Vector spaces abstract the arithmetic properties of addition and scalar multiplication of vectors. In we know how to add vectors and to multiply vectors by scalars. Indeed, it is straightforward to verify that each of the eight properties listed in Table ?? is valid for vectors in . Remarkably, sets that satisfy these eight properties have much in common with . So we define:

The vector mentioned in (A3) in Table ?? is called the zero vector.


Table 1: Properties of Vector Spaces: suppose and .




(A1) Addition is commutative



(A2) Addition is associative



(A3) Additive identity exists



(A4) Additive inverse exists



(M1) Multiplication is associative



(M2) Multiplicative identity exists



(D1) Distributive law for scalars



(D2) Distributive law for vectors




When we say that a vector space has the two operations of addition and scalar multiplication we mean that the sum of two vectors in is again a vector in and the scalar product of a vector with a number is again a vector in . These two properties are called closure under addition and closure under scalar multiplication.

In this discussion we focus on just two types of vector spaces: and function spaces. The reason that we make this choice is that solutions to linear equations are vectors in while solutions to linear systems of differential equations are vectors of functions.

An Example of a Function Space

For example, let denote the set of all functions . Note that functions like and are in since they are defined for all real numbers , but that functions like and are not in since they are not defined for all .

We can add two functions and by defining the function to be: We can also multiply a function by a scalar by defining the function to be: With these operations of addition and scalar multiplication, is a vector space; that is, satisfies the eight vector space properties in Table ??. More precisely:

  • Define the zero function by For every in the function satisfies: Therefore, and is the additive identity in .
  • Let be a function in and define . Then is also a function in , and Thus, has the additive inverse .

After these comments it is straightforward to verify that the remaining six properties in Table ?? are satisfied by functions in .

Sets that are not Vector Spaces

It is worth considering how closure under vector addition and scalar multiplication can fail. Consider the following three examples.

(i)
Let be the set that consists of just the and axes in the plane. Since and are in but is not in , we see that is not closed under vector addition. On the other hand, is closed under scalar multiplication.
(ii)
Let be the set of all vectors where and are integers. The set is closed under addition but not under scalar multiplication since is not in .
(iii)
Let be the closed interval in . The set is neither closed under addition () nor under scalar multiplication (). Hence the set is not closed under vector addition and not closed under scalar multiplication.

Subspaces

Note that in order for a subset of a vector space to be a subspace it must be closed under addition and closed under scalar multiplication. That is, suppose and . Then

  • , and
  • .

The -axis and the -plane are examples of subsets of that are closed under addition and closed under scalar multiplication. Every vector on the -axis has the form . The sum of two vectors and on the -axis is which is also on the -axis. The -axis is also closed under scalar multiplication as , and the -axis is a subspace of . Similarly, every vector in the -plane in has the form . As in the case of the -axis, it is easy to verify that this set of vectors is closed under addition and scalar multiplication. Thus, the -plane is also a subspace of .

In Theorem ?? we show that every subset of a vector space that is closed under addition and scalar multiplication is a subspace. To verify this statement, we need the following lemma in which some special notation is used. Typically, we use the same notation to denote the real number zero and the zero vector. In the following lemma it is convenient to distinguish the two different uses of , and we write the zero vector in boldface.

Proof
Let be a vector in and use (D1) to compute By (A4) the vector has an additive inverse . Adding to both sides yields Associativity of addition (A2) now implies A second application of (A4) implies that and (A3) implies that .

Next, we show that the additive inverse of a vector is unique. That is, if , then .

Before beginning the proof, note that commutativity of addition (A1) together with (A3) implies that . Similarly, (A1) and (A4) imply that .

To prove uniqueness of additive inverses, add to both sides of the equation yielding Properties (A2) and (A3) imply But Therefore , as claimed.

To verify that , we show that is the additive inverse of . Using (M1), (D1), and the fact that , calculate Thus, is the additive inverse of and must equal , as claimed.

Proof
We have to show that is a vector space using the operations of addition and scalar multiplication defined on . That is, we need to verify that the eight properties listed in Table ?? are satisfied. Note that properties (A1), (A2), (M1), (M2), (D1), and (D2) are valid for vectors in since they are valid for vectors in .

It remains to verify (A3) and (A4). Let be any vector. Since is closed under scalar multiplication, it follows that and are in . Lemma ?? states that and ; it follows that and are in . Hence, properties (A3) and (A4) are valid for vectors in , since they are valid for vectors in .

Examples of Subspaces of
In a sense that will be made precise all subspaces of can be written as the span of a finite number of vectors generalizing Example ??(b) or as solutions to a system of linear equations generalizing Example ??(c).

Examples of Subspaces of the Function Space

Let be the set of all polynomials in . The sum of two polynomials is a polynomial and the scalar multiple of a polynomial is a polynomial. Thus, is closed under addition and scalar multiplication, and is a subspace of .

As a second example of a subspace of , let be the set of all continuously differentiable functions . A function is in if and exist and are continuous for all . Examples of functions in are:

(i)
Every polynomial is in .
(ii)
The function is in for each constant .
(iii)
The trigonometric functions and are in for each constant .
(iv)
is twice differentiable everywhere and is in .

Equally there are many commonly used functions that are not in . Examples include:

(i)
is neither defined nor continuous at .
(ii)
is not differentiable (at ).
(iii)
is neither defined nor continuous at for any integer .

The subset is a subspace and hence a vector space. The reason is simple. If and are continuously differentiable, then Hence is differentiable and is in and is closed under addition. Similarly, is closed under scalar multiplication. Let and let . Then Hence is differentiable and is in .

The Vector Space

Another example of a vector space that combines the features of both and is . Vectors have the form where each coordinate function . Addition and scalar multiplication in are defined coordinatewise — just like addition and scalar multiplication in . That is, let be in and let be in , then

The set satisfies the eight properties of vector spaces and is a vector space. Solutions to systems of linear ordinary differential equations are vectors in .

Exercises

Verify that the set consisting of all scalar multiples of is a subspace of .
Let be the set of all matrices. Verify that is a vector space.
Let Let be the set of vectors such that . Verify that is a subspace of . Compare with .

In Exercises ?? – ?? you are given a vector space and a subset . For each pair, decide whether or not is a subspace of .

and consists of vectors in that have a in their first component.
and consists of vectors in that have a in their first component.
and consists of vectors in for which the sum of the components is .
and consists of vectors in for which the sum of the components is .
and consists of functions satisfying .
and consists of functions satisfying .
and consists of functions satisfying .

In Exercises ?? – ?? which of the sets are subspaces?

.
.
.
where is a matrix.
where is a matrix and is a fixed nonzero vector.
Let be a vector space and let and be subspaces. Show that the intersection is also a subspace of .
For which scalars do the solutions to the equation form a subspace of ?
For which scalars do the solutions to the equation form a subspace of ?
Show that the set of all solutions to the differential equation is a subspace of .
Recall from equation (??) of Section ?? that solutions to the system of differential equations are Use this formula for solutions to show that the set of solutions to this system of differential equations is a vector subspace of .