Vector space: properties – Serlo

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In this chapter we will consider some properties of vector spaces which can be derived directly from the definition of a vector space. So every vector space must satisfy these properties, no matter how abstract or high-dimensional.

Overview[Bearbeiten]

In this section we use again the operation symbols "" and "" to distinguish the vector addition and the scalar multiplication with the field addition "" and the field multiplication "".

We want to derive simple properties and rules from the eight axioms of the vector space. Since we have demanded in the axioms only the existence of the zero vector and the additive inverse, the following questions arise first: Is the zero vector unique or are there several zero vectors? Is the inverse element of addition unique or can there be more than one? The answer to both questions is:

  • In every vector space there is exactly one zero vector . So nope, there cannot be more than one zero vector in a vector space.
  • The inverse with respect to addition is unique. So for every vector there is exactly one other vector with .

Further statements that we will prove in the following are:

  • For every we have that: .
  • For every we have that: .
  • From it follows that or .
  • For all and all we have that: .

In the following, we denote by a vector space over a field .

Uniqueness of the zero vector[Bearbeiten]

Proving uniqueness of the zero vector (YouTube video (German) by the YouTube channel Maths CA).

Theorem (Uniqueness of the zero vector)

In every vector space , the zero vector is unique

Proof (Uniqueness of the zero vector)

Suppose we had two vectors and with the zero vector property, i.e. and satisfy for all vectors the equation respectively.

If we take the first equation and plug in for the concrete vector , and if we analogously, in the second equation plug in for the vector , then we get and .

Because of the commutative law of vector addition and accordingly

Thus .

We have shown in total that in a vector space two vectors with the null property are equal. Thus, the zero vector is unique.

Inverses are unique[Bearbeiten]

Proving the uniqueness of inverses (YouTube video (German) by the YouTube channel Maths CA).

Theorem (Inverses are unique)

The inverse with respect to addition is unique. That means, for every vector there is exactly one vector with .

Proof (Inverses are unique)

Let . We assume that and are two additive inverses to . Let thus and . We now show that and that there can be only one additive inverse.


Thus the two additive inverses and are identical.

Hint

In the proof above, we wrote and , respectively, for the additive inverse of . However, we have seen that there is exactly one additive inverse. This is normally denoted by and we will use this notation for the additive inverse in the following sections.

Scaling by zero results in the zero vector [Bearbeiten]

Scaling by zero results in the zero vector - proof (YouTube video (German) by the YouTube channel Maths CA).

Theorem (Scaling by zero results in the zero vector)

For every we have that: .

Proof (Scaling by zero results in the zero vector)

So .

Question: Can a vector be multiplied with a scalar in such a way that the original vector is the result?

The answer is yes, because if we choose , then according to the associative law of scalar multiplication, we have

Now is only well-defined if . But this is always true, since . If there was then which is excluded by the premise .

This shows why it is useful to define vector spaces over a field and not something weaker (like a ring). This is because in fields there is a multiplicative inverse to every non-zero element. So there is an inverse with to every with . Thus, the fact that is a field guarantees that every scaling (stretching) can be scaled back by the inverse scaling (with the inverse stretching factors) such that .

Scaling the zero vector again gives the zero vector[Bearbeiten]

Scaling the zero vector again gives the zero vector - proof (YouTube video (German) by the YouTube channel Maths CA).

Theorem (Scaling the zero vector again gives the zero vector)

For all we have that: .

Proof (Scaling the zero vector again gives the zero vector)

Let be the neutral element of vector addition and arbitrary. We have that:

So .

Scalar multiplication leaves no zero divisors[Bearbeiten]

Proof that scalar multiplication leaves no zero divisors (YouTube video (German) by the YouTube channel Maths CA).

Theorem

Let and be such that . Then, either or (or both).

In already proved theorems we have shown that in the case or the equation is satisfied. Now we show conversely that it follows from that or .

Proof

Let . If , then there exists a with . Thus:

So from we get . Thus or must hold, and we cannot at the same time have and .

Scaling by a negative scalar[Bearbeiten]

Scaling a vector by a negative scalar (YouTube video (German) by the YouTube channel Maths CA).

Theorem (Scaling by a negative scalar)

For all and all we have that

Proof (Scaling by a negative scalar)

By the scalar distributive law for scalar multiplication we have on one hand:

This shows that is the additive inverse of . Thus . On the other hand, according to vectorial distributive law.

This shows that is the additive inverse of , so it is equal to . Thus, in total

Hint

From the above theorem, we immediately get . This follows from (where we have used ).