\[\textbf{Proof logarithm}\] \[ \begin{aligned} a^b + 1 &= a^b + a^0 \\ b^a + 1 &= b^a + b^0 \\ a^x a^y &= a^{x + y} \qquad \text{By } \log \text{definition} \\ \log_a (a^x a^y) &= x + y \\ \log_a a^x a^y &= x + y = \log_a a^x + \log_a a^y \\ \Rightarrow \log a^x a^y &= \log a^x + \log a^y \end{aligned} \]

\[\textbf{Ring}\] $\text{Let a, b, c } \in \mathbf{R}$ $\text{There are two binary operations: addition and multiplication. They satisfy}$
\[ \text{Associative Law} \] \[ a \times b \times c = a \times (b \times c) \] \[ \text{Distritutive Law} \] \[a \times (b + c) = a \times b + a \times c \] \[\text{Additive inverse} \] \[\text{For all a in $\mathbf{R}$, there exists b such that}\] \[a + b = 0 \] \[ \text{Multiplicative identity} \] \[ \forall a \in \mathbb{R}, \quad \exists 1 \in \mathbb{R}\text{ such as }\] \[1 \times a = a \times 1\]

\[\textbf{Ideal}\] Let $(\mathbf{I}, +)$ is the subgroup of $(\mathbf{R}, +)$ Let $a \in \mathbf{I}$ and $r \in \mathbf{R}$, then $a \cdot r, r \cdot a \in \mathbf{I}$, $\mathbf{I}$ is called ideal of $\mathbf{R}$
For example:
$2\mathbb{Z}$ is the ideal of $\mathbb{Z}$

Proof:
\[ \text{let } a \in \mathbb{Z}, r = 2k \in 2\mathbb{Z} \quad \text{ where } a, k \in \mathbb{Z} \\ a \cdot r, r \cdot a = 2a \cdot k \in 2\mathbb{Z} \\ \implies 2\mathbb{Z} \text{ is ideal of } \mathbb{Z} \\ \] Following screenshot can represent the ring $\mathbf{R} = (\mathbb{Z}, \mathbb{Z})$ and and ideal $\mathbf{I} = (2\mathbb{Z}, 2\mathbb{Z})$ blue dots

\[\textbf{Left Ideal}\] if $(I, +)$ is subgroup of $(R, +)$ and $\forall r \in R, \forall x \in I$ , $r x \in I$, then $(I, +)$ is called the left ideal of $(R, +)$
Similarly, $(I, +)$ is called right idea of $(R, +)$ if $\forall r \in R, \forall x \in I, xr \in I$

\[\textbf{Principle Ideal}\] The ideal is generated by one element from $R$, $R$ is communtative $\textbf{Ring}$. if $a \in R$ then $Ra \in I$, then $\mathbb{I}$ is principle ideals
$\left< 2 \right>$, $2\mathbb{Z}$ is principle ideal.

Integral domain is commutative \textbf{Ring} $R$, if $a, b \in R$ and $a, b \neq 0$ then $ab \neq 0$

An integral domain is called Euclidean if there exists function $f: R\backslash\{0\} \rightarrow \mathbb{N} \text{ satisfies the two properties:}$
1. $f(a) < f(ab) \text{ for all nonzero } a, b \in R$
2. $\forall a, b \in R \text{ with } b \neq 0 , \text{ there exists } q, r \text{ such that } a = q*b + r \text{ where } f(r) < f(q)$

\[ \textbf{Ring homomorphism} \] Let $\phi$ is a function between two rings $R$, then $\phi$ is a $\mathit{ring}$ homomorphism if
for all $a \in R$ and $b \in R$
\[\phi(a+b) = \phi(a) + \phi(b)\] \[\phi(ab) = \phi(a)\phi(b)\] and \[\phi(1) = 1\] Ideal
Let $R$ be a ring and let $I$ is additive subgroup of $R$, then $I$ is called an ideal of $R$ and write $I \triangleleft R$
if $\forall a \in I$ and $\forall r \in R $, and $ ar \in I$ and $ra \in I$

