Serlo: EN: Field as a vector space

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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 $\lambda \cdot x\in ℝ$.

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  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.

The field is a vector space over itself

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