1  Fields and Spaces

Fields

A field is a set \mathbb{F}, equipped with two operations addition + and multiplication \cdot, obeying the rules (axioms) listed below.

Axioms of fields

For all x, y, and z in the field \mathbb{F} (\forall x, y, z \in \mathbb{F}), we have:

  • Closure under addition and multiplication:

    x + y \in \mathbb{F},

    x \cdot y \in \mathbb{F}.

  • Commutativity of addition and multiplication:

    x + y = y + x,

    x \cdot y = y \cdot x.

  • Associativity of addition and multiplication:

    (x + y) + z = x + (y + z),

    (x \cdot y) \cdot z = x \cdot (y \cdot z).

  • Distributive property of multiplication:

    (x + y) \cdot z = x \cdot z + y \cdot z.

  • There is an element in \mathbb{F} called “zero” 0 \in \mathbb{F} such that

    x + 0 = x,

    and there is another element in \mathbb{F} called “one” 1 \in \mathbb{F}, 1 \neq 0, such that

    x \cdot 1 = x.

  • For each x \in \mathbb{F}, there is an element in \mathbb{F} called addictive inverse x_{I} of x such that

    x + x_{I} = 0,

    and if x \neq 0, there is an element in \mathbb{F} called multiplicative inverse x^{-1} of x such that

    x \cdot x^{-1} = 1.

Properties of fields

  • Zero and one are unique: there is only one “zero” and one “one” in any field \mathbb{F}.

    We prove “zero” is unique by contradiction.

    Suppose there are two different “zero”s 0_{0} and 0_{1}.

    Due to the definition of “zero”,

    \begin{aligned} 0_{1} + 0_{0} & = 0_{1} & [0_{0} \text{ is "zero"}] \\ 0_{0} + 0_{1} & = 0_{0} & [0_{1} \text{ is "zero"}] \\ \end{aligned}

    Due to the Commutativity axiom,

    \begin{aligned} 0_{1} + 0_{0} & = 0_{0} + 0_{1} \\ 0_{1} & = 0_{0} \\ \end{aligned}

    which contradicts to the fact that 0_{0} and 0_{1} (1_{0} and 1_{1}) are different. Thus, there is only a unique “zero” in \mathbb{F}.

    The proof that the “one” in any \mathbb{F} is unique is the same as above by replacing every addition + with multiplication \cdot and 0_{0}, 0_{1} with 1_{0}, 1_{1}.

  • Addictive and multiplicative inverse of every element are unique: there is only one addictive inverse and multiplicative inverse of every element (other than “zero” for multiplicative inverse) in any field \mathbb{F}.

    We prove the addictive inverse is unique by contradiction.

    Suppose there are two different addictive inverses of x: x_{1} and x_{2}.

    By following the definition of addictive inverse,

    (x + x_{1}) + x_{2} = 0 + x_{2} = x_{2}.

    Also,

    \begin{aligned} (x + x_{1}) + x_{2} & = x_{2} \\ x + (x_{1} + x_{2}) & = x_{2} & [\text{associativity}] \\ x + (x_{2} + x_{1}) & = x_{2} & [\text{commutativity}] \\ (x + x_{2}) + x_{1} & = x_{2} & [\text{associativity}] \\ x_{1} & = x_{2} \end{aligned}

    which contradicts to the fact that x_{1} and x_{2} are different. Thus, the addictive inverse of every element in any field is unique.

    The proof that the multiplicative inverse of every element in any field is unique is the same as above by replacing every addition + with multiplication \cdot.

  • For every x in \mathbb{F}, x \cdot 0 = 0

    Consider

    \begin{aligned} x \cdot 0 + x \cdot 0 & = x \cdot (0 + 0) & [\text{distributive property}] \\ & = x \cdot 0 & [\text{definition of } 0] \end{aligned}

    By the definition of addictive inverse, there must exist an addictive inverse y of x \cdot 0 in \mathbb{F} such that

    \begin{aligned} x \cdot 0 + y & = 0 \\ (x \cdot 0 + x \cdot 0) + y & = 0 \\ x \cdot 0 + (x \cdot 0 + y) & = 0 & [\text{associativity}] \\ x \cdot 0 + 0 & = 0 & [\text{addictive inverse}] \\ x \cdot 0 & = 0 & [\text{definition of 0}] \end{aligned}

  • x_{I} = 1_{I} \cdot x

    \begin{aligned} (x + x_{I}) + 1_{I} \cdot x & = 1_{I} \cdot x \\ (x_{I} + x) + 1_{I} \cdot x & = 1_{I} \cdot x \\ x_{I} + (x + 1_{I} \cdot x) & = 1_{I} \cdot x \\ x_{I} + (1 \cdot x + 1_{I} \cdot x) & = 1_{I} \cdot x \\ x_{I} + (1 + 1_{I}) \cdot x & = 1_{I} \cdot x \\ x_{I} + 0 \cdot x & = 1_{I} \cdot x \\ x_{I} + 0 & = 1_{I} \cdot x \\ x_{I} & = 1_{I} \cdot x \end{aligned}

Vector spaces

A vector space, defined over a field \mathbb{F}, is a non-empty set \mathcal{V} (whose members are called vectors), equipped with two operations: vector addition + and scalar multiplication \cdot, obeying the rules (axioms) listed below.

