten]

We consider the subset of the -vector space . Visually this subset is a line. We want to find out, whether is a subspace of . So by definition, we need to show that the set together with the operations and satisfy all vector space axioms. We proceed as in the article Proofs for vector spaces. That means, we prove that the vector addition and the scalar multiplication are well defined and that eight axioms hold.

First we have to show that the two operations are well-defined. The crucial point here is whether we really "land in the subspace again" under addition and scalar multiplication. More precisely, the addition in is a map with . Our new addition arises by first restricting the domain of definition to the subset . We get a map . So the range of values remains the same for now. However, to make the set a vector space, we need a map , so after addition, we "land again in the subspace". So, is the image of contained in ? What we would like to show is:

For all we have that also .

This property is also called *completeness under addition*.

Analogously one can derive a criterion for the well-definedness of the scalar multiplication:

For all and we have that also .

This property is called *completeness under scalar multiplication*.

We now check both properties in our concrete example:

First, the addition. Let . That is, exist such that and . Then . If we set , we have that . So .

Now the scalar multiplication. Let as we just did, and let . Then we have . If we set , we have that . So .

Hence, the completeness under addition and scalar multiplication are indeed valid. So the vector space operations are well defined. We note that here we have worked very concretely with the definition of the set . More specifically, we have used that every element of is of the form .

Now we check the eight vector space axioms.

First, the four axioms for addition:

Associative law of addition: Let . We must show that . Since is the constrained version of , we must show . This follows from the associative law for the vector space . Note that since , we also have , so the relations in are indeed valid.

The commutative law for addition in can be traced back to the commutative law in .

Existence of a neutral element: We must show that an element exists such that for all . Since is a vector space, it contains a zero vector with for all . In particular, this holds for all . Since addition in is only the restriction of addition in , it suffices to show, That . For then we can set . The element is more precisely the vector . This can be written as and thus lies in . So there is indeed a neutral element of addition within .

Existence of additive inverse in : Let . We must show that there exists a , such that . We know that holds in . So if we can show , we are done: then we can choose . We know that holds. Furthermore, we have already shown that is complete under scalar multiplication, so indeed follows.

The four axioms of scalar multiplication can also be traced back to the corresponding properties of . This works similarly to the first two axioms of addition. We use that all relevant equations hold analogously in if one expresses the operations in by those in So we indeed have a subspace.

In order to show that the operations and are well-defined, we needed to establish the properties of completeness formulated above. For this we have worked closely with the definition of . Furthermore, for the third axiom of addition, we had to show that the neutral element of addition in is also an element of . Again, we worked concretely with the definition of . The axiom for the existence of the inverse with respect to addition could be traced back to the completeness of scalar multiplication. For all other axioms we could use that the analogous axioms in hold.

So, in total, we actually **only** needed to show **three things**:

- The completeness of with respect to addition.
- The completeness of with respect to scalar multiplication

For these we had to work with the definition of and . The above arguments that these three properties suffice should be true for every vector space and all subsets of . So, in the general case, it should be enough to prove these three properties (and it actually is).

But first, we demonstrate that the three rules are necessary. That is, we show that none of the three rules can be omitted. For this we give subsets of , each of which violates exactly one of the three rules and is indeed not a subspace.

#### Counterexample: The empty set[Bearbeiten]

We first consider the empty set . This is of course a subset of .

Let us check completeness with respect to of addition, , it is satisfied. This is because all statements about the empty set (missing) trivially always hold. In the same way, the completeness of scalar multiplication is satisfied.

However, the third rule is violated: . That is simply because the empty set contains by definition no elements. The property cannot be derived in general from the completeness of addition and of scalar multiplication.

Now, is not a vector space, because contains no element, in particular no neutral element of addition. Accordingly, cannot be a subspace. Be aware that the property cannot in general be derived from the completeness properties and must be, in principle, separately checked. (However, this step is often seen as obvious)

#### Counterexample: Vectors with integer entries[Bearbeiten]

Our second example shows that scalar multiplication is actually needed: take the set of integer vectors . If we identify the vectors with points in , we get some kind of "lattice":

This set is obviously a subset of and again the question arises whether it is a subspace. In contrast to the first example now the zero vector is contained in . All other axioms of vector addition are also valid. The sum of two vectors from is again in .

