Serlo: EN: Subspace

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

• $0_V\in U$.
• For all $v,u\in U$ we have that $v+u\in U$.
• For all $u\in U$ and for all $\lambda \in K$ we have that $\lambda \cdot u\in U$.

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

Motivation

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 $(\mathbb{Z},+)$ can be seen as a subgroup of the set of rational numbers $(\mathbb{Q},+)$, which is again a subgroup of the real numbers $(ℝ,+)$. 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.

Subspace criterion Vorlage:Anker

Derivation of the criterion

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

Checking the vector space axioms for an example

We consider the subset $U=\{(x,2x{)}^T\mid x\in ℝ\}$ of the $ℝ$-vector space $V={ℝ}^2$. Visually this subset is a line. We want to find out, whether $U$ is a subspace of $V={ℝ}^2$. So by definition, we need to show that the set $U$ together with the operations ${+}_U$ and ${\cdot }_U$ 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 $V$ is a map $V\times V\to V$ with $(u,v)\mapsto u+v$. Our new addition arises by first restricting the domain of definition $V\times V$ to the subset $U\times U$. We get a map ${+}^{\prime }:U\times U\to V$. So the range of values remains the same for now. However, to make the set $U$ a vector space, we need a map ${+}^{\prime }:U\times U\to U$, so after addition, we "land again in the subspace". So, is the image of ${+}^{\prime }$ contained in $U$? What we would like to show is: Vorlage:Important This property is also called completeness under addition.

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

This property is called completeness under scalar multiplication.

We now check both properties in our concrete example:

First, the addition. Let $u,u^{\prime }be\in U$. That is, $x,y\in ℝ$ exist such that $u=(x,2x{)}^T$ and $u^{\prime }=(y,2y{)}^T$. Then $u+u^{\prime }=(x+y,2(x+y){)}^T$. If we set $z:=x+y$, we have that $u+u^{\prime }=(z,2z{)}^T$. So $u+u^{\prime }\in U$.

Now the scalar multiplication. Let $u=(x,2x{)}^T\in U$ as we just did, and let $\lambda \in K$. Then we have $\lambda \cdot u=(\lambda x,2\lambda x{)}^T$. If we set $z:=\lambda x$, we have that $\lambda \cdot u=(z,2z{)}^T$. So $\lambda \cdot u^{\prime }\in U$.

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 $U$. More specifically, we have used that every element of $U$ is of the form $(x,2x{)}^T$.

Now we check the eight vector space axioms.

First, the four axioms for addition:

Associative law of addition: Let $u,v,w\in U$. We must show that $u{+}_U(v{+}_{U}w)=(u{+}_{U}v){+}_{U}w$. Since ${+}_U$ is the constrained version of $+$, we must show $u+(v+w)=(u+v)+w$. This follows from the associative law for the vector space $V$. Note that since $U\subseteq V$, we also have $u,v,w\in V$, so the relations in $V$ are indeed valid.

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

Existence of a neutral element: We must show that an element $0_{{}_U}\in U$ exists such that $0_{{}_U}{+}_{U}u=u{+}_U{0}_{{}_U}=u$ for all $u\in U$. Since $V$ is a vector space, it contains a zero vector with $0+v=v+0=v$ for all $v\in V$. In particular, this holds for all $u\in U$. Since addition in $U$ is only the restriction of addition in $V$, it suffices to show, That $0\in U$. For then we can set $0_{{}_U}=0$. The element $0\in V={ℝ}^2$ is more precisely the vector $(0,0{)}^T$. This can be written as $(0,2\cdot 0{)}^T$ and thus lies in $U$. So there is indeed a neutral element of addition within $U$.

Existence of additive inverse in $U$: Let $u\in U$. We must show that there exists a $u^{\prime }\in U$, such that $u{+}_U{u}^{\prime }=0$. We know that $u+(-u)=0$ holds in $V$. So if we can show $-u\in U$, we are done: then we can choose $u^{\prime }=-u$. We know that $-u=(-1)\cdot u\in U$ holds. Furthermore, we have already shown that $U$ is complete under scalar multiplication, so indeed $-u\in U$ follows.

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

In order to show that the operations ${+}_U$ and ${\cdot }_U$ are well-defined, we needed to establish the properties of completeness formulated above. For this we have worked closely with the definition of $U$. Furthermore, for the third axiom of addition, we had to show that the neutral element of addition in $V$ is also an element of $U$. Again, we worked concretely with the definition of $U$. 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 $V$ hold.