Example
$R = (\mathbb{N}, +)$ and $I = (2k, +) \quad k \in \mathbb{N}$
\[\text{Let I be a kernal of } \phi, \text{ then I is an ideal of R} \] Let $a \in I$ and $r \in R$, then $\phi(ra) = \phi(r)\phi(a)$
$I$ is kernal of $\phi$
$\Rightarrow \phi(a) = 0 \therefore \phi(ra) = 0, \therefore ra \in I$
Let $\phi: \mathbb{C} \rightarrow \mathbb{C}$ be the map send a complex number to its complex conjugate. Then $\phi$ is an automorphism of $\mathbb{C}$. $\phi$ is its own inverse.

\begin{equation} \begin{aligned} \phi(z) &= \overline{z}\\\\ \phi(z_1 + z_2) &= \overline{z_1 + z_2}\\ \overline{z_1 + z_2} &= \overline{z_1} + \overline{z_2}\\ \phi(z_1 z_2) &= \overline{z_1 z_2}\\ \overline{z_1 z_2} &= \overline{z_1} \cdot \overline{z_2} \nonumber\\ \phi(\phi(z)) &= z
\end{aligned} \end{equation}
Let $\phi: \mathbb{R}[x] \rightarrow \mathbb{R}[x]$ be the map that send $f(x)$ to $f(x+1)$.
Then $\phi$ is an automorphism of \mathbb{R}[x]. The inverse map sends $f(x)$ to $f(x-1)$

\[ \textbf{ Semigroup } \] Semigroup is a set $S$ and a binary operator $\otimes \colon S \times S \rightarrow S$ that satisfies associative property
\[ \forall \text{ a, b, c} \in S \text{ such as } a \otimes b\otimes c = a \otimes (b \otimes c) \]

\[ \textbf{ Monoid } \] A monoid is a triple $(S, \otimes, \overline{1})$
1. $\otimes$ is closed associative binary operator on the set $S$
2. $\overline{1}$ is identity element for $\otimes$
$\forall\quad a, b, c \in S$ such as
\[ a \otimes b \otimes c = a \otimes (b \otimes c) \] \[ a \otimes \overline{1} = \overline{1} \otimes a = a \]

abstract class SemiGroup[A]{
    def add(x: A, y:A):A
}

abstract class Monoid[A] extends SemiGroup[A]{
    def unit: A
}

object MyMonoid extends App{
      implicit object StringMonoid extends Monoid[String]{
          def add(x: String, y:String):String = x concat y
          def unit: String = ""
      }

      implicit object IntMonoid extends Monoid[Int]{
          def add(x: Int, y:Int):Int = x + y
          def unit: Int = 0
      }
      def sum[A](xs: List[A])(implicit m: Monoid[A]): A = {
          if(xs.isEmpty) m.unit 
          else m.add(xs.head, sum(xs.tail))
      }
      println(sum(List(1, 2, 3)))
      println(sum(List("a", "b", "c")))
}

$\text{Let a, b, c} \in \mathbf{G}$
There is binary operation * and satisfy
Closure Law
$ a*b \in \mathbf{G} $
Associative Law
$ a*b*c = a*(b*c)$
Identity
$\exists \mathit{e} \in \mathbf{G} \text{ such that } \mathit{e}*a = a*\mathit{e} \in \mathbf{G}$
Inverse
$ \text{If a } \in \mathbf{G}, \exists a^{-1} \in \mathbf{G} \text{ such that } a*a^{-1} = e $

Given a group $(G, \otimes)$
1. $H$ is the subset of $G$
2. $H$ forms a group under the same binary operation as $G$
$(\mathbb{Z}, +)$
$(\mathbb{2Z}, +)$

Coming soon

Coming soon

Given group $(G, \oplus)$ and $(H, \otimes)$,for all $a, b \in G$
If there is function $\phi : G \rightarrow H$
If $\phi(a \oplus b) = \phi(a) \otimes \phi(b)$, then $\phi$ is group homomorphism from $(G, \oplus)$ to $(H, \otimes)$

