Field as a vector space – "Math for Non-Geeks"

Let $K$ be a field. We now consider $K$ as a vector space over itself.

Introduction

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

We also know the vector space ${ℝ}^2$. The vectors in ${ℝ}^2$ have the form $(x,y{)}^T$ with $x,y\in ℝ$. We can get ${ℝ}^2$ from ${ℝ}^3$ 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 ${ℝ}^3$, 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 $ℝ$. 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 $ℝ$ (just like vectors), because for all $x,y\in ℝ$ we have $x+y\in ℝ$ and for all $\lambda \in ℝ$ and $x\in ℝ$ it holds that $\lambda \cdot x\in ℝ$.

• 
  Addition of the vectors $1$ and $2$ on the real line
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  Scalar multiplication of the vector $\sqrt{2}$ with the scalar $1.5$ on the real line

Now our field $ℝ$ should be an $ℝ$-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 $⊞:K\times K\to K$ on $K$ by

$(a,b)\mapsto a⊞b:=a+b.$

Similarly, we define a scalar multiplication $⊡:K\times K\to K$ via

$(\lambda ,a)\mapsto \lambda ⊡a:=\lambda \cdot a.$

The field is a vector space over itself

Theorem ($K$ is a vector space)

$(K,⊞,⊡)$ 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.

Proof step: Associativity of addition

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

Then:

$\begin{array}{cc}(x⊞y)⊞z& =\\ [0.3em]& \downarrow \text{definition of }⊞\\ [0.3em]& =(x+y)⊞z\\ [0.3em]& \downarrow \text{definition of }⊞\\ [0.3em]& =(x+y)+z\\ & \downarrow \text{associativity of addition in }K\\ [0.3em]& =x+(y+z)\\ & \downarrow \text{definition of }⊞\\ [0.3em]& =x⊞(y⊞z).\end{array}$

This shows the associativity of the addition.

Proof step: Commutativity of addition

Let $x,y\in K$.

Then:

$\begin{array}{cc}x⊞y& =\\ [0.3em]& \downarrow \text{definition of }⊞\\ [0.3em]& =x+y\\ [0.3em]& \downarrow \text{commutativity of the addition in }K\\ [0.3em]& =y+x\\ [0.3em]& \downarrow \text{definition of }⊞\\ [0.3em]& =y⊞x.\end{array}$

This shows the commutativity of the addition.

Proof step: Neutral element of addition

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

Let $x\in K$. Then: $x⊞0_{⊞}=x+0_{⊞}=x+0_K=x.$

Thus, we have shown that $0_{⊞}=0_K\in K$ 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 $x⊞y=0_K$ .

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

Then: $x⊞y=x+y=x+(-x)=0_K.$

Thus, we have shown that for any $x\in K$ there is a $y\in K$ with $x⊞y=0_K$.

Proof step: Scalar distributive law

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

Then:

$\begin{array}{cc}(\lambda +\mu )⊡x& =\\ [0.3em]& \downarrow \text{definition of }⊡\\ [0.3em]& =(\lambda +\mu )\cdot x\\ & \downarrow \text{distributive law in }K\\ [0.3em]& =\lambda \cdot x+\mu \cdot x\\ [0.3em]& \downarrow \text{definition of }⊞\\ [0.3em]& =(\lambda \cdot x)⊞(\mu \cdot x)\\ [0.3em]& \downarrow \text{definition of }⊡\\ [0.3em]& =(\lambda ⊡x)⊞(\mu ⊡x).\end{array}$

Thus, the scalar distributive law is shown.

Proof step: Vectorial distributive law

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

Then:

$\begin{array}{cc}\lambda ⊡(x⊞y)& =\\ & \downarrow \text{definition of }⊞\\ [0.3em]& =\lambda ⊡(x+y)\\ [0.3em]& \downarrow \text{definition of }⊡\\ [0.3em]& =\lambda \cdot (x+y)\\ & \downarrow \text{distributive law in }K\\ [0.3em]& =\lambda \cdot x+\lambda \cdot y\\ [0.3em]& \downarrow \text{definition of }⊞\\ [0.3em]& =(\lambda \cdot x)⊞(\lambda \cdot y)\\ [0.3em]& \downarrow \text{Definition of }⊡\\ [0.3em]& =(\lambda ⊡x)⊞(\lambda ⊡y).\end{array}$

Thus, the vectorial distributive law is shown.

Proof step: Associativity of multiplication

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

Then:

$\begin{array}{cc}(\lambda \cdot \mu )⊡x& =\\ [0.3em]& \downarrow \text{definition of }⊡\\ [0.3em]& =(\lambda \cdot \mu )\cdot x\\ [0.3em]& \downarrow \text{associative law for multiplication in }K\\ [0.3em]& =\lambda \cdot (\mu \cdot x)\\ [0.3em]& \downarrow \text{definition of }⊡\\ [0.3em]& =\lambda ⊡(\mu \cdot x)\\ [0.3em]& \downarrow \text{definition of }⊡\\ [0.3em]& =\lambda ⊡(\mu ⊡x).\end{array}$

This shows the associative law for multiplication.

Proof step: Unitary law

Let $x\in K$.

Then: $1_K⊡x=1_K\cdot x=x.$

Thus, we have shown the unitary law.

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