# 30 Vector Spaces

Vector spaces are sets of vectors that must satisfy ten axioms. Consider the following definition from Larson and Valvo [16].

**Definition VI.4**

Let be a set on which two operations (vector addition and scalar multiplication) are defined. If the listed axioms are satisfied for every

**u**,

**v**, and

**w**in and every scalar (real number) and , then is called a

**vector space**.

- is in .
- has a zero vector
**0**such that for every**u**in , . - For every
**u**in , there is a vector in denoted by**-u**such that . - is in .
- .
- .
- .
- 1(
**u**)=**u**.

The first column gives the axioms for addition. The vector space must be closed under addition, contain the additive identity (denoted by **0**), and additive inverses. Furthermore, addition must be commutative and associative. The second column gives the rules for scalar multiplication. The vector space must be closed under scalar multiplication, and any vector multiplied by one must equal the vector itself. Furthermore, the associative property and distributive properties (distributing a scalar and distributing a vector) should hold.

Vectors come in a variety of forms. We already discussed vectors that can occur in -dimensional space. But vectors can also be functions, polynomials, and matrices [16]. Thus, certain sets of functions, polynomials, or matrices can be vector spaces.

Let’s work through the process of showing that a set of vectors is a vector space. This problem is one I completed for Elementary Linear Algebra and comes from Larson and Falvo [16].

**Example 62**

Show that the set with the standard operations (vector addition and scalar multiplication) is a vector space.

*Proof.*

Consider the vectors , , and in .

Vector addition is closed because

which is an element of since is a real number. Vector addition is also commutative because

and associative because

The zero vector is in because 0 is a real number, and for every vector in ,

For every vector in , there exists , which is in because is a real number, such that

Scalar multiplication is closed because for every scalar ,

which is in because is a real number. Both distributive properties hold because

and

where and are scalars and **u** and **v** are vectors in .

Scalar multiplication is also associative because

Finally,

Because all ten axioms are satisfied, is a vector space.

Vector spaces also have subspaces, which are defined by Larson and Falvo as follows [16].

**Definition VI.5**

A nonempty subset of a vector space is called a

**subspace**of if is a vector space under the operations of addition and scalar multiplication defined in .

In a vector space , there are always the two operations of vector addition and scalar multiplication. However, how these operations are defined may differ from one vector space to another [16].

If we have a vector space and take a subset of those vectors, that subset is a subspace if it satisfies all of the axioms of a vector space for vector addition and scalar multiplication, which are defined the same way as they are in .

A special type of subspace of a vector space is known as the span of a subset of .

#### Spanning Sets

Larson and Falvo provide us with the following definitions [16].

**Definition VI.6**

A vector

**v**in a vector space is called a

**linear combination**of the vectors in if

**v**can be written in the form

where are scalars.

**Definition VI.7**

If is a set of vectors in a vector space , then the

**span**of is the set of all linear combinations of the vectors in ,

Let be a subset of a vector space . That is, every vector in is also in . The span of , denoted span(), is the set of all vectors that can be “built” by adding together scalar multiples of the vectors in . Furthermore, span() is a subspace of [16].

Sometimes, span() is equal to . When this occurs, we say that is a spanning set of and that spans [16]. Let’s look at an example of how to show that a subset of a vector space spans . This is a problem I completed for Elementary Linear Algebra and comes from Larson and Falvo [16].

**Example 63**

Show that the set spans .

*Proof.*

To prove this, we need to show that there exist scalars , , and such that for every vector in ,

By scalar multiplication and vector addition, we have

This is a system of equations which has a unique solution if and only if the determinant of the coefficient matrix (matrix consisting of only the coefficients) is not equal to zero [16]. So, we need to find the determinant of

By Theorem VI.3 in chapter 28, the determinant is equal to the sum of the products of each entry with its cofactor (see definition VI.2 in chapter 28) for every entry in whichever row or column we choose. Let’s go with the first column.

We need to find cofactors , , and by first finding the minors , and . Deleting the first row and first column of the coefficient matrix gives us

with determinant

which means

Deleting the second row and first column of the coefficient matrix gives us

with determinant

which means

Deleting the third row and first column of the coefficient matrix gives us

with determinant

which means

So, the determinant of is

This means that our system of equations has a unique solution for all real numbers , , and . This is because the determinant of the coefficient matrix is not equal to zero if and only if the system of equations has a unique solution [16]. Thus, for every vector in , we can find , , and such that

Therefore, spans .

We will now discuss a specific type of subset that spans a vector space, which is referred to as a basis for that vector space.

#### Basis

In essence, a basis of a vector space is the largest set of independent vectors such that every vector in can be written as a linear combination of vectors in . Consider the following definition from Larson and Falvo [16].

**Definition V1.8**

A set of vectors is a vector space is called a

**basis**for if the following conditions are true.

- spans .
- is linearly independent.

The condition that spans ensures that every vector in can be formed by some linear combination of vectors in . However, it is possible to have more vectors in a spanning set than are actually needed. This can happen if one of the vectors in a spanning set can itself be written as a linear combination of the other vectors in the spanning set. Thus, the condition that is linearly independent ensures that we have the minimum number of vectors needed in order to span because a set of vectors is linearly independent if none of the vectors can be written as a linear combination of the other vectors [16]. The formal definition is given by Larson and Falvo as follows [16].

**Definition VI.9**

A set of vectors in a vector space is called

**linearly independent**if the vector equation

has only the trivial solution If there are also nontrivial solutions, then is called **linearly dependent**.

Let’s now look at a problem that I completed for Elementary Linear Algebra that comes from Larson and Falvo [16].

**Example 64**

Is the set of vectors

a basis for ?

**Solution**

By example 63, we know that spans . So, all we have to do is see whether the vectors in are linearly independent. By definition VI.9, the set of vectors in are linearly independent if and only if the equation

has only the trivial solution. From our equation, we have the following system

As we saw in Example 63, this system of equations has a unique solution because the determinant of the coefficient matrix is not zero. Because the solution is unique and , and is a solution, , and is the only solution. Thus, the set of vectors in are linearly independent by definition VI.9. Therefore, is a basis for by definition VI.8.

In our next section, we will discuss row and column spaces of matrices.