Nevertheless, the is not a subspace of , because is *not* complete with respect to scalar multiplication. For instance, and , but is not contained in . Thus does not satisfy all vector space axioms and is therefore not a subspace.

The completeness with respect to scalar multiplication can also *not* be derived from the other two properties. If we want to prove that is a subspace, we must always show that for every and for every scalar , we also have .

#### Counterexample: A cross of coordinate axes [Bearbeiten]

We have already seen in both examples of above that every subspace contains the zero vector and is closed under scalar multiplication. Finally, we want to look at a third and last example, which satisfies the above two conditions, but still does not satisfy all vector space axioms. For this we choose the *axis cross*, the set formed by union of the two lines through the origin and . So we consider the subset . Illustrated in the plane as points, the set looks like an infinite "cross":

Is a subspace? Obviously, the zero vector is contained in . Moreover, we have that for any and that also is an element of . Thus is complete under scalar multiplication. Nevertheless, is not a subspace. To see this, we choose the vectors and . Then, we have , but for the sum we have that .

So the completeness under the vector space addition can *not* be derived from the other properties. That means we always have to check the completeness under the vector space addition to prove that is a subspace.

### Statement and proof of the criterion [Bearbeiten]

We considered an example that a subset of is a subspace if it satisfies the following three properties:

- completeness with respect to addition,
- completeness with respect to scalar multiplication, and.
- .

We have seen examples of subsets of where one of these properties was not satisfied in each case and which also do not form a subspace of . So we assume that these three properties are necessary and sufficient for a subset to be a subspace. This is the theorem of the subspace criterion, which we will now prove.

**Theorem** (subspace criterion)

A subset of a -vector space with vector addition and scalar multiplication is a subspace exactly if the following three conditions hold:

- .
- For all we have that .
- For all and for all we have that .

In other words: A subset of a vector space is a subspace if it contains the zero element and is complete with respect to vector addition and scalar multiplication.

**Proof** (subspace criterion)

The theorem contains an *if and only if"*, which means that we have to show *two implications*. One direction is: Every subspace satisfies conditions 1, 2 and 3. The other direction can be formulated as follows: Any subset of the vector space that satisfies conditions 1, 2 and 3 must be a subspace.

**Proof step:** Every subspace satisfies the conditions 1, 2 and 3.

Let be any subspace of . Then by definition , is also a vector space. Thus for all axioms from the definition of a vector space hold.

In particular, this also means that the operations and restricted to are well-defined. But this is only another formulation of the conditions 2) and 3).

Moreover, since U together with forms an abelian group, contains the neutral element . That is, it we also have that condition 1 holds.

**Proof step:** If conditions 1, 2 and 3 are satisfied, the subset must be a subspace.

Let be a subset of for which the three conditions hold. We need to show that is a vector space. To do this, we show that satisfies *all* properties from the definition of a vector space.

From condition 1, it follows that is a non-empty set. From conditions 2 and 3 we can deduce that the of on restricted links and are well-defined. So we have a non-empty set with an operation (vector addition) and an operation (scalar multiplication).

Now we have to prove that together with forms an abelian group and the axioms of scalar multiplication hold. We may use that is a vector space. From this, by we can conclude:

*associative law:*For all we have that:*commutative law:*For all we have that:*scalar distributive law:*For all and all we have that:*vectorial distributive law:*For all and all we have that:*associative law for scalars:*For all and all we have that:*Neutral element of scalar multiplication:*For all and for (the neutral element of multiplication in ) we have that: . The 1 is also called*neutral element of scalar multiplication*.

It remains to establish the axioms *existence of a neutral element* and *existence of an inverse element*. The former is given by condition 1. So it suffices to show that for every there exists a such that . But now, for any condition 3 implies

So is the inverse element to and contained in .