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

• The completeness of $U$ with respect to addition.
• The completeness of $U$ with respect to scalar multiplication
• $0\in U$

For these we had to work with the definition of $U$ and $V$. The above arguments that these three properties suffice should be true for every vector space $V$ and all subsets $U$ of $V$. 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 $V={ℝ}^2$, each of which violates exactly one of the three rules and is indeed not a subspace.

Counterexample: The empty set

We first consider the empty set $U=\mathrm{\varnothing }$. This is of course a subset of ${ℝ}^2$.

Let us check completeness with respect to of addition, $\mathrm{\forall }u,u^{\prime }\in U:u+u^{\prime }\in U$, 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: $0\notin U$. That is simply because the empty set contains by definition no elements. The property $0\notin U$ cannot be derived in general from the completeness of addition and of scalar multiplication.

Now, $U$ is not a vector space, because $U$ contains no element, in particular no neutral element of addition. Accordingly, $U$ cannot be a subspace. Be aware that the property $0\in U$ 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 entriesVorlage:Anker

Our second example shows that scalar multiplication is actually needed: take the set of integer vectors $U:={\mathbb{Z}}^2$. If we identify the vectors with points in ${ℝ}^2$, we get some kind of "lattice":

${\mathbb{Z}}^2$ as a subset ${ℝ}^2$

This set is obviously a subset of ${ℝ}^2$ and again the question arises whether it is a subspace. In contrast to the first example now the zero vector $0=(0,0{)}^T$ is contained in $U={\mathbb{Z}}^2$. All other axioms of vector addition are also valid. The sum of two vectors from ${\mathbb{Z}}^2$ is again in ${\mathbb{Z}}^2$.

Nevertheless, the ${\mathbb{Z}}^2$ is not a subspace of ${ℝ}^2$, because ${\mathbb{Z}}^2$ is not complete with respect to scalar multiplication. For instance, $v=(1,0{)}^T\in {\mathbb{Z}}^2$ and $\lambda =\frac{1}{2}\in ℝ$, but $\lambda \cdot v={(\frac{1}{2},0)}^T$ is not contained in ${\mathbb{Z}}^2$. Thus ${\mathbb{Z}}^2$ 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 $U$ is a subspace, we must always show that for every $u\in U$ and for every scalar $\lambda \in K$, we also have $\lambda \cdot u\in U$.

Counterexample: A cross of coordinate axes Vorlage:Anker

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 $G:=\{(0,t{)}^T:t\in ℝ\}$ and $H:=\{(t,0{)}^T:t\in ℝ\}$. So we consider the subset $U:=G\cup H\subseteq {ℝ}^2$. Illustrated in the plane as points, the set looks like an infinite "cross":

The axis cross

Is $U$ a subspace? Obviously, the zero vector $0$ is contained in $U$. Moreover, we have that for any $v\in U$ and $\lambda \in ℝ$ that also $\lambda v$ is an element of $U$. Thus $U$ is complete under scalar multiplication. Nevertheless, $U$ is not a subspace. To see this, we choose the vectors $v_1=(1,0{)}^T$ and $v_2=(0,1{)}^T$. Then, we have $v_1,v_2\in U$, but for the sum we have that $v_1+v_2=(1,1{)}^T\notin U$.

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 $U$ is a subspace.

Statement and proof of the criterion Vorlage:Anker

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

• completeness with respect to addition,
• completeness with respect to scalar multiplication, and.
• $0\in U$.

We have seen examples of subsets $U$ of $V$ where one of these properties was not satisfied in each case and which also do not form a subspace of $V$. 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.

How to prove that a set is a subspace Vorlage:Anker

General proof structure

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 $U$ is a subspace of a $K$-vector space $V$? 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 $U\subseteq V$. Then, to show that $U$ is a subspace, we need to check the three properties from the criterion. So, in total, we need to show the following four statements:

1. $U\subseteq V$
2. $0\in U$.
3. For all $v,u\in U$ we have that $v+u\in U$.
4. For all $u\in U$ and for all $\lambda \in K$ we have that $\lambda \cdot u\in U$.

Mathe für Nicht-Freaks: Vorlage:Hinweis What do proofs of these statements look like? The proof structure of these statements looks like this: Vorlage:Liste

Finding a proof idea Vorlage:Anker

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

Mathe für Nicht-Freaks: Vorlage:Aufgabe We want to apply the subspace criterion to $U$. To do this, we check the premises of the theorem according to the above scheme.

