Proofs for vector spaces – Serlo

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In this chapter, we will demonstrate how to prove that a set with suitable operations forms a vector space.

General proof structure[Bearbeiten]

By definition, a vector space over a field is a set with two operations of addition and , of scalar multiplication, satisfying a list of axioms. These are listed in the article vector space. They are four axioms for addition and four axioms for scalar multiplication, so 8 in total.

So if we want to show that a set forms a vector space, we first have to define some operations and and then prove that the axioms are satisfied. In defining the operations, note that the sum of two vectors and the product of a scalar with a vector again give vectors from the set , i.e., for all and we have that . This is called completeness and is an important part of the well-definedness of the operations! Then we work off the axioms in the order given within the definition.

Now we want to demonstrate the whole thing with an example. As an example we choose the polynomial space of polynomials of degree less than or equal to (for a fixed ). We prove that those polynomials form a vector space.

Definition: polynomial space[Bearbeiten]

First we need to precisely define the polynomial space, i.e., the set of our vectors.

Definition (Polynomial space)

Let be a field and let be the set of all polynomials of degree , which read

On this set we introduce two operations, an addition and a multiplication by scalars from :

Definition (Addition and scalar multiplication on the polynomial space)

We define the addition as follows:

The scalar multiplication works very similar:

We want to point out that the sums on the right-hand side of the map run again only from to . So we get again polynomials which have at most degree and so we actually end up in . This is important to obtain well-defined operations and acting on .

The polynomial space is a vector space[Bearbeiten]

We now want to show that polynomials indeed form a vector space:

Theorem (polynomials form a vector space)

Let be a field and . Then, is a -vector space.

So we need to establish the 8 vector space axioms:

  • together with the operation forms an abelian group (missing). That is, the following axioms are satisfied:
    1. associative law: For all we have that:
    2. commutative law: For all we have that:
    3. Existence of a neutral element: There exists an element such that for all we have that .
    4. Existence of an inverse element: For every there exists an element such that we have .
  • In addition, the following axioms of scalar multiplication must be satisfied:
    1. scalar distributive law: For all and all we have that:
    2. vectorial distributive law: For all and all we have that:
    3. associative law for scalars: For all and all we have that:
    4. neutral element of scalar multiplication: For all and for (the neutral element of multiplication in ) we have that: .

We will now prove each of these steps individually.

Associativity of addition[Bearbeiten]

We start with the associativity of addition. This follows from the associativity of addition in .

Proof (Associativity of addition)

Let .

Then, we have:

This shows the associativity of addition.

Commutativity of addition[Bearbeiten]

Now follows the commutativity of addition. As above, this follows from the commutativity of addition in :

Proof (Commutativity of addition)

Let .

Then, we have:

This shows the commutativity of addition.

Neutral element of addition[Bearbeiten]

Now we have to prove that a zero exists, i.e. a neutral element with respect to addition. To do this, we must first find a candidate. There is an "obvious" one here: the zero polynomial . This is indeed the neutral element of the polynomial addition:

Proof (0 is the neutral element of addition)

Let .

Then, we have:

Since was arbitrarily chosen, is the neutral element with respect to addition.

Inverse with respect to addition[Bearbeiten]

The next step is the existence of an additive inverse. Here again there is an obvious choice: For a , the additive inverse is given by :

Proof ( is an inverses of )

We use the above notation for and . Then,

Thus is the inverse of .

Distributive law[Bearbeiten]

The proofs of the two distributive laws follow from the distributive law in and go similarly, so we show only the second one here:

Proof (Distributive law)

Let .

Then, we have:

This shows the distributive law for in .

Associative law with respect to multiplication[Bearbeiten]

Next we have to establish the associative law with respect to scalar multiplication. This follows (similarly to the first two axioms) from the associativity of multiplication in :

Proof (Associative law with respect to multiplication)

Let .

Then, we have:

This shows the associative law for scalar multiplication.

Unit property[Bearbeiten]

And finally, we have to establish the unit property below:

Proof (Unit property)

Let .

Then, we have:

So satisfies the unit property.

We have established all 8 vector space axioms, and hence the polynomial space is a vector space.