Vector space

Mathematical definition

Set $\mathbf{V}$ is a vector space over $\mathbb{F}$ if:

1. Are defined operations

• • of addition: $\mathbf{V} \times \mathbf{V} \to \mathbf{V}$
  • of scalar multiplication: $\mathbb{F} \times \mathbf{V} \to \mathbf{V}$

2. Are satisfied the properties

• • Commutativity: $\forall u,v\in\mathbf{V},\; u+v=v+u$;
  • Associativity: $\forall u,v,w\in\mathbf{V},\;\forall a,b\in\mathbb{F},\; (u+v)+w=u+(v+w)\;\land\;(ab)v=a(bv)$;
  • Additive identity: $\exists \vec 0\in\mathbf{V},\;\forall v\in\mathbf{V},\; \vec 0+v=v$;
  • Additive inverse: $\forall v\in\mathbf{V},\;\exists w\in\mathbf{V},\; v+w=0$;
  • Multiplicative identity: $\forall v\in\mathbf{V},\; 1v=v$;
  • Distributivity: $\forall u,v\in\mathbf{V},\;\forall a,b\in\mathbb{F},\; a(u+v)=au+av\;\land\;(a+b)u=au+bu$.

Elementary properties

For vector space $V$ over $\mathbb{F}$:

The $\vec 0$ is unique

$$\exists!\,\vec 0\in V,\;\forall v\in V,\;\vec 0+v=v.$$

Proof: Let $\vec 0,\vec 0_1\in V$ satisfy $\forall v\in V,\;\vec 0+v=v$ and $\forall v\in V,\;\vec 0_1+v=v$. Then $$\vec 0_1=\vec 0+\vec 0_1=\vec 0.$$ Hence $\vec 0=\vec 0_1$.

The additive inverse for each vector is unique

$$\forall v\in V,\;\exists!\,w\in V,\;v+w=\vec 0.$$

Proof: Let $w,w'\in V$ satisfy $v+w=\vec 0$ and $v+w'=\vec 0$. Then $$w=w+\vec 0=w+(v+w')=(w+v)+w'=\vec 0+w'=w'.$$ Hence $w=w'$.

Scalar $0$ mutiplies by any vector results in $\vec 0$ (the vector zero)

$$\forall v\in V,\;0v=\vec 0.$$

Proof: For $v\in V$, $$0v=(0+0)v=0v+0v.$$ Adding the additive inverse of $0v$, $-0v$, to both sides, we obtain $$\vec 0=0v.$$

Any scalar multiples of $\vec 0$ is $\vec 0$

$$\forall a\in\mathbb{F},\;a\vec 0=\vec 0.$$

Proof: For $a\in\mathbb{F}$, $$a\vec 0=a(\vec 0+\vec 0)=a\vec 0+a\vec 0.$$ Adding the additive inverse of $a\vec 0$, $-a\vec 0$ to both sides, we obtain $$\vec 0=a\vec 0.$$

Scalar $-1$ multiples by any vector results in the additive inverse of the vector

$$\forall v\in V,\;(-1)v=-v.$$

Proof: For $v\in V$, $$v+(-1)v=1v+(-1)v=(1+(-1))v=0v=\vec 0,$$ which shows that $(-1)v$ is the additive inverse $-v$ of $v$.