Concreate example
Given $G(\mathbb{R}, +)$ and $H(\mathbb{R}, *)$, then $\phi(x) = e^x$ is homomorphism from $G(\mathbb{R}, +)$ to $H(\mathbb{R}, *)$
\begin{align*} \forall &a, b \in \mathbb{R} \\ \phi(a + b) &= e^{a + b} \\ \phi(a)*\phi(b) &= e^{a}*e^{b} = e^{a+b} \\ \Rightarrow \phi(a_1 + b_1) &= \phi(a_2)*\phi(b_2) \\ \Rightarrow \phi(x) &= e^{x} \text{ is homomorphism from } G(\mathbb{R}, +) \text{ and } H(\mathbb{R}, *) \\ \end{align*} Note: $\phi$ will map the identity of $G$ to identity of $H$ automatically, and we can derive it, Why Not?
\begin{align*} &\forall a, b \in \mathbb{R} \\ &\text{Let a = 0, b = 0} \\ &\phi(0 + 0) = e^{0 + 0} = 1 \\ &\phi(0)*\phi(0) = e^{0}*e^{0} = 1*1 = 1 \\ &\Rightarrow \phi(0_G) = 1_H \\ &\Rightarrow \phi \text{ map the identity } 0_G \in G \text{ to the identity} 1_H \in H \quad \square \end{align*}

\[\textbf{Vector Space}\] $\text{Let }\vec{u}, \vec{v}, \vec{w} \in \vec{V} \text{ and scalars } \alpha, \beta \in \mathbb{F}$
Closure
$\vec{u} + \vec{v} \text{ and } \in \vec{V}$
Associative Law
$\vec{u} + \vec{v} + \vec{w} = \vec{u} + (\vec{v} + \vec{w})$
Commutative Law
$\vec{u} + \vec{v} = \vec{v} + \vec{u} $
Identity element of addition
$\exists \vec{0} \in \vec{V} \text{ such that } \forall \vec{u} \in \mathbb{V}$
Inverse element of addition
$\exists -\vec{u} \text{ such that } \vec{u} + (-\vec{u}) = \vec{0}$
Identity element of scalar multiplication
$\exists \mathit{1} \in \mathbb{F} \text{ such that } \mathit{1}\vec{u} = \vec{u}$
Distributivity of scale multiplication with respect to vector addition
$\alpha(\vec{u} + \vec{v}) = \alpha\vec{u} + \alpha\vec{v}$
Distributivity of scale multiplication with respect to field addition
$(\alpha + \beta)\vec{u} = \alpha\vec{u} + \beta\vec{u}$

\[ \textbf{Linear Transformations} \] \begin{aligned} & \mbox{A function } \mathit{T}: \mathbb{R}^n \rightarrow \mathbb{R}^m \mbox{ is called linear transformation, if it satisfies} \\ & \mathit{T} ( \mathbf{u} + \mathbf{v} ) = \mathit{T}(\mathbf{u}) + \mathit{T}(\mathbf{v}) \quad \forall \; \mathbf{u} \,, \mathbf{v} \in \mathbb{R}^n\\ & \mathit{T} ( \lambda \mathbf{u} ) = \lambda \mathit{T}(\mathbf{u}) \quad \mbox{all scalars } \lambda \\ \end{aligned}

\[\textbf{Euclidean Space}\] Let $V \in \mathbb{R}^n$ is vector space and an inner product is a function $ \left< , \right>: V \times V \rightarrow \mathbb{R}$, that is $(x, y) \rightarrow \left< x \,, y \right>$ for $x, y \in V$, which satisfied following axioms: \begin{equation} \begin{aligned} \langle ax \,, y \rangle &= a \langle x \,, y \rangle \\ \langle x \,, by \rangle &= b \langle x \,, y \rangle \\ \langle x + y \,, z \rangle &= \langle x \,, z \rangle + \langle y \,, z \rangle \\ \langle x\,, y + z \rangle &= \langle x \,, y \rangle + \langle x \,, z \rangle \\ \langle y \,, x \rangle &= \langle x \,, y \rangle \\ \langle x \,, x \rangle &> 0 \quad x \neq 0 \quad \text{positive definite}\\ \end{aligned} \end{equation} If the inner product is defined as $ \left< u, v \right> = u^{T}v $ , we have Euclidean Structure $(\mathbb{R}^{n}, u^{T}v)$ or $(\mathbb{R}^{n}, \circ)$

