% This is lecture no. 1 of MH1201/: Linear Algebra II.

\chapter{Vector Spaces}%
\label{cha:vector_spaces}

\section{Vector Space Axioms}%
\label{sec:vector_space_axioms}

\begin{mdframed}
	We define $\mathsf{V}$ as a non-empty set of vectors and $F$ as a field of scalars like $\mathbb{R},\ \mathbb{C}$.
	Vectors can be added:
	\[
		u + v
		\qquad
		u,\ v\in \mathsf{V}
	.\]
	Vectors can be multiplied by scalars:
	\[
		\lambda u
		\qquad
		u\in \mathsf{V},\ \lambda \in F
	.\]
	\begin{note}
		Vector addition or scalar multiplication in general will \textbf{not} yield another vector in $\mathsf{V}$.
	\end{note}
\end{mdframed}

\bigbreak
\noindent
A vector space follows the following axioms:
\begin{enumerate}
	\item Closure under vector addition:
		\[
		u + v \in \mathsf{V},\quad \forall u, v \in \mathsf{V}
		.\]
	\item Closure under scalar multiplication:
		\[
		\lambda u \in \mathsf{V},\quad \forall u\in \mathsf{V},\lambda\in F
		.\]
	\item Vector addition is commutative:
		\[
		u+v=v+u,\quad \forall u,v\in \mathsf{V}
		.\]
	\item Vector addition is associative:
		\[
			(u+v)+w=u+(v+w),\quad \forall u,v,w\in \mathsf{V}
		.\]
	\item There is a \textbf{unique} zero vector:
		\[
			\exists z\in \mathsf{V} \text{ s.t } v+z=z+v=v,\quad \forall v\in \mathsf{V}
		.\]
	\item There are \textbf{unique} negative vectors:
		\[
			\forall v\in \mathsf{V},\ \exists w \in \mathsf{V} \text{ s.t. } v+w=0
		.\]
	\item Multiplying with the scalar 1 has no effect:
		\[
		1v = v,\ \forall v\in \mathsf{V}
		.\]
	\item Scalar multiplication is associative:
		\[
			(\lambda\mu)v=\lambda(\mu v),\ \forall \lambda,\mu\in F,\forall v\in \mathsf{V}
		.\]
	\item Vector addition is distributive:
		\[
			\lambda(u+v)= \lambda u + \lambda v,\ \forall \lambda\in F,\forall u,v\in \mathsf{V}
		.\]
	\item Addition of scalars is distributive:
		\[
			(\lambda+\mu)v=\lambda v+\mu v,\ \forall \lambda,\mu\in F,\forall v\in \mathsf{V}
		.\]
\end{enumerate}

To prove that $\mathsf{V}$ is a vector space, we must show that all ten axioms are satisfied.
To prove that $\mathsf{V}$ is not a vector space, we must show that at least one condition is not satisfied.

\section{Valid Vector Spaces}%
\label{sec:valid_vector_spaces}
\begin{exmp}
	\label{fn}
	The set of all $n$-tuples with entries from a field $F$ is denoted by $F^{n}$:
	\[
		\{(a_1,a_2,\dotsc,a_n): a_i\in F\}
	.\]
	if $u=(a_1,a_2,\dotsc,a_n)\in F^{n}$, $v=(b_1,b_2,\dotsc,b_n)\in F^{n}$, and $\lambda\in F$ then:
	\[
		u+v=(a_1+b_1,a_2+b_2,\dotsc,a_n+b_n);\quad \lambda u=(\lambda a_1,\lambda a_2,\dotsc,\lambda a_n)
	.\]
\end{exmp}
\begin{exmp}
	$M_{m\times n}(F)$, the set of all $m\times n$ matrices with entries from $F$:
	\[
		\left\{
			\begin{bmatrix}
				a_{11} & a_{12} & \cdots & a_{1n} \\
				a_{21} & a_{22} & \cdots & a_{2n} \\
				\vdots & \vdots & \ddots & \vdots \\
				a_{m1} & a_{m2} & \cdots & a_{mn} \\
			\end{bmatrix}
			: a_{ij}\in F
		\right\}
	.\]
	The rows of the preceding matrix are regarded as vectors in $F^{n}$ and the columns are regarded as vectors in $F^{m}$.
	If $A,B\in M_{m\times n}(F)$ and $\lambda\in F$, then:
	\[
		(A+B)_{ij}=a_{ij}+b_{ij};\quad (\lambda A)_{ij}=\lambda a_{ij},\ 
		(1\leqslant i\leqslant m,\ 1\leqslant j\leqslant n)
	.\]
\end{exmp}
\begin{exmp}
	$\mathcal{F}(S,\mathbb{R})$, the set of all functions from $S$ to $\mathbb{R}$, where $S$ is any non-empty set:
	\[
		\{f:S\mapsto \mathbb{R}\}
	.\]
	If $f,g\in\mathcal{F}(S,\mathbb{R})$ and $\lambda\in\mathbb{R}$, then:
	\[
		(f+g)(s)=f(s)+g(s);
		\quad
		(\lambda f)(s)=\lambda[f(s)],\ \forall s\in S
	.\]
	These operations are common in calculus.
\end{exmp}
\begin{exmp}
	$P(\mathbb{R})$, the set of all \textbf{polynomials} with coefficients from $\mathbb{R}$:
	\[
		\{a_{n}x^{n}+\dotsb+a_1x+a_0:a_{i}\in\mathbb{R}\}
	.\]
	If $f(x)=a_nx^{n}+\dotsb+a_1x+a_0$, $g(x)=b_{m}x^{m}+\dotsb+b_1x+b_0$, $m\leqslant n$, and $\lambda\in\mathbb{R}$, then:
	\begin{align*}
		f(x)+g(x)&=(a_n+b_{n})x^{n}+\dotsb+(a_1+b_1)x+(a_0+b_0)\\
		\lambda f(x)&=\lambda a_nx^{n}+\lambda a_{n-1}x^{n-1}+\dotsb+\lambda a_1x+\lambda a_0
	\end{align*}
\end{exmp}