**Hint**

Instead of in some mathematical texts also is required. Both requirements are equivalent (if one adds the other two conditions 2 and 3): If there is a , then because of the completeness of scalar multiplication also must be contained in .

**Hint**

Another equivalent formulation of the criterion is:

A non-empty subset is a subspace if for any two vectors also every linear combination , with an arbitrary lies in .

It is easy to convince ourselves of the equivalence of both formulations:

Since is not empty, there exists some , and therefore also lies in (point 1). With also is contained in (point 3). Finally, with , we also have that is in (point 2).

Conversely, is not empty by 1. Further, if it contains some elements and , then by means of 3, also and must lie in and from 2 we finally get .

## How to prove that a set is a subspace [Bearbeiten]

### General proof structure[Bearbeiten]

Before we examine the procedure in more detail with an example, it is useful to understand the general proof structure. How can we show that a set is a subspace of a -vector space ? We can use the subspace criterion that we just established. In order for us to use the criterion we must first check the preconditions. The theorem requires that . Then, to show that is a subspace, we need to check the three properties from the criterion. So, in total, we need to show the following four statements:

- .
- For all we have that .
- For all and for all we have that .

**Hint**

We can also replace the second statement "" by "". If we add the conditions 3. and 4. the two statements are equivalent.

What do proofs of these statements look like? The proof structure of these statements looks like this:

- Proof of "": Let . Then, we have that , since ...
- Proof of "": Let be the zero vector. Then, we have that , since ...
- Proof of "": Let be arbitrary. We have that ... and hence .
- Proof of "": Let and be arbitrary. Since ... we know that .

### Finding a proof idea [Bearbeiten]

We consider an easy example problem, in order to get an idea for the proof:

**Exercise**

Let and . Show that: is a subspace of the -vector space .

We want to apply the subspace criterion to . To do this, we check the premises of the theorem according to the above scheme.

- : let . By definition of , there is some with . Since is a vector space, it follows that .
- : We have seen in "Vector space: properties" that for every vector we have . So we have that also . Thus we get .
- completeness of addition: Let . By definition of , there exist with and . Since , we can add them: . Because of we finally obtain .
- completeness of scalar multiplication: Let and let . By definition of there is a with . Since we can multiply it with : . Because of it follows that .

This shows that all conditions hold, so by the subspace criterion, is indeed a subspace of .

### Writing down the proof[Bearbeiten]

Now we write down the proof, by generalizing the simple example to "any vector space":

**Proof**

We check the premises of the theorem according to the above scheme.

- : let . Then there exists a with . Hence, .
- : because of , we have .
- completeness of addition: let . Then there exist with and . We calculate: .
- completeness of scalar multiplication: let and let . Then there is a with . We calculate .

This shows that all the preconditions hold. So it follows from the subspace criterion that is a subspace of .

## Examples and counterexamples for subspaces[Bearbeiten]

### Examples [Bearbeiten]

In the following, we will look at first examples to consolidate our idea of subspaces and to avoid misinterpretations. We will also use the subspace criterion.

#### Trivial subspaces[Bearbeiten]

In every -vector space , there are tow "trivial subspaces":

**Example** (Trivial subspaces)

Let be a vector space. First, the zero vector space is always a subspace of . This is a vector space and since , we also have .

On the other hand, the complete vector space is also a subspace of . Finally we have that and is a vector space.

Since and are *always* subspaces for every vector space , they are called *trivial subspaces*.

the following example with shows that sometimes, only the *trivial* subspaces are subspaces:

**Example** (subspaces of )

as an -vector space has only the trivial subspaces and , as one can easily verify:

Let be a subspace with . We want to show . Since , there exists a real number . Because is complete under scalar multiplication, we have that for all it holds that . Hence, .

**Warning**

Although is a subfield of , the set is not a subspace of -vector space . For instance, for we have that the scalar multiple .