\[\textbf{Euclidean Structure}\] Euclidean Structure is defined by Inner product \[ \langle \vec{u}, \vec{v} \rangle = \sum_{k=1}^{n} u_{k} v_{k}\] Length function is defined by the norm of Inner product \[ \| \langle \vec{u}, \vec{u} \rangle \| = \sqrt{ \sum_{k=1}^{n} u_{k}^2 }\] Distance function is called Euclidean metric. The formula expresses a special case of Pythagorean Theorem. \[ d(u - v) = \| u - v \| = \sqrt{ \sum_{k=1}^{n} (u_{k}-v_{k})^2 }\] The angle between $\vec{u}$ and $\vec{u}$ is given by \[ \beta = \arccos \frac{ \langle \vec{u}, \vec{v} \rangle }{\|\vec{u}\| \|\vec{v}\| } \]

\[\textbf{Affine Space}\] An affine space is a set of points that admits free transitive action of a vector space $\vec{V}$ That is, there is a map $X \times \vec{V} \rightarrow X:(x, \vec{v}) \mapsto x + \vec{v}$,
called translation by a vector $\vec{v}$, such that
1. Addition of vectors corresponds to composition of translation, i.e., for all $x \in X \text{ and } \vec{u}, \vec{v} \in \vec{V}, (x + \vec{u}) + \vec{v} = x + (\vec{u} + \vec{v})$
2. The zero vector $\vec{0}$ acts as the identity vector, i.e., for all $x \in X, x + \vec{0} = x$
3. The action is transitive, i.e., for all $x, y \in X, \text{ exists } \vec{v} \in \vec{V} \text{ such that } y = x + \vec{v}$
4. The dimension of X is the dimension of vector space translations, $\vec{V}$

Or There is unique map
$X \times X \rightarrow \vec{V}:(x, y) \mapsto y - x \text{ such that } y = x + (y - x) \text{ for all }x, y \in X$
It furthermore satifies
1. For all $x, y, z \in X, z - x = (z - y) + (y - x)$
2. For all $x, y, \in X$ and $\vec{u}, \vec{v} \in \vec{V}$, $ (y + \vec{v}) - (x + \vec{u}) = (y - x) + (\vec{v} - \vec{u})$
3. For all $x \in X, x - x = \vec{0}$
4. For all $x, y \in X, y - x = -(x - y)$

\[ \textbf{Affine Space from linear system equation} \] Consider an $(m \times n)$ linear sytem equations
$\sum_{k=1}^{n} a_{i k} x_{k} = c_{i}, (1 \leq i \leq m) \quad\quad\quad \text{(1)}$
where $d = n - rank(M), c_{i} \ne \vec{0} \in \mathbb{R}^{m}$
When the system has at least one solution $x_{p}$ then the full set of solution is a d-dimension affine space
$A \subset \mathbb{R}^{n}$
Since $x_{p} \in A, \text{ we can declare point } x_{p} \text{ as origin of A and then introduct A coordinates as follows:homogenous system}$
$\sum_{k=1}^{n} a_{i k} x_{k} = \vec{0} \quad (1 \leq i \leq m)$
$\Rightarrow dim(\ker(M)) = d \quad \text{(Rank Theorem)}$
$\Rightarrow \text{(1) has d-linear independent solution } \vec{b_{j}} \in \mathbb{R}^{n} \quad\quad (1 \leq j \leq d)$
Affine Space $A$ can be written as
$A = \Big\{ x_{p} + \sum_{j=1}^{d}\alpha_{j}\vec{b_{j}} \quad \mid \quad \alpha_{j} \in \mathbb{R} \quad\quad (1 \leq j \leq d)\Big\} $
$\text{The } \alpha_{j} \text{ can be served as coordinates in A, so that A looks as it were a d-dimension coordiate space.}$
$\text{But note that addition(+) in the space refers to the chosen point } x_{p}, \text{ and not to the origin of the base vector space}$

