# Vector Space

DEFINITION1: Let $K$ be a set with at least two distinct elements, $+:K\times{K}\rightarrow{K}$ and $\cdot:K\times{K}\rightarrow{K}$ be two functions. If the following conditions hold, then $\left( K,+,\cdot \right)$ is called a field:

F1) $\left( a+b \right) +c=a+ \left( b+c \right)$ for all $a,b,c\in{K}$,

F2) $a+b=b+a$ for all $a,b\in{K}$,

F3) there exists ${0}\in{K}$ such that $a+0=a$ for all $a\in{K}$,

F4) for each $a\in{K}$, there exists $b\in{K}$ such that $a+b=0$,

F5) $\left( ab \right) c=a \left( bc \right)$ for all $a,b,c\in{K}$,

F6) $ab=ba$ for all $a,b\in{K}$,

F7) $a \left( b+c \right)=ab+ac$ for all $a,b,c\in{K}$,

F8) there exists ${1}\in{K}$ such that $1a=a$ for all $a\in{K}$,

F9) for all $a\in K \setminus \left\{0\right\}$, there exists $b\in{K}$ such that $ab=1$.

EXAMPLE1: Each of the sets $\mathbb{Q}$ Rational Numbers, $\mathbb{R}$ Real Numbers and $\mathbb{C}$ Complex Numbers is a field with usual operations of addition and multiplication.

EXAMPLE2: Let $p$ be a prime number, $k\in{\mathbb{Z}}$$k+p\mathbb{Z}=\left\{k+px : x\in{\mathbb{Z}} \right\}$ and $\mathbb{Z}_{p}=\left\{ k+p\mathbb{Z} : k\in{\mathbb{Z}} \right\}$. For $k,l\in \mathbb{Z}$, we define $\left( k+p\mathbb{Z} \right) +_{p} \left( l+p\mathbb{Z} \right) := \left( k+l \right) +p\mathbb{Z}$ and $\left( k+p\mathbb{Z} \right) \cdot_{p} \left( l+p\mathbb{Z} \right) :=kl+p\mathbb{Z}$. Then, $\left( \mathbb{Z}_{p},+_{p},\cdot_{p} \right)$ is a field with $p$ elements.

EXAMPLE3: The set of Integers $\mathbb{Z}$ satisfies the conditions F1 to F8. However, $2\in\mathbb{Z}$ but there is not an integer satisfying the equality $2k=1$. So, it does not satisfy F9 i.e., it is not a field.

DEFINITION2: Let $K$ be a field, $X$ be a set, $+:X\times{X}\rightarrow{X}$ and $\cdot:K\times{X}\rightarrow{X}$ be two functions. If the following conditions hold, then $\left( X, K,+,\cdot \right)$ is called a vector space or linear space:

L1) $\left( x+y \right) +z=x+ \left( y+z \right)$ for all $x,y,z\in{X}$,

L2) $x+y=y+x$ for all $x,y\in{X}$,

L3) there exists $\theta\in{X}$ such that $x+\theta=x$ for all $x\in{X}$,

L4) for each $x\in{X}$, there exists $y\in{X}$ such that $x+y=\theta$,

L5) $a \left( x+y \right) =ax+ay$ for all $a\in{K}$ and $x,y\in{X}$,

L6) $\left( a+b \right) x=ax+bx$ for all $a,b\in{K}$ and $x\in{X}$,

L7) $a \left( bx \right) = \left( ab \right) x$ for all $a,b\in{K}$ and $x\in{X}$,

L8) $1x=x$ for all $x\in{X}$.

Instead of the expression "$\left( X, K,+,\cdot \right)$ is a vector space", we can say "$X$ is a $K$-vector space". Each element of $X$ is called a vector and each element of $K$ is called a scalar. If $K=\mathbb{R}$, then $X$ is called a real vector space, and if $K=\mathbb{C}$, then $X$ is called a complex vector space. The element $\theta$ in the condition L3 is called zero vector. Zero vector is denoted by $\theta$ and the zero of the field $K$ is denoted by $0$. A vector space is never empty because $\theta\in X$. In the condition L4, the vector $y\in X$ satisfying the condition $x+y=\theta$ for each vector $x\in X$ is called the additive inverse of $x$.

EXAMPLE4: $\mathbb{R}$ is a $\mathbb{R}$-vector space. Indeed, every field is a vector space over itself.

EXAMPLE5: We define $\mathbb{R}^{n}=\left\{ \left( x_{1},x_{2},\dots,x_{n} \right)\,\vert\,x_{i}\in\mathbb{R},\,i=\overline{1,n} \right\}$, where $n\in \mathbb{N}$. Vector addition and scalar multiplication are defined as follows:

$+:\mathbb{R}^{n}\times{\mathbb{R}^{n}}\rightarrow{\mathbb{R}^{n}}$,

$\left( x_{1},x_{2},\dots,x_{n} \right) + \left( y_{1},y_{2},\dots,y_{n} \right) := \left( x_{1}+y_{1},x_{2}+y_{2},\dots,x_{n}+y_{n} \right)$

for all $\left( x_{1},x_{2},\dots,x_{n} \right) , \left( y_{1},y_{2},\dots,y_{n} \right) \in{\mathbb{R}^{n}}$

and

$\cdot:\mathbb{R}\times{\mathbb{R}^{n}}\rightarrow{\mathbb{R}^{n}}$,

$\lambda{ \left( x_{1},x_{2},\dots,x_{n} \right) }:= \left( \lambda{x_{1}},\lambda{x_{2}},\dots,\lambda{x_{n}} \right)$

for all $\lambda\in\mathbb{R}$ and $\left( x_{1},x_{2},\dots,x_{n} \right) \in{\mathbb{R}^{n}}$.

