# Span – Serlo

In linear algebra, the span of a subset of a vector space over a field is the set of all linear combinations with vectors from and scalars from . The span is often called the linear hull of or the span of .

The span forms a subspace of the vector space , namely the smallest subspace that contains .

## Derivation of the span[Bearbeiten]

### Generating vectors of the -plane[Bearbeiten]

We consider the vector space and restrict to the -plane. I.e., the set of all vectors of the form with :

Each vector of this plane can be written as a linear combination of the vectors and :

With the set of these linear combinations, every point of the -plane can be reached. In particular, the two vectors and lie in the -plane. Furthermore, all linear combinations of the two vectors and lie in the -plane. This is because the component of the two vectors under consideration is and thus the third component of the linear combination of the vectors must also be .

In summary, we can state: Every vector of the -plane is a linear combination of and . Every linear combination of these two vectors is also an element of this plane. So the vectors and *generate* the -plane. Or as a mathematician would say, they **span** the -plane (like two rods spanning a side of a tent).

The -plane is a subspace of the vector space . We call this subspace . Our two vectors span the subspace (=plane) . So we write

We say that "* is substpace generated by the two vectors and *" or that "* is the linear hull of the two vectors and * or even better: ** is the span of the two vectors and **.

Are these generating vectors unique? The answer is no, because the plane can also be spanned by two vectors like and :

There is hence also

Thus, the two vectors spanning a plane are not necessarily unique.

Intuitively, we can think of the span of vectors as the set of all possible linear combinations that can be built from these vectors. In our example this means

Another intuition is the following: The span of a set describes the vector space where all combinations of directions represented by elements from are merged.

### The span of even monomials[Bearbeiten]

We now examine a slightly more complicated example: Consider the vector space of polynomials over . Let . The elements from are the monomials , , , ans so on. In other words, all monomials that have an even exponent. For odd exponents, however . We consider , the set of all linear combinations with vectors of . For example is an element in . In particualr, is a subspace of since it contains polynomials.

Further, the set is not empty, since it contains for instance .

Let us now consider two polynomials . By construction of , and consist exclusively of monomials with an even exponent. Thus, of addition of and also results in a polynomial with exclusively even exponents. The set is therefore closed with respect to addition.

The same argument gives us completeness with respect to scalar multiplication. Thus the set is a subspace of the vector space of all polynomials. As we will see later, it is even the smallest subspace that contains .

## Definition of the span[Bearbeiten]

Above, we found out that the span of a set is the set of all linear combinations with vectors from . Intuitively, the span is the subspace resulting from the union of all directions given by vectors from . Now, we make this intuition mathematically precise.

**Definition** (Span of a set)

Let be a vector space over the field . Let be a non-empty set. We define the *span* of as the set of all vectors from which can be represented as a finite linear combination of vectors from and denote it as :

For the empty set we define:

Alternatively, one can call the span of a set the *generated subspace* or *linear hull*.

**Hint**

The sum always has only finitely many summands, even if M is infinite.

**Hint**

Occasionally, the notation is also used for the span. The advantage of this notation is that it is clear which field defines the vector space. It makes a difference which field we use as a basis. For the example for we have that , but . It can be shown that and .

## Example[Bearbeiten]

**Example** (Line through the origin is a certain span)

Let . We consider the set as a subset of the vector space . The span is the straight line through the origin pointing in the direction of the vector .

**Example** (plane through the origin as a span)

Let and be two vectors from . The span of these two vectors is the -plane. The following transformation shows

## Overview: Properties of the span[Bearbeiten]

Let be a -vector space, , subsets of and a subspace of . Then, we have

- For a vector we have
- If , then
- From one can usually not conclude
- is a subspace of
- For a subspace we have
- is the smallest subspace of including

## Properties of the span[Bearbeiten]

### The span of a vector in [Bearbeiten]

For a vector we have that . For the zero vector the span again consists only of the zero vector, so . If holds, then is exactly the set of elements that lie on the line through the origin im direction of the vector .

### Span preserves subsets[Bearbeiten]

**Theorem** (Span preserves subsets)

Let be a -vector space and let . If , then also .

**Proof** (Span preserves subsets)

Since and is an element in the span of every set, we have .

Thus we can assume without loss of generality that . We consider any element . By the definition of the span, vectors and exist such that . Because of we have for all with that . Hence also . Consequently, .

**Hint**

The converse of the above theorem does not hold true in general! By this we mean: From we *cannot* conclude .

A possible counterexample is:

Here,

Thus , since in both cases we get exactly the multiples of the vector . Since the two subsets are equal, we have in particular , but . Therefore, the converse of the theorem cannot hold true in general.

### The set is contained in its span[Bearbeiten]

**Theorem** ( is contained in its span)

Let ein -vector space and . Then, there is .

