Vector Space

Definition: Vector Space

A vector space $(V,F,+,\cdot )$ consists of a non-empty set $V$ and a field $F$ which are equipped with two operations - a vector addition $+:V\times V\to V$ and a scalar multiplication $\cdot :V\times F\to V$ - which have the following properties:

• Commutativity: $\mathbf{u}+\mathbf{v}=\mathbf{v}+\mathbf{u}\forall \mathbf{u},\mathbf{v}\in V$

• Associativity I: $\mathbf{u}+(\mathbf{v}+\mathbf{w})=(\mathbf{u}+\mathbf{v})+\mathbf{w}\forall \mathbf{u},\mathbf{v},\mathbf{w}\in V$

• Associativity II: $(\lambda \mu )\mathbf{u}=\lambda (\mu \mathbf{u})\forall \lambda ,\mu \in F,\forall \mathbf{u}\in V$

• Distributivity I: $\lambda (\mathbf{u}+\mathbf{v})=\lambda \mathbf{u}+\lambda \mathbf{v}\forall \mathbf{u},\mathbf{v}\in V,\forall \lambda \in F$

• Distributivity II: $(\lambda +\mu )\mathbf{v}=\lambda \mathbf{v}+\mu \mathbf{v}\forall \lambda ,\mu \in F,\forall \mathbf{v}\in V$

• Existence of a zero vector: There is an element $0\in V$ such that $\mathbf{v}+0=\mathbf{v}\forall \mathbf{v}\in V$

• Existence of the identity element: There is an element $1\in F$ such that $1\cdot \mathbf{u}=\mathbf{u}\forall \mathbf{u}\in V$

• Existence of vector inverses: For every $\mathbf{v}\in V$ there is a $-\mathbf{v}\in V$ such that $\mathbf{v}+(-\mathbf{v})=0$

Notation: Vector Subtraction

For any two vectors $\mathbf{u}$ and $\mathbf{v}$, where $-\mathbf{v}$ is the vector inverse of $\mathbf{v}$, we denote $\mathbf{u}+(-\mathbf{v})$ as simply $\mathbf{u}-\mathbf{v}$.

Theorem: Uniqueness of the Zero Vector

Every vector space has exactly one zero vector.

PROOF

Suppose that the vector space $(V,F,+,\cdots )$ had two zero vectors $0,0^{\prime }\in V$, i.e. $\mathbf{v}+0=\mathbf{v}$ $\mathbf{v}+0^{\prime }=\mathbf{v}$ for all $\mathbf{v}\in V$.

It follows then that

$$0+0^{\prime }=0$$ $$0^{\prime }+0=0^{\prime }$$

But vector addition is commutative and so $0+0^{\prime }=0^{\prime }+0$ which means that $0=0^{\prime }$.

Theorem: Uniqueness of Vector Inverses

For every vector $\mathbf{v}\in V$ in a vector space $(V,F,+,\cdot )$ there is exactly one inverse vector $-\mathbf{v}\in V$ such that

$$\mathbf{v}+(-\mathbf{v})=0$$

PROOF

Suppose that there was another vector ${\mathbf{v}}^{\prime }\in V$ such that

$$\mathbf{v}+{\mathbf{v}}^{\prime }=0$$

It then follows that

$$-\mathbf{v}+(\mathbf{v}+{\mathbf{v}}^{\prime })=-\mathbf{v}+0=-\mathbf{v}$$

and so

$${\mathbf{v}}^{\prime }=(-\mathbf{v}+\mathbf{v})+{\mathbf{v}}^{\prime }=-\mathbf{v}+(\mathbf{v}+{\mathbf{v}}^{\prime })=-\mathbf{v}+0=-\mathbf{v}$$