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:


\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}}



\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}}.


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.

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