\[ \textbf{Affine space and linear system} \] The solution set $\mathit{K}$ of any system $\mathbf{A}\mathbf{x}=\mathbf{b}$ of $m$ linear equations in $n$ unknowns is an affine space, namely a coset of $\ker{T_{A}}$ represented by a particular solution $\mathbf{s} \in \mathbb{R}^{n}$ \[ \mathit{K} \in \mathbf{s} + \ker{T_{A}} \] $\mathbf{Proof}$: If $\mathbf{s} \,, \mathbf{w} \in \mathbf{K}$, then $\mathbf{A}(\mathbf{s} - \mathbf{w}) = \mathbf{A}\mathbf{s} - \mathbf{A}\mathbf{w} = \mathbf{b} - \mathbf{b} = \mathbf{0}$ so that $\mathbf{s} - \mathbf{w} \in \ker{T_{A}}$. Now let $\mathbf{k} = \mathbf{s} - \mathbf{w} \in \ker{T_{A}}$. Then \[ \mathbf{w} = \mathbf{s} + \mathbf{k} \in \ker{T_{A}} \] Hence $\mathbf{K} \subseteq \mathbf{s} + \ker{T_{A}}$. To show the conversion inclusion, suppose $\mathbf{w} \in \mathbf{s} + \ker{T_{A}}$. Then $\mathbf{w} = \mathbf{s} + \mathbf{K}$ for some $\mathbf{k} \in \ker{T_{A}}$. But then \[ \mathbf{A}\mathbf{w} = \mathbf{A}(\mathbf{s} + \mathbf{k}) = \mathbf{A}\mathbf{s} + \mathbf{A}\mathbf{k} = \mathbf{b} + \mathbf{0} = \mathbf{b} \] so $\mathbf{w} \in \mathit{K}$, and $\mathbf{s} + \ker{T_{A}} \subseteq \mathit{K}$. Thus, $\mathit{K} = \mathbf{s} + \ker{T_{A}} \quad \square$

\[\textbf{If } \gcd(a, b) = 1 \textbf{ and } a \vert bc\] \[ \textbf{Prove } a \vert c \] $\gcd(a, b) = 1 $
$\Rightarrow \exists m, n \in \mathbf{N} \quad ma+nb = 1$
$\Rightarrow mac + nbc = c$
$\Rightarrow ak = bc \quad k \in \mathbf{N} \because a \vert bc \quad $
$\Rightarrow mac + n(ak)=c \quad (ak=bc) $
$\Rightarrow a(mc + nk) = c$
$\Rightarrow a \vert c $

\[\textbf{Theorem 1}\] The image of transformation is spanned by the image of the any basis of its domain.
For $T:\vec{V} \rightarrow \vec{W}, \text{ if } \beta=\{ \vec{b_1},\vec{b_2},...,\vec{b_n} \} \text{ is a basis of }\vec{V}, \text{ then }T(\beta) = \{ T(\vec{b_1}), T(\vec{b_2}), ... ,T(\vec{b_n})\} \text{ spans the image of }T$

For all $\vec{v} \in \vec{V}, \vec{v} = \alpha_1\vec{b_1} + \alpha_2\vec{b_2} + ... + \alpha_n\vec{b_n}$
$\Rightarrow T(\vec{v}) = T(\alpha_1\vec{b_1} + \alpha_2\vec{b_2} + ... + \alpha_n\vec{b_n})$
$\Rightarrow T(\vec{v}) = \alpha_1 T(\vec{b_1}) + \alpha_2 T(\vec{b_2}) + ... + \alpha_n T(\vec{b_n})$
$\Rightarrow \{ T(\vec{b_1}), T(\vec{b_2}),...,T(\vec{b_n})\} \text{ spans the image of }T$

