Definition:Vector Space

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Let $\left({K, +_K, \times_K}\right)$ be a division ring.

Let $\left({G, +_G}\right)$ be an abelian group.

Let $\left({G, +_G, \circ}\right)_K$ be a unitary $K$-module.

Then $\left({G, +_G, \circ}\right)_K$ is a vector space over $K$ or a $K$-vector space.

That is, a vector space is a unitary module whose scalar ring is a division ring.

If $\times_K$ is commutative, then $\left({K, +_K, \times_K}\right)$ is by definition a field.

In that case, the scalar ring of $\left({G, +_G, \circ}\right)_K$ is called the scalar field of $\left({G, +_G, \circ}\right)_K$.

Vector Space Axioms

The vector space axioms consist of the abelian group axioms:

\((G0)\)   $:$   Closure Axiom      \(\displaystyle \forall \mathbf x, \mathbf y \in G:\) \(\displaystyle \mathbf x +_G \mathbf y \in G \)             
\((G1)\)   $:$   Associativity Axiom      \(\displaystyle \forall \mathbf x, \mathbf y, \mathbf z \in G:\) \(\displaystyle \left({\mathbf x +_G \mathbf y}\right) +_G \mathbf z = \mathbf x +_G \left({\mathbf y +_G \mathbf z}\right) \)             
\((G2)\)   $:$   Identity Axiom      \(\displaystyle \exists \mathbf 0 \in G: \forall \mathbf x \in G:\) \(\displaystyle \mathbf 0 +_G \mathbf x = \mathbf x = \mathbf x +_G \mathbf 0 \)             
\((G3)\)   $:$   Inverse Axiom      \(\displaystyle \forall \mathbf x \in G: \exists \left({-\mathbf x}\right) \in G:\) \(\displaystyle \mathbf x +_G \left({-\mathbf x}\right) = \mathbf 0 \)             
\((C)\)   $:$   Commutativity Axiom      \(\displaystyle \forall \mathbf x, \mathbf y \in G:\) \(\displaystyle \mathbf x +_G \mathbf y = \mathbf y +_G \mathbf x \)             

together with the properties of a unitary module:

\((U1)\)   $:$     \(\displaystyle \forall \lambda \in K: \forall \mathbf x, \mathbf y \in G:\) \(\displaystyle \lambda \circ \left({\mathbf x +_G \mathbf y}\right) = \lambda \circ \mathbf x +_G \lambda \circ \mathbf y \)             
\((U2)\)   $:$     \(\displaystyle \forall \lambda, \mu \in K: \forall \mathbf x \in G:\) \(\displaystyle \left({\lambda + \mu}\right)\circ \mathbf x = \lambda \circ \mathbf x +_G \mu \circ \mathbf x \)             
\((U3)\)   $:$     \(\displaystyle \forall \lambda, \mu \in K: \forall \mathbf x \in G:\) \(\displaystyle \lambda \circ \left({\mu \circ \mathbf x}\right) = \left({\lambda \cdot \mu}\right) \circ \mathbf x \)             
\((U4)\)   $:$     \(\displaystyle \forall \mathbf x \in G:\) \(\displaystyle 1_K \circ \mathbf x = \mathbf x \)             


Let $V$ be a vector space.

Any element $v$ of $V$ is called a vector.

Zero Vector

The identity of $\left({G, +_G}\right)$ is usually denoted $\mathbf 0$, or some variant of this, and called the zero vector.

Note that on occasion it is advantageous to denote the zero vector differently, for example by $e$, or $0_V$ or $0_G$, in order to highlight the fact that the zero vector is not the same object as the zero scalar.

Also known as

A vector space is also sometimes called a linear space, especially when discussing the real vector space $\R^n$.

The notation $\left({G, +_G, \circ, K}\right)$ can also be seen for this concept.

Also defined as

Some sources insist that $\left({K, +_K, \times_K}\right)$ needs to be a field, not just a division ring, for this definition to be valid.

Also see

As a vector space is also a unitary module, all the results which apply to modules, and to unitary modules, also apply to vector spaces.