Spaces, Vector Spaces, Degree of Polynomial, Fields.
In mathematics, a space is a set with some added structure.
A vector space (or linear space) $V$ over a field $F$ consists of a set where two operations (vector addition and scalar multiplication) are defined and satisfies the following properties:
- $\forall x, y \in V, x + y = y + x$
(commutativity of addition) - $\forall x, y, z \in V, (x + y) + z = x + (y + z)$
(associativity of addition) - $\exists 0 \in V$ s.t. $\forall x \in V, x + 0 = x$
(existence of identity element for addition, i.e null vector) - $\forall x \in V, \exists y \in V $ s.t. $x + y = 0$
(existence of inversion for addition) - $\forall x \in V, 1x=x$
(existence of identity element for multiplication) - $\forall a,b \in F, \forall x \in V, (ab)x=a(bx)$
- $\forall a \in F, \forall x, y\in V, a(x+y)=ax+ay$
- $\forall a,b \in F, \forall x \in V, (a+b)x = ax+ bx$
$x+y$ is called the sum of $x$ and $y$; $ax$ the product of $a$ and $x$.
Elements of the field $F$ are called scalars;
elements of the vector space $V$ are called vectors.
Prove that $\R^n / \R$ is a vector space. Notice that $\R^\infty$ is not included in this case, $\R^\infty$ is the set of sequences.
Proof scratch:
1.$\vec{v}+\vec{u}=\vec{u}+\vec{v}$
2.$(\vec{u}+\vec{v})+\vec{w} = \vec{u}+(\vec{v}+\vec{w}) $
3.$\exists \vec{0},\vec{u}+\vec{0}=\vec{u}$
4.$1\in \R, 1 \cdot \vec{u}=\vec{u}$
5.$a, b \in \R, a(b\vec{u})= (ab)\vec{u}$
6.$a,b \in \R, (a+b)\vec{u}=a\vec{u}+b\vec{u}$
7.$a\in \R, \vec{u},\vec{v}, a(\vec{u} + \vec{v})=a\vec{u} + a\vec{v}$
8.$\vec{u},-\vec{u}, \vec{u} + \vec{v}=\vec{0}$
$\mathbb{P}_n(x)=$set of all polynomials of degree$\leq n$ with real coefficient and in variable $x$.
vector addition:
With $f(x), g(x) \in \mathbb{P}_n(x)$
$f(x) = a_0+ a_1x+ … + a_nx^n,$
$g(x) = b_0 +b_1x+ … + b_nx^n.$
$f(x)+g(x):=(a_0+b_0)+(a_1 +b_1)x+…+(a_n+b_n)x^n$
scalar multiplication:
$c\in F = \R, cf(x) := ca_0 + (ca_1)x + … + (ca_n)x^n$
we claim $\mathbb{P}_n(x)$ is a vector space on $\R$ with respect to these two operations, because all the properties stands with defined vector addition and scalar multiplicaiton.
The highest term’s index with non-zero coefficient.
What is the degree of zero polynomial?
undefined (some book states that the degree of zero polynomial is $-1$)
what can you fill in the boxes below so that the equation stands?
$0=\square x^\square$
$\R^\infty=\set{a_1, a_2, a_3,..|a_i \in \R}=\set{(a_n)_{n\geq1}, a_n\in \R}$
$(a_n) + (b_n) = (a_n+b_n)$
$\lambda \in \R, \lambda (a_n) = (\lambda a_n)$
$\R^\infty/\R$ is a vector space.
$M_{m\times n}(\R)$ = set of all $m \times n$ matrices with real entries.
$A, B \in M_{m \times n}(\R), A=(a_{ij})_{m\times n}, B=(b_{ij})_{m\times n}$
$A+B=(a_{ij}+b_{ij})_{m\times n}$
$\lambda \in \R,\lambda A =(\lambda a_{ij})_{m\times n}$
$M_{m\times n}(\R)/\R$ is a vector space.
Why matrix multiplication works that way?
$\R^2 / \R$
$(a_1, a_2)+(b_1, b_2) = (a_1+b_1, a_2+b_2)$
$\lambda (a_1, a_2) = (\lambda a_1, -a_2)$
$1(a_1, a_2) \neq (a_1, a_2)$ (VS 5 does not stand)
NOT A VECTOR SPACE
$\R^4 / \R$
$(a_1, a_2)+(b_1, b_2) = (a_1+b_1+1, a_2+b_2)$
$\lambda (a_1, a_2) = (\lambda a_1, \lambda a_2)$
$\begin{align*}\lambda ((a_1, a_2) + (b_1, b_2)) &= \lambda(a_1 + b_1 + 1, a_2 + b_2) \br &= (\lambda a_1 + \lambda b_1 + \lambda, \lambda a_1 + \lambda b_1)\end{align*}$ $\lambda(a_1, a_2) + \lambda(b_1, b_2) = (\lambda a_1 + \lambda b_1 + 1, \lambda b_1 + \lambda b_2)$
(VS 7 does not stand)
NOT A VECTOR SPACE
$V=\set{a}, F=\R$
$a+a=a$
$\lambda a = a$
Oui! C’est un vector space!
Same as $V=\set{0}, F=\R$. Isomorphism.
$\P_n(x) = \set{\text{polynomial of degree} \leq n}$ is a vector space over $\R$.
$\P(x) = \set{\text{all polynomials of every possible degree}}$ is a vector space over $\R$.
Infinity is not a valid degree. As we discussed before, degree must be a number, but infinity is not.
$P(x)$ is also a vector space
$x,y,z \in V$, if $x + z = y+z$, then $x= y$.
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The zero vector (or null vector) $0$ described in VS3 is unique.
i.e. if $0, 0'$ both satisfy (VS3) then $0=0'$.
i.e. $\forall x \in V, x + 0=x, x + 0'=x$, then $0=0'$.
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The addtictive inverse (vector) described in VS4 is unique.
The additive inverse of $x$ can be denoted by $-x$.
In any vector space $V$, the following statements are true:
- $\forall x \in V, 0 \in F, 0x = 0.$
- $\forall a \in F, \forall x \in V, (-a) x = -(ax) = a(-x).$
- $\forall a \in F, 0 \in V, a0=0$.
Please be careful what is the symbol ‘0’ is referring to. When $0 \in F$, it is a scalar; when $0 \in V$, it is a vector.
A field $F$ is a set on which two operations addition $+$ and multiplication $\cdot$ are defined so that following conditions are satisfied
- $a+b=b+a, a\cdot b= b\cdot a$
(commutativity of addition and multiplication) - $(a+b)+c=a+(b+c), (a\cdot b)\cdot c = a\cdot (b\cdot c)$
(associativity of addition and multiplication) - $\exists 0 \in F, 1\in F$ s.t. $\forall a, 0+a=a, 1\cdot a = a$
(existence of identity elements for addition and multiplication) - $a\in F, \exists c \in F$ s.t. $a+c=0$
$b \in F, b \neq 0, \exists d \in F$ s.t. $b\cdot d = 1$
(existence of inversion for addition and multipliation) - $a \cdot (b+c) = a\cdot b + a\cdot c$
(distributivity of multiplication over addition)
The elements $0$ and $1$ are called **identity elements** for addition and multiplication, respectively.
The elements elements $c$ and $d$ referred to in (F 4) are called an **additive inverse** for $a$ and a **multiplicative inverse** for $b$, respectively.
$\R$ is a field. $\C$ is a field. $\Q$ is a field.
$\Z$ is NOT a field. (F4)
Every field is a vector space over itself.