#### Line through the origin[Bearbeiten]

In this example, we consider a straight line in that passes through the origin. Let the equation of the line be given by . So we can write down the straight line as a set of points:

**Exercise**

Show that is a subspace of the -vector space .

**Proof**

We want to use the subspace criterion above. Because of we have that . Let , i.e. and . From this we obtain and hence . Let and . Since we have that also and hence . So we have shown that all the conditions of the subspace criterion are satisfied. Thus, is a subspace of .

**Alternative proof**

We can also see in another way that is a subspace. To do this, we consider the characterization of :

Now we recall the section "How to prove that a set is a subspace", where we saw that such subsets form subspaces. These subsets were of the form with some . In this example .

#### A subspace of [Bearbeiten]

In the following exercise we consider a plane in which passes through . We show that this plane always forms a subspace of .

**Exercise** (plane within )

Let . Show that is a subspace of .

**Proof** (plane within )

For the proof we have to show that the subspace criterion is satisfied. We divide this proof into three steps:

**Proof step:**

We have that , since .

**Proof step:** is complete with respect to addition

Consider two vectors and from . By definition of , we have that and . Now, under addition,

So

Hence , which establishes completeness of addition.

**Proof step:** is complete with respect to scalar multiplication

Let and let . So we have that . Consequently,

So we have:

Thus and hence, is also complete under scalar multiplication.

We have proved the conditions of the subspace criterion and thus shown that is a subspace.

#### A subspace of the polynomials[Bearbeiten]

Let us now turn to a slightly more abstract example, namely the polynomial vector space. We show that the subset of polynomials of degree less or equal is a subspace:

**Theorem** (polynomials of degree )

Let . Then is a subspace of the vector space of polynomials .

**Proof** (polynomials of degree )

We must show that the three conditions of the subspace criterion hold:

**Proof step:**

We have , so

**Proof step:** is complete with respect to addition

Let . Then . Thus, we can find with , such that and holds. So we arrive at , which means that the degree . So indeed,

**Proof step:** is complete with respect to scalar multiplication

Let and . Then . Thus, we can find with , such that holds. So we arrive at , which implies for our degree that . So indeed, .

Now all three subspace criteria are fulfilled, and hence, is indeed a subspace.

### Counterexamples[Bearbeiten]

We have already seen above three examples for subsets of which do not form a subspace. For a better understanding we now also consider counterexamples for other vector spaces.

#### Line that does NOT pass through the origin[Bearbeiten]

**Example**

which describes a subspace of . Now we displace it up by distance one and get the following set:

#### Bounded subset of [Bearbeiten]

**Example**

We consider the -vector space . Let

is not a subspace of . This is because is not closed under scalar multiplication. We know . But , because .

Alternatively, we can show that is not complete with respect to addition. For example, and , but because .

#### Graph of a non-linear function[Bearbeiten]

**Example**

Now, consider the -vector space . Let

The set is not a subspace of , because it is not complete with respect to addition. To see this, we consider the two elements . Then, we have , since . Intuitively, the set fails to be a subspace, because it is a "curved surface" and not a plain one through the origin.

#### polynomials with degree is not a subspace[Bearbeiten]

**Example**

As a more abstract example, consider now the polynomial vector space for any field . Let

We show that does not form a subspace of since it is not complete with respect to addition. To see this, consider the two elements . Then, we have , since .

## Other criteria for subspaces[Bearbeiten]

We will now learn about three criteria that make proofs easier in many cases. For this we will anticipate and use the notion of a linear map.

### Kernel of a linear map[Bearbeiten]

In the examples for subspaces, we considered the following sets:

We proved above that and are subspaces of and , respectively. The two sets are defined according to the same principle. The subspaces contain all vectors that satisfy certain conditions. The conditions are

These look very similar. Both conditions tell us that some expression in and or in and should be zero. This expression is linear in and , respectively. That is, both formulas can also be written down as linear maps:

With these, we can rewrite our subspaces as

Thus , as well as , is the kernel of a linear map. One can show, in general, that the kernel of a linear map is always a subspace.

