Definition

DEFINITION 3.1.1 (Vector Space)   A vector space over ${\mathbb{F}},$ denoted $V({\mathbb{F}}),$ is a non-empty set, satisfying the following axioms:

1. VECTOR ADDITION: To every pair ${\mathbf u}, {\mathbf v}\in V$ there corresponds a unique element ${\mathbf u}\oplus {\mathbf v}$ in $V$ such that
   1. ${\mathbf u}\oplus {\mathbf v}= {\mathbf v}\oplus {\mathbf u}$ (Commutative law).
   2. $({\mathbf u}\oplus {\mathbf v}) \oplus {\mathbf w}= {\mathbf u}\oplus ({\mathbf v}\oplus {\mathbf w})$ (Associative law).
   3. There is a unique element ${\mathbf 0}$ in $V$ (the zero vector) such that ${\mathbf u}\oplus {\mathbf 0}= {\mathbf u},$ for every ${\mathbf u}\in V$ (called the additive identity).
   4. For every ${\mathbf u}\in V$ there is a unique element $-{\mathbf u}\in V$ such that ${\mathbf u}\oplus (-{\mathbf u}) = {\mathbf 0}$ (called the additive inverse).

      $\oplus$ is called VECTOR ADDITION.

2. SCALAR MULTIPLICATION: For each ${\mathbf u}\in V$ and $\alpha \in {\mathbb{F}},$ there corresponds a unique element $\alpha \odot {\mathbf u}$ in $V$ such that
   1. $\alpha \cdot ( \beta \odot {\mathbf u}) = (\alpha \beta) \odot {\mathbf u}$ for every $\alpha, \beta \in {\mathbb{F}}$ and ${\mathbf u}\in V.$
   2. $1 \odot {\mathbf u}= {\mathbf u}$ for every ${\mathbf u}\in V,$ where $1 \in {\mathbb{R}}.$

3. DISTRIBUTIVE LAWS: RELATING VECTOR ADDITION WITH SCALAR MULTIPLICATION
   For any $\alpha, \beta \in {\mathbb{F}}$ and ${\mathbf u}, {\mathbf v}\in V,$ the following distributive laws hold:
   1. $\alpha \odot ({\mathbf u}\oplus {\mathbf v}) = (\alpha \odot {\mathbf u}) \; \oplus \; (\alpha \odot {\mathbf v}).$
   2. $(\alpha + \beta) \odot {\mathbf u}= (\alpha \odot {\mathbf u}) \; \oplus \; (\beta \odot {\mathbf u}).$

Note: the number 0 is the element of ${\mathbb{F}}$ whereas ${\mathbf 0}$ is the zero vector.

Remark 3.1.2   The elements of ${\mathbb{F}}$ are called SCALARS, and that of $V$ are called VECTORS. If ${\mathbb{F}}= {\mathbb{R}},$ the vector space is called a REAL VECTOR SPACE. If ${\mathbb{F}}= {\mathbb{C}},$ the vector space is called a COMPLEX VECTOR SPACE.

We may sometimes write $V$ for a vector space if ${\mathbb{F}}$ is understood from the context.

Some interesting consequences of Definition 3.1.1 is the following useful result. Intuitively, these results seem to be obvious but for better understanding of the axioms it is desirable to go through the proof.

THEOREM 3.1.3   Let $V$ be a vector space over ${\mathbb{F}}.$ Then

1. ${\mathbf u}\oplus {\mathbf v}= {\mathbf u}$ implies ${\mathbf v}= {\mathbf 0}.$
2. $\alpha \odot {\mathbf u}= {\mathbf 0}$ if and only if either ${\mathbf u}$ is the zero vector or $\alpha = 0.$
3. $(-1) \odot {\mathbf u}= - {\mathbf u}$ for every ${\mathbf u}\in V.$

Proof. Proof of Part 1.
For ${\mathbf u}\in V,$ by Axiom 1d there exists $-{\mathbf u}\in V$ such that $-{\mathbf u}\oplus {\mathbf u}= {\mathbf 0}.$

Hence, ${\mathbf u}\oplus {\mathbf v}= {\mathbf u}$ is equivalent to

$$-{\mathbf u}\oplus ({\mathbf u}\oplus {\mathbf v}) = -{\mathbf u}... ... 0}\oplus {\mathbf v}= {\mathbf 0}\Longleftrightarrow {\mathbf v}= {\mathbf 0}.$$

Proof of Part 2.
As ${\mathbf 0}= {\mathbf 0}\oplus {\mathbf 0}$ , using the distributive law, we have

$${\alpha}\odot {\mathbf 0}= {\alpha}\odot ({\mathbf 0}\oplus {\mathbf 0}) = ({\alpha}\odot {\mathbf 0}) \; \oplus \; ({\alpha}\odot {\mathbf 0}).$$

Thus, for any ${\alpha}\in {\mathbb{F}}$ , the first part implies $\alpha \odot {\mathbf 0}= {\mathbf 0}$ . In the same way,

$$0 \odot {\mathbf u}= (0 + 0) \odot {\mathbf u}= (0 \odot {\mathbf u}) \; \oplus (0 \odot {\mathbf u}).$$

Hence, using the first part, one has $0 \odot {\mathbf u}= {\mathbf 0}$ for any ${\mathbf u}\in V.$

Now suppose $\alpha \odot {\mathbf u}= {\mathbf 0}.$ If $\alpha = 0$ then the proof is over. Therefore, let us assume $\alpha \neq 0$ (note that $\alpha$ is a real or complex number, hence $$\frac{1}{\alpha}$$ exists and

$${\mathbf 0}= \frac{1}{\alpha} \odot {\mathbf 0}= \frac{1}{\alpha}... ...frac{1}{\alpha} \; \alpha ) \odot {\mathbf u}= 1 \odot {\mathbf u}= {\mathbf u}$$

as $1 \odot {\mathbf u}= {\mathbf u}$ for every vector ${\mathbf u}\in V.$

Thus we have shown that if ${\alpha}\neq 0$ and $\alpha \odot {\mathbf u}= {\mathbf 0}$ then ${\mathbf u}= {\mathbf 0}.$

Proof of Part 3.
We have ${\mathbf 0}= {\mathbf 0}{\mathbf u}= (1 + (-1)){\mathbf u}= {\mathbf u}+ (-1) {\mathbf u}$ and hence $(-1) {\mathbf u}= - {\mathbf u}.$ height6pt width 6pt depth 0pt