\section{Invalid Vector Spaces}%
\label{sec:invalid_vector_spaces}

\begin{exmp}
	Let $\mathsf{V}=\{(a_1,a_2):a_1,a_2\in\mathbb{R}\}$, $F=\mathbb{R}$.
	For $(a_1,a_2)$, $(b_1,b_2)\in \mathsf{V}$ and $\lambda\in\mathbb{R}$, define the following:
	\begin{align*}
		(a_1,a_2)+(b_1,b_2)&=(a_1+b_1,a_2-b_2)\\
		\lambda(a_1,a_2)&=(\lambda a_1,\lambda a_2)
	\end{align*}
	Condtions are not satisfied, vector addition is not commutative:
	\begin{align*}
		(a_1,a_2)+(b_1,b_2)&=(a_1+b_1,a_2-b_2)\\
		(b_1,b_2)+(a_1,a_2)&=(a_1+b_1,b_2-a_2)\\
		(a_1,a_2)+(b_1,b_2)&\neq(b_1,b_2)+(a_1,a_2),\ \forall a_2,b_2\in\mathbb{R}
	.\end{align*}
\end{exmp}
\begin{exmp}
	Let $\mathsf{V}=\{(a_1,a_2):a_1,a_2\in\mathbb{R}\}$, $F=\mathbb{R}$.
	For $(a_1,a_2)$, $(b_1,b_2)\in \mathsf{V}$ and $\lambda\in\mathbb{R}$, define the following:
	\begin{align*}
		(a_1,a_2)+(b_1,b_2)&=(a_1+b_1,0)\\
		\lambda(a_1,a_2)&=(\lambda a_1,0)
	\end{align*}
	Conditions are not satisfied, checking if there is a \textbf{unique} zero vector $z=(b_1,b_2)$ such that:
	\[
		u+z=u,\ \forall u=(a_1,a_2)\in \mathsf{V}
	.\]
	Since:
	\[
		\text{LHS}=u+z=(a_1,a_2)+(b_1,b_2)=(a_1+b_1,0)
	.\]
	the given equality becomes $(a_1+b_1,0)=(a_1,a_2)$, $\forall(a_1,a_2)\in \mathsf{V}$, which is equivalent to:
	\[
		z = (b_1,b_2),\ \forall (b_1,b_2)\in \mathsf{V}\text{ s.t. } b_1=0
	.\]
	which is not unique.
\end{exmp}

\section{Elementary Consequences}%
\label{sec:elementary_consequences}

The following are elementary consequences of the definition of a vector space:

\begin{mdframed}
	\noindent
	Cancelation Law for vector addition:
	\[
		\forall u,v,w\in \mathsf{V},\quad u+w=v+w\implies u=v
	.\]
\end{mdframed}
\begin{mdframed}
	Properties of Scalar Multiplication in any Vector Space $\mathsf{V}$:
	\begin{align*}
		0_{f}u&=0_{v},\ \forall u\in \mathsf{V},\ 0_f\in F,\ 0_v\in \mathsf{V}\\
		(-\lambda)u&=-(\lambda u)=\lambda(-u),\ \forall\lambda\in F,\ \forall u\in \mathsf{V}\\
		\lambda 0&=0,\ \forall\lambda\in F,\ 0\in \mathsf{V}
	\end{align*}
\end{mdframed}