• $U\subseteq {ℝ}^2$: let $v\in U$. By definition of $U$, there is some $\lambda \in ℝ$ with $v=\lambda \cdot u$. Since ${ℝ}^2$ is a vector space, it follows that $v=\lambda \cdot u\in {ℝ}^2$.
• $0\in U$: We have seen in "Vector space: properties" that for every vector $v\in {ℝ}^2$ we have $0\cdot v=0$. So we have that also $0\cdot u=0$. Thus we get $0\cdot U$.
• completeness of addition: Let $v,w\in U$. By definition of $U$, there exist $\lambda ,\mu \in ℝ$ with $v=\lambda \cdot u$ and $w=\mu \cdot u$. Since $v,w\in {ℝ}^2$, we can add them: $v+w=\lambda \cdot u+\mu \cdot u=(\lambda +\mu )\cdot u$. Because of $\lambda +\mu \in ℝ$ we finally obtain $v+w\in U$.
• completeness of scalar multiplication: Let $v\in U$ and let $\mu \in ℝ$. By definition of $U$ there is a $\lambda \in ℝ$ with $v=\lambda \cdot u$. Since $v\in {ℝ}^2$ we can multiply it with $\mu$: $\mu \cdot v=\mu \cdot (\lambda \cdot u)=(\mu \cdot \lambda )\cdot u$. Because of $\mu \cdot \lambda \in ℝ$ it follows that $\mu \cdot v\in U$.

This shows that all conditions hold, so by the subspace criterion, $U$ is indeed a subspace of ${ℝ}^2$.

Writing down the proof

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

Examples and counterexamples for subspaces

Examples Vorlage:Anker

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

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

the following example with $V=ℝ$ shows that sometimes, only the trivial subspaces are subspaces:

Line through the origin

In this example, we consider a straight line $U$ in ${ℝ}^2$ that passes through the origin. Let the equation of the line be given by $y=2x$. So we can write down the straight line as a set of points: Vorlage:Einrücken Mathe für Nicht-Freaks: Vorlage:Aufgabe Mathe für Nicht-Freaks: Vorlage:Beweis

A subspace of ${ℝ}^3$

In the following exercise we consider a plane in ${ℝ}^3$ which passes through $0$. We show that this plane always forms a subspace of ${ℝ}^3$.

A subspace of the polynomials

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 $n$ is a subspace:

Counterexamples

We have already seen above three examples for subsets of ${ℝ}^2$ which do not form a subspace. For a better understanding we now also consider counterexamples for other vector spaces.

Other criteria for subspaces

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

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

Vorlage:Einrücken We proved above that $U_1$ and $U_2$ are subspaces of ${ℝ}^2$ and ${ℝ}^3$, respectively. The two sets are defined according to the same principle. The subspaces contain all vectors that satisfy certain conditions. The conditions are Vorlage:Einrücken These look very similar. Both conditions tell us that some expression in $x$ and $y$ or in $a,b$ and $c$ should be zero. This expression is linear in $x,y$ and $a,b,c$, respectively. That is, both formulas can also be written down as linear maps: Vorlage:Einrücken With these, we can rewrite our subspaces as Vorlage:Einrücken

Thus $U_1$, as well as $U_2$, 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

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.

If we look at the calculation again, we realize that the $x$-values $0,2$ and $4$ did not matter at all. We could have chosen other ones, as well. The proof goes the same way for the statement:

Let $r,s,t\in ℝ$ (and these numbers may or may not be different). The map $f:V\to {ℝ}^3$, $f(p)=(p(r),p(s),p(t){)}^T$ is linear and the image of $f$ is a subspace of ${ℝ}^3$.

We know that $U=f(V)\subseteq {ℝ}^3$ is a subspace. We also find an explicit representation for $U$: a polynomial $p\in V$ has the form $p(x)=ax+b$ for $a$ and $b\in ℝ$. Moreover, $p(0)=b$ and $p(2)=2a+b$ and $p(4)=4a+b$.

The subspace $U$ thus has the form Vorlage:Einrücken So it is a plane in ${ℝ}^3$.

Span of vectors

We will later prove a general theorem that every product of a subset $M\subseteq V$ is a subspace of $V$.

This allows us to shorten one of the proofs above:

We proved above that for $u\in {ℝ}^2$ and $U:=\{\lambda \cdot u|\lambda \in ℝ\}$ the set $U$ is a subspace of the $ℝ$-vector space ${ℝ}^2$.

The set $U$ is exactly the span of the set $M=\{u\}$ in the vector space $V={ℝ}^2$. The span of $M$ is exactly all linear combinations of elements from $M$. In our case, these are even the multiples of $u$. Therefore $U$ is a subspace of ${ℝ}^2$.

Exercises

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