Vector spaces¶

Field¶

Field

A Field \(F\) is a commutative division ring with \(0\neq 1\). that is, it is a commutative ring and every elements except \(0\) in \(F\) has a multiplicative inverse in \(F\).

What is a field?

Is \(\mathbb{Z}\), the set of all integers, a field?
Is \(\mathbb{Z}/2\mathbb{Z}\) a field? Find a sufficient condition for \(\mathbb{Z}/m\mathbb{Z}\) to be a field, where \(m\in\mathbb{N},m\ge2\).

Ans

No. \(2\) has no multiplicative inverse in \(\mathbb{Z}\).
Yes. If \(m\) is a prime, then it is a field by the Lagrange theorem of subgroups.

Characteristic of a ring

Let \(R\) be a ring. The characteristic of \(R\), denoted by \(\text{char}(R)\), is the number

\[ \text{char}(R)=\min\left\{n\in\mathbb{Z}_{\ge0}\,\middle|\,\underbrace{1+1+\dots+1}_{n}=0\right\}. \]

Vector space¶

Vector space

Let \((V,+)\) be an abelian group and \(F\) be a field.

A vector space is \((V,+,\cdot)\) where \(\cdot:F\times V\to V\) is a group action that satisfies

\[ \begin{cases} c_1(v_1+v_2)&=c_1v_1+c_1v_2\\ (c_1+c_2)v_1&=c_1v_1+c_2v_1 \end{cases} \]

for all \(c_1,c_2\in F,v_1,v_2\in V\).

Module (generalized vector spaces)

Reqires same conditions except that \(F\) can be a ring.

Examples¶

Example

Assume that the following uses component wise addition and component wise scalar multiplication.

\(F^n\) is a vector space over \(F\) where \(F\) is a field.
\(\mathbb{R}^n\) is a vector space over \(\mathbb{Q}\), and it is infinite dimensional.
\(\mathbb{C}^n\) is a vector space over \(\mathbb{R}\) and its dimension is \(2n\).
\(\mathbb{Q}\) is not a vector space over \(\mathbb{R}\).

Polynomials

Let \(F\) be a field. The polynomials in \(x\) with coefficients in \(F\) is the set

\[ F[x]:=\left\{a_nx^n+a_{n-1}x^{n-1}+\dots+a_1x+a_0\,\middle|\,a_i\in F,n\in\mathbb{Z}_{\ge0}\right\}. \]

The degree of a polynomial \(f(x)\in F[x]\), denoted by \(\deg(f)\), is the largest integer \(n\) such that \(a_n\neq0\).

Define addition between \(\displaystyle f(x)=\sum_{i=0}^n a_ix^i,g(x)=\sum_{i=0}^m b_ix^i\in F[x]\), where \(n=\deg(f)\ge m=\deg(g)\), as

\[ f(x)+g(x)=a_nx^n+a_{n-1}x^{n-1}+\dots+(a_m+b_m)x^m+(a_{m-1}+b_{m-1})x^{m-1}+\dots+(a_0+b_0). \]

Define scalar multiplication between \(c\in F,f(x)\in F[x]\) by

\[ c\cdot f(x)=\sum_{i=0}^{n}(ca_n)x^n. \]

Then \(F[x]\) is a vector space over \(F\). In linear algebra for undergraduates, the textbooks often write this as \(\mathsf{P}(F)\).

Additionally, given \(k\in\mathbb{Z}_{\ge0}\), \(\mathsf{P}_k(F)\) is the set that contains all polynomials with degree less or equal to \(k\), and inherits the addition and scalar multiplication from \(\mathsf{P}(F)\), which becomes a vector space over \(F\), and is a subspace of \(\mathsf{P}(F)\).

Sequences

Let \(F\) be a field. Let \(V:=\{\{a_n\}_{n=1}^{\infty}:a_n\in F\}\), the set of all sequences in \(F\).

Define addition \(\{a_n\}_{n=1}^{\infty}+\{b_n\}_{n=1}^{\infty}:=\{a_n+b_n\}_{n=1}^{\infty}\).

Define scalar multiplication \(c\{a_n\}_{n=1}^{\infty}:=\{ca_n\}_{n=1}^{\infty}\).

Then \(V\) is a vector space over \(F\).

Usage

Let \(\{f_n\}_{n=0}^{\infty}\) be the fibonacci sequence, i.e., \(f_0=0,f_1=1\) and \(f_{n}=f_{n-1}+f_{n-2}\) for \(n\ge 2\).

Let \(W\subset V\) be defined by

\[ W:=\{\{w_n\}\,\mid\,w_n=w_{n-1}+w_{n-2},w_0,w_1\in\mathbb{R}\} \]

Then it is a subspace of \(V\). \(\dim(W)=2\) because it is isomorphic to \(\mathbb{R}^2\) by sending \(w_0\) to \(a_1\), \(w_1\) to \(a_2\) for all \(\{w_n\}\in W,(a_1,a_2)\in\mathbb{R}^2\), for example.

Therefore, the linearly independent set \(\left\{\{\varphi^n\}_{n=0}^{\infty},\left\{\left(\dfrac{1}{\varphi}\right)^n\right\}_{n=0}^{\infty}\right\}\subset W\) spans \(W\), where \(\varphi=\frac{1+\sqrt{5}}{2}\).

Hence, we can find unique \(\alpha,\beta\in\mathbb{R}\), such that

\[ \alpha\varphi^n+\beta\frac{1}{\varphi^n}=f_n \]

for all \(n\in\mathbb{N}\). Specifically, we can solve the equations

\[ \begin{cases} \alpha+\beta&=f_0\\ \alpha\varphi+\frac{\beta}{\varphi}&=f_1 \end{cases} \]

and we obtain the Binet's formula:

\[ f_n=\frac{1}{\sqrt{5}}\left(\varphi^{n}-\frac{1}{\varphi^n}\right). \]

Note

The sequence space is useful for solving homogeneous recurrence sequences, in fact, the closed form formula of recurrence relations is related to the Jordan normal form of \(A\) when solving for

\[ \begin{pmatrix}a_{n+k}\\a_{n+k-1}\\\vdots\\a_{n}\end{pmatrix}=A^n\begin{pmatrix}a_{k}\\a_{k-1}\\\vdots\\a_{1}\end{pmatrix}. \]

This is why they are in a linear combination of geometric series, and the ratio may be complex numbers (irrational numbers) even if the original sequence is real (positive integers).

Functions

Let \(S\) be a nonempty set and \(F\) be a field. Let

\[ \mathcal{F}(S,F^n):=\left\{f\mid f:S\to F\text{ is a function}\right\} \]

Define addition between \(f,g\in\mathcal{F}(S,F)\) by \((f+g)(x)=f(x)+g(x),\forall x\in S\).

Define scalar multiplication for \(c\in F, f\in\mathcal{F}(S,F)\) by \((c\cdot f)(x)=cf(x),\forall x\in S\).

Then \(\mathcal{F}(S,F)\) is a vector space over \(F\).

Note that the codomain of the function space must be a set with addition and scalar multiplication defined.

Continuous functions

Let \(A\subset F^m\) and let \(\mathsf{C}(A,F^n)\) be the subset of \(\mathcal{F}(A,F^n)\) that contains all continuous functions from \(A\) to \(F^n\).

Then it is a vector space over \(F\).

Differentiable functions

Analytic functions

Riemann Integrable functions