# Field as a vector space – Serlo

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• ↳ Project "Serlo"
• ↳ Linear algebra
Contents "Linear algebra"
• Vector spaces • Introduction: Vector space • Vector space • Vector space: properties • Proofs for vector spaces • Field as a vector space • Coordinate spaces • Sequence spaces • Function spaces • Subspace • Cosets of a subspace • Quotient space • Linear combinations, generators and bases • Linear maps • Matrices Let $K$ be a field. We now consider $K$ as a vector space over itself.

## Introduction

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

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

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

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

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

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

## Definition of the vector space structure

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

Definition (Vector space structure on $K$ )

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

$(a,b)\mapsto a\boxplus b:=a+b.$ Similarly, we define a scalar multiplication $\boxdot \colon K\times K\to K$ via

$(\lambda ,a)\mapsto \lambda \boxdot a:=\lambda \cdot a.$ ## The field is a vector space over itself

Theorem ($K$ is a vector space)

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

How to get to the proof? ($K$ is a vector space)

We proceed as in the article Proofs for vector spaces.

Proof ($K$ is a vector space)

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

Let $x,y,z\in K$ .

Then:

{\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 $x,y\in K$ .

Then:

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

Let $x\in K$ . Then:

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

Proof step: Inverse with respect to addition

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

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

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

Proof step: Scalar distributive law

Let $\lambda ,\mu \in K$ and $x\in K$ .

Then:

{\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 $\lambda \in K$ and $x,y\in K$ .

Then:

{\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 $\lambda ,\mu \in K$ and $x\in K$ .

Then:

{\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 $x\in K$ .

Then:

$1_{K}\boxdot x=1_{K}\cdot x=x.$ Thus we have shown the unitary law.

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