Vector spaces

Definition

Definition : Vector space

Let $E$ be a non empty set, with two operations : an internal operation (or law) called addition and noted $+$:

$$\begin{array}{ccc} E \times E & \longrightarrow & E \\ (u, v) & \longrightarrow & u + v \end{array}$$

and an external operation (or law) called multiplication and noted $\cdot$ :

$$\begin{array}{ccc} \mathbb{K} \times E & \longrightarrow & E \\ (\lambda, u) & \longrightarrow & \lambda \cdot u \end{array}$$

We say that $(E, +, \cdot)$ is a vector space on $\mathbb{K}$ or a $\mathbb{K}$-vector space if the following properties are satisfied :

• The addition on $E$ is :
  1. commutative : $u + v = v + u$ for all $(u, v) \in \mathbb{K}^2$
  2. associative : $(u + v) + w = u + (v + w)$ for all $(u, v, w) \in \mathbb{K}^3$
  3. own a "zero" called null vector : there exists $\vec{0}$ such that $\vec{u} + \vec{0} = \vec{u}$, for all $\vec{u} \in E$
  4. Every element has an opposite : for all $u \in E$, there is an element denoted $-u$ such that $u + (-u) = 0$
• External multiplication verifies :
  1. $1 \cdot u = u$ pour tout $u \in E$
  2. $\lambda \cdot (\mu \cdot u) = (\lambda\mu) \cdot u$ for all $(\lambda, \mu) \in \mathbb{K}^2, u \in E$
• Multiplication is distributive over addition :
  1. $\lambda \cdot (u + v) = \lambda \cdot u + \lambda \cdot v$ for all $\lambda \in \mathbb{K}$, for all $(u, v) \in E^2$
  2. $(\lambda + \mu) \cdot u = \lambda \cdot u + \mu \cdot u$ for all $(\lambda, \mu) \in \mathbb{K}^2$, for all $u \in E$

Results

Proposition : Vector space axioms

If $(E, +, \cdot)$ is a $\mathbb{K}$-vector space, then :

1. Its null vector is unique
2. Every vector of $E$ has a unique opposite
3. $\lambda \cdot 0 = 0$ for all $\lambda \in \mathbb{K}$
4. $0 \cdot u = 0$ for all $u \in E$
5. If $\lambda \cdot u = 0$ then $\lambda = 0$ or $u = 0$
6. $-u = (-1) \cdot u$ for all $u \in E$

Proof

1. If $0_E'$ is an other null vector, then with property (3) of vectors spaces, $0_E$ + $0_E' = 0_E'$ but also $0_E$ + $0_E' = 0_E$ hence $0_E = 0_E'$.
2. If $u$ own two opposites denoted $v$ and $w$, then $v = v + 0 = v + (u + w) = (v + u) + w = 0_E + w = w$ (by successively applying (3) (4) (2) (4) (3))
3. We have $\lambda \cdot 0_E = \lambda \cdot (0_E + 0_E) = \lambda \cdot 0_E + \lambda \cdot 0_E$ by successively applying the vector spaces properties (3) and (7), so, adding the opposite of $\lambda \cdot 0_E$, we found $0_E = \lambda \cdot 0_E$.
4. We write $0_E \cdot u = (0_E + 0_E) \cdot u = 0_E \cdot u + 0_E \cdot u$ by applying (8) and we also conclude that $0_E \cdot u = 0_E$
5. We suppose that $\lambda \cdot u = 0_E$ with $\lambda \neq 0_E$, and then we show $u = 0_E$. We have $u = 1 \cdot u = (\lambda^{-1}\lambda) \cdot u = \lambda^{-1} \cdot (\lambda \cdot u) = \lambda^{-1} \cdot 0_E = 0_E$
6. We remark that $u + (-1) \cdot u = 1 \cdot u + (-1) \cdot u = (1 + (-1)) \cdot u = 0_E \cdot u = 0$ and we conclude thanks to the opposite unicity of $u$ that we have demonstrated in (2).