### Image of a linear map[Bearbeiten]

Just as with the kernel, we can show in general that the image of a linear map is always a subspace. This sometimes allows us to find simpler proofs that a given set is a subspace.

**Example** (Linear map on the polynomials of degree smaller or equal to 1)

We consider an example of a linear map of , defined on the vector space of real polynomials of degree smaller or equal to 1 and mapping into .

We define a map that assigns to a polynomial in vector space the vector of its function values at positions , and . So we define the linear map by

We recall how the addition and scalar multiplication of polynomials is defined: For two polynomials the sum is defined via for all . For a scalar and a polynomial the scalar multiplication is given by .

First, we prove that is indeed a linear map. For this we have to prove the additivity and the homogeneity. So we choose and .

**additivity:**

**homogeneity:**

Thus we know that is a subspace of .

If we look at the calculation again, we realize that the -values and did not matter at all. We could have chosen other ones, as well. The proof goes the same way for the statement:

Let (and these numbers may or may not be different). The map , is linear and the image of is a subspace of .

We know that is a subspace. We also find an explicit representation for : a polynomial has the form for and . Moreover, and and .

The subspace thus has the form

So it is a plane in .

### Span of vectors[Bearbeiten]

We will later prove a general theorem that every product of a subset is a subspace of .

This allows us to shorten one of the proofs above:

We proved above that for and the set is a subspace of the -vector space .

The set is exactly the span of the set in the vector space . The span of is exactly all linear combinations of elements from . In our case, these are even the multiples of . Therefore is a subspace of .

## Exercises[Bearbeiten]

**Exercise** (A subspace of ?)

Is a subspace of ?

**How to get to the proof?** (A subspace of ?)

The equation is equivalent to . These equations are the diagonals in . This set is just a "45-degree-rotated" version of the axis cross in this example and hence NOT a subspace. Again, the set is complete with respect to scalar multiplication, but not complete with respect to addition. We may take a vector from each of the first and second bisectors to find a counterexample.

**Solution** (A subspace of ?)

The set is not a subspace, because and are in . But is not in .

**Exercise** (A subspace of? (Teil II))

Is a subspace of ?

**How to get to the proof?** (A subspace of? (Teil II))

After the last exercise we are somewhat warned that powers can be critical in the conditions, as they "curve surfaces". But is equivalent for real numbers to (our surface is flat), and all elements of are thus the multiples of the vector .

**Solution** (A subspace of? (Teil II))

is a subspace of . Since we have

According to a Example from above, is a subspace of .

**Exercise** (A subspace of ?)

Is a subspace of ?

**How to get to the proof?** (A subspace of ?)

We check the subspace criterion, as described above in this example.

**Solution** (A subspace of ?)

Since we consider a subset of , we have .

The zero vector satisfies both conditions: For we have , and the second component of the zero vector is zero. So is not empty.

Now we have to show that for also holds. To do this, we again check both conditions separately.

If and , then . So again has the required form.

If the second components of and are zero, then this is also true for the second component of .

Now the last thing we have to prove is that with and we also have .

We see that , so it has again the required form. With the second component of is also zero.

Thus is indeed a subspace of .

**Solution** (A subspace of ? (Alternative solution))

Later we have more methods available and can perform the subspace proof with more abstract methods.

We see that , where is the span of and . The span is always a subspace.

is the kernel of the linear map , . The kernel of a linear map is also always a subspace.

In the next section we prove that the intersection of two subspaces is again a subspace, so we can identify that is a subspace of .

**Exercise** (A subspace of ?)

Let be any field and . Let . We define . What conditions must satisfy for to be a subspace of ?

**Solution** (A subspace of ?)

We first assume that is a subspace of . Then must hold. But according to the definition of we have that . So we know that can be a vector space only if .

We now want to check whether is indeed a subspace in the case . For this, we use the subspace criterion and assume :

For we have that . Hence, .

If and in , then we have that and . Thus,, so .

If , then we have that . For every we have that also , so .

Thus the conditions of the subspace criterion hold and is a subspace of for .