Axioms of vector spaces

For all u, v, and w in the vector space \mathcal{V} (\forall u, v, w \in \mathcal{V}) and for all \alpha and \beta in the field \mathbb{F}, we have

  • Closure under vector addition and scalar multiplication

u + v \in \mathcal{V},

\alpha \cdot v \in \mathcal{V}.

  • Commutativity of vector addition:

    u + v = v + u.

    Note that there is no requirement of the commutativity of scalar multiplication in the definition of the vector space.

  • Associativity of vector addition and scalar multiplication:

    (u + v) + \mathbf{w} = u + (v + \mathbf{w}),

    (\alpha \cdot \beta) \cdot v = \alpha \cdot (\beta \cdot v).

    Note that in the left hand side of the second equation, the first dot is field multiplication while the second one is the scalar multiplication, but the in the right hand side both dots are scalar multiplication.

  • Distributive property of scalar multiplication:

    \alpha \cdot (u + v) = \alpha \cdot u + \alpha \cdot v,

    (\alpha + \beta) \cdot v = \alpha \cdot v + \beta \cdot v.

  • There is an element in \mathcal{V} called “zero” vector 0 \in \mathcal{V} such that

    v + 0 = v,

    and the definition of “one” in the field 1 \in \mathbb{F} is applied in the vector space

    1 \cdot v = v.

    Note that there is no requirement for the existence of “one” vector in the definition of the vector space.

  • For each v \in \mathcal{V}, there is an element in \mathcal{V} called addictive inverse vector v_{I} of v such that

    v + v_{I} = 0.

    Note that there is no requirement for the scalar multiplicative inverse in the definition of the vector space.

Linear combination

Let \mathcal{V} be a vector space over the field \mathbb{F}. Given a set of vectors v_{1}, \dots, v_{n} \in \mathcal{V} and a set of field elements \alpha_{1}, \dots, \alpha_{n} \in \mathbb{F}, the vector u is a linear combination of v_{1}, \dots, v_{n} with \alpha_{1}, \dots, \alpha_{n} as coefficients if

u = \sum_{i=1}^{n} \alpha_{i} v_{i}.

Subspaces

Given a vector space \mathcal{V} over a field \mathbb{F}, a subspace \mathcal{W} of \mathcal{V} is a non-empty subset of \mathcal{V} (\mathcal{W} \subseteq \mathcal{V}) that follows the closure axioms.

For all u and v in the subspace \mathcal{W} (\forall u, v \in \mathcal{W}) and for all \alpha in the field \mathbb{F} (\forall \alpha \in \mathbb{F}), we have

u + v \in \mathcal{W},

\alpha \cdot v \in \mathcal{W}.

Properties of subspace

Let \mathcal{U} and \mathcal{V} be the subspaces of a vector space over \mathbb{F}.

  • 0 \in \mathcal{W}.

    Since the subspace \mathcal{W} cannot be empty, there is at least an element v \in \mathcal{W}.

    Since 1 \in \mathbb{F} and 1_{I} \in \mathbb{F},

    1_{I} \cdot v = v_{I} \in \mathcal{W},

    according to the axiom of the closure under scalar multiplication.

    According to the axiom of the closure under vector addition,

    v + v_{I} = 0 \in \mathcal{W}.

    Thus, “zero” vector must be in \mathcal{W}.

  • \mathcal{W} is also a vector space.

    By virtue of the closure axioms, all axioms of the vector space are obeyed by \mathcal{W}.

    Since all elements in \mathcal{W} are also in the vector space \mathcal{V}, the axioms of

    • closure

    • commutativity

    • associativity

    • distributive property

    • 1 \cdot v = v

    in \mathcal{W} follow directly from those in \mathcal{V}

    The existence of “zero” vector and addictive inverse are proved in the proof above.

  • The subspace

    \mathcal{W} + \mathcal{U} = \left\{ w + u \mid w \in \mathcal{W}, u \in \mathcal{U} \right\}

    is also a subspace.

    Since 0 \in \mathcal{W} and 0 \in \mathcal{U}, then 0 \in \mathcal{W} + \mathcal{U}. Thus, \mathcal{W} + \mathcal{U} is non-empty.

    Suppose a, b \in \mathcal{W} + \mathcal{U}, then there exists elements a_{1} \in \mathcal{W}, a_{2} \in \mathcal{U} such that a = a_{1} + a_{2} and b_{1} \in \mathcal{W}, b_{2} \in \mathcal{U} such that b = b_{1} + b_{2}.