\[\textbf{Rank Theorem} \] If the domain is finite dimension, then the dimension of domain is the sum of rank and nullity of the transformation
$\text{Let } T:\vec{V} \rightarrow \vec{W} \text{ be a linear transformation },\text{let n be the dimension of }\vec{V},$
$\text{let k be nullity of }T \text{ and let k be the rank of }T$
$\text{Show } n = k + r$
$\text{Let }\beta = \{ \vec{b_1}, \vec{b_2},...,\vec{b_k}\} \text{ be the basis of kernal of }T, \text{ the basis can be extended to } \gamma = \{ \vec{b_1}, \vec{b_2},...,\vec{b_k}, \vec{b_{k+1}},...,\vec{b_n}\}$
$\text{let }\vec{v} \in \vec{V} \Rightarrow \vec{v} = \alpha_1 \vec{b_1} + \alpha_2 + \vec{b_2} +,..., + \alpha_k \vec{b_k} + \alpha_{k+1}\vec{b}_{k+1}+,...,+\alpha_{n}\vec{b_n}$
$\text{Let }T(\vec{v}) = T(\alpha_1 \vec{b_1} + \alpha_2 + \vec{b_2} +,..., + \alpha_k \vec{b_k} + \alpha_{k+1}\vec{b}_{k+1}+,...,+\alpha_{n}\vec{b_n}) = \vec{0}$
$\Rightarrow \vec{v} = \alpha_1 \vec{b_1} + \alpha_2 + \vec{b_2} +,..., + \alpha_k \vec{b_k} + \alpha_{k+1}\vec{b}_{k+1}+,...,+\alpha_{n}\vec{b_n} \in \ker(T) \quad\quad \text{(1)}$
$\because \vec{v} = \sigma_1 \vec{b_1} + \sigma_2 + \vec{b_2} +,..., + \sigma_k \vec{b_k} \in \ker(T) \quad\quad \text{(2)}$
$(1) - (2) \Rightarrow \vec{0} = (\alpha_1-\sigma_1)\vec{b_1} + (\alpha_2 - \sigma_2)\vec{b_2}+,...,+ (\alpha_k - \sigma_k)\vec{b_k}+ \alpha_{k+1}\vec{b}_{k+1}+,...,+\alpha_{n}\vec{b_n} $
$\because \vec{b}_{1}, \vec{b}_{2},...,\vec{b}_{k},\vec{b}_{k+1}, \vec{b}_{k+2},...,\vec{b_n} \text{ are linearly independent}$
$\therefore \alpha_{k+1}, \alpha_{k+2}, ... , \alpha_{n} \text{ are all zero} \quad\quad \text{(3)}$
$T(\vec{v}) = T(\alpha_1 \vec{b_1}) + T(\alpha_2 \vec{b_2}) +,..., + T(\alpha_k \vec{b_k}) + T(\alpha_{k+1}\vec{b}_{k+1})+,...,+T(\alpha_{n}\vec{b_n}) = \vec{0}$
$T(\vec{v}) = \alpha_1 T(\vec{b_1}) + \alpha_2 T(\vec{b_2}) +,..., + \alpha_k T(\vec{b_k}) + \alpha_{k+1}T(\vec{b}_{k+1})+,...,+\alpha_{n}T(\vec{b_n}) = \vec{0}$
$\because \beta = \{ \vec{b_1}, \vec{b_2},...,\vec{b_k}\} \text{ is the basis of kernal of }T$
$\therefore T(\vec{b_1}) = \vec{0},..., T(\vec{b_k}) = \vec{0}$
$\therefore T(\vec{v}) = \alpha_{k+1}T(\vec{b}_{k+1})+,...,+\alpha_{n}T(\vec{b_n}) = \vec{0} \quad\quad \text{(4)}$
$\text{(3) and (4)} \Rightarrow \{ T(\vec{b}_{k+1}), T(\vec{b_{k+2}}), ... , T(\vec{b_{n}}) \} \text{ are linearly independent}$
$\Rightarrow \dim(\vec{V}) = \text{ nullity(T) } + \text{ rank(T) } \text{ or }$
$\Rightarrow \dim(\vec{V}) = \dim(\ker(T)) + \dim(\text{img(T)}) $
$\Rightarrow n = k + r \quad \square$