Then, $\mathbb{R}^{n}$ is an $\mathbb{R}$-vector space. In general, $K^{n}=\left\{ \left( x_{1},x_{2},\dots,x_{n} \right)\,\vert\,x_{i}\in R,\,i=\overline{1,n} \right\}$ is a $K$-vector space when $K$ is a field.

EXAMPLE6: If $K,L$ are two fields, $K\subset L$, and $n\in \mathbb{N}$, then $L^{n}$ is a $K$-vector space. Accordingly, $\mathbb{C}^{n}$ is an $\mathbb{R}$-vector space, $\mathbb{C}^{n}$ is a $\mathbb{Q}$-vector space and $\mathbb{R}^{n}$ is a $\mathbb{Q}$-vector space.

EXAMPLE7: If $K,L$ are two fields and $K\subsetneqq L$, then $K$ is not an $L$-vector space.

SOLUTION: In order that $K$ is an $L$-vector space, the scalar product ($\cdot$) must be defined from $L\times K$ to $K$. We will show that this fails. $K\subsetneqq L$ implies that there exists $\alpha\in L$ such that $\alpha\notin K$. Since $1\in K$, then $\left( \alpha,1 \right) \in L\times K$. By $\cdot:L\times K\rightarrow K$, we get that $\alpha.1$ must be in the field $K$, i.e., $\alpha\in K$. This is a contradiction. Consequently, $K$ is not an $L$-vector space.

By Example7, $\mathbb{Q}$ is not an $\mathbb{R}$-vector space and $\mathbb{R}$ is not a $\mathbb{C}$-vector space.

EXAMPLE8: Let $T$ be a set, $K$ be a field and

$X=\left\{ f \,\vert\,f:T\rightarrow K \text{ is a function}\right\}$.

Vector addition and scalar multiplication are defined as follows:

For all $f,g\in X$$\left( f+g \right) \left( t \right) :=f \left( t \right) +g \left( t \right)$ for each $t\in T$.

For all $f\in X$ and $\lambda\in K$$\left( \lambda f \right) \left( t \right) :=\lambda f \left( t \right)$ for each $t\in T$.

Then, $X$ is a $K$-vector space and denoted by $X=K^{T}$.

Consider $K=\mathbb{R}$ and $T=\{1,2,\dots,n\}$, then we have $X=\mathbb{R}^{n}$.

Similarly, if $T=\mathbb{N}$ and $K=\mathbb{R}$, we have $X=\mathbb{R}^{\mathbb{N}}$ i.e., $X$ is the space of all real sequences and denoted by $S$. We can write $S= \left\{ \left( x_{n} \right) \,\vert\,x_{n}\in{\mathbb{R}}\, ,n\in{\mathbb{N}} \right\}$. If we take $K=\mathbb{C}$ instead of $\mathbb{R}$, we obtain $S=\mathbb{C}^{\mathbb{N}}= \left\{ \left( x_{n} \right) \,\vert\,x_{n}\in{\mathbb{C}}\, ,n\in{\mathbb{N}} \right\}$.

EXAMPLE9: Let $K$ be a field and $K\left[ x \right]$ be the set of all polynomials with coefficients in $K$ i.e.,

$K\left[ x \right] = \left\{ f\left( x \right) =a_{n}x^{n}+a_{n-1}x^{n-1}+\cdots+a_{1}x+a_{0}\,\vert\, a_{i}\in{K},\, i=\overline{0,n},\, n\in \mathbb{N} \right\}$.

Vector addition and scalar multiplication are defined as follows:

Let $\lambda\in K$ and the polynomials $f\left( x \right)=a_{n}x^{n}+a_{n-1}x^{n-1}+\cdots+a_{1}x+a_{0}$ and $g\left( x \right)=b_{m}x^{m}+b_{m-1}x^{m-1}+\cdots+b_{1}x+b_{0}$ be in $K\left[ x \right]$. Assume that the relations $s=\max\left\{ n,m \right\}$, $i>n\Rightarrow{a_{i}=0}$ and $i>m\Rightarrow{b_{i}=0}$ hold.

$\left( f+g \right) \left( x \right) := \left( a_{s}+b_{s} \right) x^{s}+ \left( a_{s-1}+b_{s-1} \right) x^{s-1}+\cdots+ \left( a_{1}+b_{1} \right) x+ \left( a_{0}+b_{0} \right)$.

$\left( \lambda{f} \right) \left( x \right) :=\lambda{a_{n}}x^{n}+\lambda{a_{n-1}}x^{n-1}+\cdots+\lambda{a_{1}}x+\lambda{a_{0}}$.

Then, $K\left[ x \right]$ is a $K$-vector space.

PROPOSITION1: Let $K$ be a field, $X$ be a $K$-vector space. Then,

a) The zero element $\theta\in X$ is unique.

b) The inverse element of each vector $x\in X$ is unique (the unique element satisfying the equality $x+y=\theta$ for fixed vector $x$ is denoted by $y=-x$).

c) $0x=\theta$ for each $x\in X$.

d) $\lambda \theta=\theta$ for each $\lambda\in K$.

e) $\left( -1 \right) x=-x$ for each $x\in X$.

f) $\lambda{x}=\theta\Leftrightarrow{\lambda=0\lor{x=\theta}}$.

PROOF:

DEFINITION3: Let $K$ be a field, $X$ be a $K$-vector space, and $x,y\in X$. The difference of the vectors $x$ and $y$ is denoted by $x-y$ and defined as

$x-y:=x+\left( -y \right)$,

where $-y$ is the additive inverse of the vector $y$.