Vector Space

A Vector space is an abstract framework which can be used to define multiple abstract objects as vectors, with constrained operations. A vector space is a set of anything, matrices, functions, or even numbers, as long as it has two operations strictly defined:

• Vector addition
• Scalar multiplication

A vector space is defined using a Field. We first define (or fix) a field, $\mathbb{F}$, with elements in the field being referred to as scalars. For example, if we want to work with real scalars, we fix $\mathbb{F}=\mathbb{R}$.

Definition §

Fix a field of scalars, $\mathbb{F}$. A vector space over $\mathbb{F}$ is defined to be a non-empty set, $V$, with two Binary Operations - Vector addition and scalar multiplication:

$$V_{\mathbb{F}}=(V,F,+,\times )$$

• $V$ = Set of vectors
• $F$ = Set of scalars, i.e. set used to define the field
• $+$ = Addition defined for vectors: $+:V\times V\to V$
• $\times$ = Multiplication defined with vectors and scalars: $F\times V\to V$

Note that we can let addition and multiplication do whatever we want, as long as it operates and returns on the right sets, and follows the axioms. Then, the $V_{\mathbb{F}}$ must follow the following axioms:

Axioms of a Vector Space ^theorem

Existence of Bases ^theorem

Let $V_{\mathbb{F}}$ be a vector space. Then:

• $V_{\mathbb{F}}$ must contain a basis
• Any set that spans $V_{\mathbb{F}}$ has some subset that is a basis of $V_{\mathbb{F}}$: $$V_{\mathbb{F}} = \text{span}(S) \iff \text{$S$ contains a basis for $V_{\mathbb{F}}$}

* Any [linearly independent](Linear%20Dependence.md) set in $V_{\mathbb{F}}$ can be extended to form a basis for $V_{\mathbb{F}}$. #TODO https://thetawise.ai/answers/Every-linearly-independent-subset-of-V-can-be-extended-to-a-basis.

Proof: $0_F\times \vec{v}=\vec{0}$ §

#todo One of the properties we take for granted is that the zero vector is the result of scalar multiplying any vector, $\vec{v}\in V$ by the field zero, $0_F\in F$.

We’ll use the 8 axioms above, along with the 10 field axioms:

$$LHS=0_F\times \vec{v}$$

We can use F4, where $0_F^{\prime }$ is the additive inverse to $0_F$

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

$$\therefore LHS=(0_F+0_F^{\prime })\times \vec{v}$$

Bad practice

It’s generally bad practice to write this as $a-b$ instead of $a+a^{\prime }$ , however tempting it might be. But I’m not the boss of you.

We can use distributivity of scalar multiplication with respect to field addition to expand:

$$=(0_F\times \vec{v})+(0_F^{\prime }\times \vec{v})$$

And use distributivity of scalar multiplication with respect to vector addition backwards:

$$=0_F\times (\vec{v}+\vec{v})$$

Resources §

• Proofs using the vector space axioms - Matthew Towers