# Field as a vector space – Serlo

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Let ${\displaystyle K}$ be a field. We now consider ${\displaystyle K}$ as a vector space over itself.

## Introduction

From school we already know the vector space ${\displaystyle \mathbb {R} ^{3}}$ over the field ${\displaystyle \mathbb {R} }$. The vectors in ${\displaystyle \mathbb {R} ^{3}}$ have the form ${\displaystyle (x,y,z)^{T}}$ with ${\displaystyle x,y,z\in \mathbb {R} }$. We can consider the vectors in a 3-dimensional coordinate system. Since ${\displaystyle \mathbb {R} ^{3}}$ is a vector space, we can add and scale vectors.

We also know the vector space ${\displaystyle \mathbb {R} ^{2}}$. The vectors in ${\displaystyle \mathbb {R} ^{2}}$ have the form ${\displaystyle (x,y)^{T}}$ with ${\displaystyle x,y\in \mathbb {R} }$. We can get ${\displaystyle \mathbb {R} ^{2}}$ from ${\displaystyle \mathbb {R} ^{3}}$ by deleting one of the coordinates ${\displaystyle x,y,z}$ (e.g., the last one). Illustratively, we then go from the 3-dimensional coordinate system to the ${\displaystyle xy}$ plane. So when omitting a coordinate from ${\displaystyle \mathbb {R} ^{3}}$, the vector space structure is conserved. What happens if we delete another coordinate?

For example, if we omit the second coordinate of ${\displaystyle (x,y)}$, only ${\displaystyle x}$ remains and we get an element in ${\displaystyle \mathbb {R} }$. Illustratively, we thus go from the ${\displaystyle xy}$ plane to the ${\displaystyle x}$ axis. Again, when deleting a coordinate, the vector space structure should not be broken.

We can add and scale the elements in ${\displaystyle \mathbb {R} }$ (just like vectors), because for all ${\displaystyle x,y\in \mathbb {R} }$ we have ${\displaystyle x+y\in \mathbb {R} }$ and for all ${\displaystyle \lambda \in \mathbb {R} }$ and ${\displaystyle x\in \mathbb {R} }$ it holds that ${\displaystyle \lambda \cdot x\in \mathbb {R} }$.

Now our field ${\displaystyle \mathbb {R} }$ should be an ${\displaystyle \mathbb {R} }$-vector space. Visually, this vector space is the number line.

We can apply this idea to an arbitrary field ${\displaystyle K}$, since also in an arbitrary ${\displaystyle K}$ we can add elements and multiply them by scalars in ${\displaystyle K}$. Therefore, we conjecture that ${\displaystyle K}$ is a ${\displaystyle K}$-vector space.

## Definition of the vector space structure

Let ${\displaystyle (K,+,\cdot )}$ be a field. Then we can define an addition and a scalar multiplication.

Definition (Vector space structure on ${\displaystyle K}$)

We define an addition ${\displaystyle \boxplus \colon K\times K\to K}$ on ${\displaystyle K}$ by

${\displaystyle (a,b)\mapsto a\boxplus b:=a+b.}$

Similarly, we define a scalar multiplication ${\displaystyle \boxdot \colon K\times K\to K}$ via

${\displaystyle (\lambda ,a)\mapsto \lambda \boxdot a:=\lambda \cdot a.}$

## The field is a vector space over itself

Theorem (${\displaystyle K}$ is a vector space)

${\displaystyle (K,\boxplus ,\boxdot )}$ is a ${\displaystyle K}$-vector space .

How to get to the proof? (${\displaystyle K}$ is a vector space)

We proceed as in the article Proofs for vector spaces.

Proof (${\displaystyle K}$ is a vector space)

So now we have to establish the eight Vector space axioms.

Let ${\displaystyle x,y,z\in K}$.

Then:

{\displaystyle {\begin{aligned}(x\boxplus y)\boxplus z&=\\[0.3em]&{\color {OliveGreen}\left\downarrow \ {\text{definition of }}\boxplus \right.}\\[0.3em]&=(x+y)\boxplus z\\[0.3em]&{\color {OliveGreen}\left\downarrow \ {\text{definition of }}\boxplus \right.}\\[0.3em]&=(x+y)+z\\[0.3em]&{\color {OliveGreen}\left\downarrow \ {\text{associativity of addition in }}K\right.}\\[0.3em]&=x+(y+z)\\[0.3em]&{\color {OliveGreen}\left\downarrow \ {\text{definition of }}\boxplus \right.}\\[0.3em]&=x\boxplus (y\boxplus z).\end{aligned}}}

This shows the associativity of the addition.

Let ${\displaystyle x,y\in K}$.

Then:

{\displaystyle {\begin{aligned}x\boxplus y&=\\[0.3em]&{\color {OliveGreen}\left\downarrow \ {\text{definition of }}\boxplus \right.}\\[0.3em]&=x+y\\[0.3em]&{\color {OliveGreen}\left\downarrow \ {\text{commutativity of the addition in }}K\right.}\\[0.3em]&=y+x\\[0.3em]&{\color {OliveGreen}\left\downarrow \ {\text{definition of }}\boxplus \right.}\\[0.3em]&=y\boxplus x.\end{aligned}}}

This shows the commutativity of the addition.

