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.
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]
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]
The span of
is a subspace of
[Bearbeiten]
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: 
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]
The span of
is the smallest subspace of
, containing
[Bearbeiten]
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]
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.
Adding elements of the span doesn't change the span[Bearbeiten]
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.
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 (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
.