    Because of closure under addition,

    a_{1} + b_{1} \in \mathcal{W},

    a_{2} + b_{2} \in \mathcal{U}.

    Thus, according to the definition of the set addition

    a + b = (a_{1} + b_{1}) + (a_{2} + b_{2}) \in \mathcal{W} + \mathcal{U}.

    Again, suppose x \in \mathcal{W} + \mathcal{U}, then there exists elements x_{1} \in \mathcal{W}, x_{2} \in \mathcal{U} such that x = x_{1} + x_{2}.

    Because of closure under scalar multiplication,

    \alpha \cdot x_{1} \in \mathcal{W} \quad \forall \alpha \in \mathbb{F},

    \alpha \cdot x_{2} \in \mathcal{U} \quad \forall \alpha \in \mathbb{F}.

    Thus, according to the definition of the set addition

    \alpha \cdot x = \alpha \cdot x_{1} + \alpha \cdot x_{2} \in \mathcal{W} + \mathcal{U} \quad \forall \alpha \in \mathbb{F}.

    Thus, \mathcal{W} + \mathcal{U} is a non-empty set that is closed for both vector addition and scalar multiplication and thus is a subspace.

  • The subspace

    \mathcal{W} \cap \mathcal{U} = \left\{ v \mid v \in \mathcal{W}, v \in \mathcal{U} \right\}

    is also a subspace.

    Since 0 \in \mathcal{W} and 0 \in \mathcal{U}, then 0 \in \mathcal{W} \cap \mathcal{U}. Thus, \mathcal{W} \cap \mathcal{U} is non-empty.

    Suppose a, b \in \mathcal{W} \cap \mathcal{U}, then because of the closure axiom of the subspaces

    a + b \in \mathcal{W}, \\ a + b \in \mathcal{U}.

    Thus, by definition,

    a + b \in \mathcal{W} \cap \mathcal{U}.

    Again, suppose x \in \mathcal{W} \cap \mathcal{U}, then because of the closure axiom of the subspaces,

    \alpha \cdot x \in \mathcal{W}, \forall \alpha \in \mathbb{F}, \\ \alpha \cdot x \in \mathcal{U}, \forall \alpha \in \mathbb{F}.

    Thus, by definition, \alpha \cdot x \in \mathcal{W} \cap \mathcal{U}, \forall \alpha \in \mathbb{F}.

    Thus, \mathcal{W} \cap \mathcal{U} is a non-empty set that is closed for both vector addition and scalar multiplication and thus is a subspace.

  • The subspace

    \mathcal{W} \cup \mathcal{U} = \left\{ v \mid v \in \mathcal{W} \mathop{\text{or}} v \in \mathcal{U} \right\}

    is a subspace if and only if \mathcal{W} \subseteq \mathcal{U} or \mathcal{U} \subseteq \mathcal{W}

    First we prove if \mathcal{U} \subseteq \mathcal{W}, then \mathcal{U} \cup \mathcal{W} is a subspace.

    Since \mathcal{U} \subseteq \mathcal{W},

    \mathcal{U} \cup \mathcal{W} = \mathcal{W}

    and thus is a subspace.

    The same argument can be made for \mathcal{W} \subseteq \mathcal{U}.

    Then we prove if \mathcal{U} \cup \mathcal{W} is a subspace, then one subspace is contained in the other by contradiction.

    Suppose \mathcal{U} \cup \mathcal{W} is a subspace, but \mathcal{U} \nsubseteq \mathcal{W} and \mathcal{W} \nsubseteq \mathcal{U}. This means there exists at least two vectors u and w such that

    u \in \mathcal{U}, u \notin \mathcal{W}

    w \in \mathcal{W}, w \notin \mathcal{U}.

    Since u, w \in \mathcal{U} \cup \mathcal{W} and closure under addition property, the vector

    v = u + w

    is also in \mathcal{U} \cup \mathcal{W}, which means it must also be in \mathcal{U} or \mathcal{W}.

    If v \in \mathcal{U}, because there must exist an addictive inverse of u \in \mathcal{U}, then

    \begin{aligned} v & = u + w \\ v + u_{I} & = u + u_{I} + w \\ w & = v + u_{I} \\ \end{aligned}

    which shows that w \in \mathcal{U} and contradicts to our assumption.

    The same contradiction can also be found when v \in \mathcal{W}.

Example: subspaces of 2-dimensional real-value column vectors

The 2-dimensional real-value column vector space \mathbb{R}^{2} has 3 types of subspaces

  1. The subspace of the zero vector only,

    \mathcal{W} = \left\{ 0 \right\}.

  2. The subspace of the vector space itself,

    \mathcal{W} = \mathbb{R}^{2}.

  3. Any “line” that goes through zero vector