Proof step: Neutral element of addition

We now have to show that there is a neutral element ${\displaystyle 0_{\boxplus }\in K}$ with respect to ${\displaystyle \boxplus }$, that is, ${\displaystyle x\boxplus 0_{\boxplus }=x}$ for all ${\displaystyle x\in K}$. It is natural to use the zero element of the field ${\displaystyle 0_{\boxplus }:=0_{K}}$ as the neutral element.

Let ${\displaystyle x\in K}$. Then:

${\displaystyle x\boxplus 0_{\boxplus }=x+0_{\boxplus }=x+0_{K}=x.}$

Thus we have shown that ${\displaystyle 0_{\boxplus }=0_{K}\in K}$ is the neutral element of the addition. In the following we will therefore simply write ${\displaystyle 0_{K}}$ for the neutral element.

Proof step: Inverse with respect to addition

Let ${\displaystyle x\in K}$. We have to show that there is a ${\displaystyle y\in K}$ such that ${\displaystyle x\boxplus y=0_{K}}$ .

It is natural to choose ${\displaystyle y}$ as the inverse in ${\displaystyle K}$ with respect to ${\displaystyle +}$, i.e., we choose ${\displaystyle y:=-x}$

Then:

${\displaystyle x\boxplus y=x+y=x+(-x)=0_{K}.}$

Thus we have shown that for any ${\displaystyle x\in K}$ there is a ${\displaystyle y\in K}$ with ${\displaystyle x\boxplus y=0_{K}}$.

Proof step: Scalar distributive law

Let ${\displaystyle \lambda ,\mu \in K}$ and ${\displaystyle x\in K}$.

Then:

{\displaystyle {\begin{aligned}(\lambda +\mu )\boxdot x&=\\[0.3em]&{\color {OliveGreen}\left\downarrow \ {\text{definition of }}\boxdot \right.}\\[0.3em]&=(\lambda +\mu )\cdot x\\[0.3em]&{\color {OliveGreen}\left\downarrow \ {\text{distributive law in }}K\right.}\\[0.3em]&=\lambda \cdot x+\mu \cdot x\\[0.3em]&{\color {OliveGreen}\left\downarrow \ {\text{definition of }}\boxplus \right.}\\[0.3em]&=(\lambda \cdot x)\boxplus (\mu \cdot x)\\[0.3em]&{\color {OliveGreen}\left\downarrow \ {\text{definition of }}\boxdot \right.}\\[0.3em]&=(\lambda \boxdot x)\boxplus (\mu \boxdot x).\end{aligned}}}

Thus the scalar distributive law is shown.

Proof step: Vectorial distributive law

Let ${\displaystyle \lambda \in K}$ and ${\displaystyle x,y\in K}$.

Then:

{\displaystyle {\begin{aligned}\lambda \boxdot (x\boxplus y)&=\\[0.3em]&{\color {OliveGreen}\left\downarrow \ {\text{definition of }}\boxplus \right.}\\[0.3em]&=\lambda \boxdot (x+y)\\[0.3em]&{\color {OliveGreen}\left\downarrow \ {\text{definition of }}\boxdot \right.}\\[0.3em]&=\lambda \cdot (x+y)\\[0.3em]&{\color {OliveGreen}\left\downarrow \ {\text{distributive law in }}K\right.}\\[0.3em]&=\lambda \cdot x+\lambda \cdot y\\[0.3em]&{\color {OliveGreen}\left\downarrow \ {\text{definition of }}\boxplus \right.}\\[0.3em]&=(\lambda \cdot x)\boxplus (\lambda \cdot y)\\[0.3em]&{\color {OliveGreen}\left\downarrow \ {\text{Definition of }}\boxdot \right.}\\[0.3em]&=(\lambda \boxdot x)\boxplus (\lambda \boxdot y).\end{aligned}}}

Thus the vectorial distributive law is shown.

Proof step: Assoziativität bezüglich Multiplikation

Let ${\displaystyle \lambda ,\mu \in K}$ and ${\displaystyle x\in K}$.

Then:

{\displaystyle {\begin{aligned}(\lambda \cdot \mu )\boxdot x&=\\[0.3em]&{\color {OliveGreen}\left\downarrow \ {\text{definition of }}\boxdot \right.}\\[0.3em]&=(\lambda \cdot \mu )\cdot x\\[0.3em]&{\color {OliveGreen}\left\downarrow \ {\text{associative law for multiplication in }}K\right.}\\[0.3em]&=\lambda \cdot (\mu \cdot x)\\[0.3em]&{\color {OliveGreen}\left\downarrow \ {\text{definition of }}\boxdot \right.}\\[0.3em]&=\lambda \boxdot (\mu \cdot x)\\[0.3em]&{\color {OliveGreen}\left\downarrow \ {\text{definition of }}\boxdot \right.}\\[0.3em]&=\lambda \boxdot (\mu \boxdot x).\end{aligned}}}

This shows the associative law for multiplication.

Proof step: Unitary law

Let ${\displaystyle x\in K}$.

Then:

${\displaystyle 1_{K}\boxdot x=1_{K}\cdot x=x.}$

Thus we have shown the unitary law.

With this we have established all eight vector space axioms and thus ${\displaystyle (K,\boxplus ,\boxdot )}$ is a ${\displaystyle K}$ vector space.