**Proof** ( is contained in its span)

If , then , and the assertion is true.

Otherwise, let be arbitrary. Then can be represented by . In particular, is a linear combination with a summand of . Thus there is , since contains all linear combinations of elements from .

This establishes the assertion .

### The span of is a subspace of [Bearbeiten]

**Theorem** (The span of is a subspace of )

is a subspace of

**Proof** (The span of is a subspace of )

If is the empty set, then by definition , and that is a subspace of . From now on we may therefore assume that is not empty.

First, it is clear that . But this is obvious according to the definition of vector space and span.

We still have to show is subspace of . In other words,we have to show that

- for two elements we have also (completeness of addition)
- (completeness of scalar multiplication)

**Proof step:**

Since is not empty, there exists at least one . Then can be written as , and therefore itself is in . So this condition is fulfilled.

**Proof step:** Completeness of addition

We show the completeness concerning the vector addition. Let . Then there are vectors and , so that and . So

Hence, is complete with respect to addition.

**Proof step:** Completeness of scalar multiplication

Establish completeness of scalar multiplication is done easily:

Thus we have proved that is a subspace of the vector space .

### The span of a subspace is again [Bearbeiten]

**Theorem** (The span of is the smallest subspace von )

The span of a subspace is again

**Proof** (The span of is the smallest subspace von )

Since is a subspace, for some vectors also all linear combinations of the are contained in . Therefore . Together with our assertion follows.

### The span of is the smallest subspace of , containing [Bearbeiten]

**Theorem** (The span of is the smallest subspace of )

Let be a -vector space and let .

Then, is the smallest subspace of including .

**Proof** (The span of is the smallest subspace of )

We already know that is a subspace. Now we show that is the smallest subspace containing .

If , the assertion is obviously true, since then .

Let be a subspace of containing . Our goal is to show that . Since this would imply that the subspace is smaller or equal to every other subspace containing .

Now if , then there are some and such that (by definition of the span).

Since is a subspace and , all linear combinations of are also contained in . This implies our assertion .

### Idempotency of the span[Bearbeiten]

**Theorem** (Idempotency of the span)

Let be a -vector space and . Then we have . This property of the span is called *idempotency*.

**Proof** (Idempotency of the span)

For we have and .

Therefore, we can now assume that is not empty.

We already know that . So it only remains to show that .

Let . Then can be written as

with and . Since for all , every can be written as a linear combination of elements in :

where and . We now write the within as a linear combination of the :

For all , the sum is an element of the field . So we obtain , which was to be shown.

**Alternative proof** (Idempotency of the span)

We know that is a subspace of , and that the span of a subspace is again.

Therefore, is again .

### Adding elements of the span doesn't change the span[Bearbeiten]

**Theorem** (Adding elements of the span doesn't change the span)

Let be a -vector space and , . Then, we have

**Proof** (Adding elements of the span doesn't change the span)

We will establish the two implications and :

**Proof step:**

The statement does always hold, since . So all that remains is to show that holds. In order to do this, we consider an element . We can write it as

with , , and . Since , one can write for all as a linear combination of elements from :

where and . Now, we plug this expression for into the formula above:

We have thus represented as a linear combination of vectors from and hence .

**Proof step:**

We show this statement using a proof by contradiction. We assume that there is some but . We now define an element , with and .

Now is a linear combination of vectors from . Thus , since . However, we also have , since . But this contradicts the assumption .

Hence our assumption is false and must hold.

**Alternative proof** (First proof step)

One can also argue as follows: we have . So also .

We have already proved that then . This set is the same as because of the idempotency of the span, so .

## Check whether vectors are inside the span[Bearbeiten]

After we have learned some properties of the span, we will show in this section how we can check whether a vector of lies within the span of or not. We will see that in order to answer this question we have to solve a linear system of equations.

**Example** (Plane and line through the origin)

Let's start with a simple example from the . We consider the line through the origin with the one-element subset of the plane . The question now is whether the vector lies in the span of . One can immediately see that

holds. In other words

Mathematically, we have to solve a system of equations. In our simple example, the exercise is to find a such that

From this equation we obtain the linear system of equations

with the obvious solution .

**Example** (Polynomials)

Let us now examine an example whose solution is not obvious at first sight. For this we consider the subset of the monomials and the polynomial . We want to show that the polynomial is not in the span of . To do this, it suffices to prove that cannot be represented as a linear combination of the monomials of . We can see this by transforming the polynomial:

We see that a summand contains the monomial , but this monomial is not contained in . Thus the polynomial is not in the span of the set .

**Example** (Vectors from )

We consider the subset of and want to prove that the vector . For this we have to show that there are coefficients such that

From this representation we get the linear system of equations

with solution , , , . Hence, we have that

and therefore .