\[\textbf{Chart}\] A $\textbf{Chart}$ on a set $M$ is a pair $(\phi, U)$ where $U$ is an open subset of $M$ and $\phi: U \rightarrow \phi(U)$ is bijection from $U$ to an open subset $\phi(U)$ in $\mathbb{R}^{m}$ An $\textbf{Atlas}$ is collection of $\mathscr{A} = \{(\phi, U)\}_{\alpha \in A}$ of charts such that the domains $U_{\alpha}$ conver $M$ \[ M = \bigcup_{\alpha \in A} U_{\alpha} \] Example:
Every open subset $U \in M$ has an Altas consisting of a single chart $(\phi, U) = (id_{U}, U)$, where $id_{U}$ denoted identity map of $U$

\[\textbf{Topological Space}\] $\textbf{A topological space}$ is pair $(X, T)$ where $X$ is a set and $U$ is subset of $X$ satifying certain axioms. $T$ is called topology
1. $\emptyset \in T$ and space $X \in T$
2. If $U_1 \in T, U_2 \in T$, then $U_1 \cap U_2 \in T$ $\textbf{ finite}$
3. If $U_i \in T$ then $\bigcup_{i \in I} U_i \in T$ $\textbf{ finite or infinite}$

\[ X = \{1, 2\} \\ T = \{ \emptyset, \{1\}, \{2\}, \{1, 2\}\} \] $(X, T)$ is topological space becase it satisfied three axioms

The elements of $T$ is called open sets,
Property 2 implies any $\mathbf{finite}$ intersection of open sets is open
Property 3 implies union of any open sets is open
Any collection of subsets of $X$ satifies above properties is called $\textbf{topology}$ on $X$
\[ \textbf{Topology} \] $\text{Let }\mathcal{M} \text{ be a set. A topology }\mathcal{Q} \text{ is a subset } \mathcal{Q} \subseteq \mathcal{P}(\mathcal{M})$ Satisfy
$1. \varnothing\subseteq \mathcal{Q}, \mathcal{M} \subseteq \mathcal{Q}$
$2. \mathcal{U} \subseteq \mathcal{Q}, \mathcal{V} \subseteq \mathcal{Q} \implies \mathcal{U} \cap \mathcal{V} \in \mathcal{Q}$
$3. \mathcal{U} \in \mathcal{Q} \implies \bigcup_{\alpha \in \mathcal{A}} \mathcal{U}_\alpha \in \mathcal{Q}$

A function $f:X \rightarrow Y$ between two topological spaces $(X, T_x)$ and $(Y, T_y)$ is called $\textbf{Homeomorphism}$ if it has the following properties:
1. $f$ is bijective
2. $f$ and $f^{-1}$ are both continuous
Take the donus to coffee cup, you can stretch and push around without rips.
Continuous and Invertable funciton from one topological space to other topoligical space. \[\textbf{Smooth}\] Given $f: U \rightarrow V$ $f$ is continuous in $U$ and all derivatives of the $f_i$ of any orders exist

\[\textbf{Diffeomorphic}\] 1. $f$ is Homeomorphic
2. $f$ and $f^{-1}$ both are differentiable in any order.(smooth or $C^{\infty}$)

Sequenently, for every $u \in \tau_x$, there is $v \in \tau_y$ such that $v = f(u)$ and $u = f^{-1}(v)$
Furthere more, since \[ f(u_1 \cap u_2) = f(u_1) \cap f(u_2) \\ f(u_1 \cup u_2) = f(u_1) \cup f(u_2) \] the equavalence extends to the structures of the spaces

Injective
$\text{if } f(x_1) = y_2 \text{ and } f(x_2) = y_2 \text{ and } y_1 = y_2, \text{ then } x_1 = x_2 $

Surjective
$\forall y \in Y \quad \exists x \in X \text{ such as } f(x) = y $

Bijective
$f \text{ is injective and surjective} $

Definition of Voronoi Diagram
For $p, q \in S$ let \[ B(p, q) = \{ x \mid d(p, x) < d(p, x) \} \] be the bisector of $p, q$, $B(p, q)$ is perpendicular to line through the center of line segment $\overline{pq}$
Given a set $S$ of n points in a plane, we wish to associate with each point $s$ a region consisting of all points in the plane closer to $s$ than any other point $s'$ in $S$. The can be described formally as \[ \mathbf{Vor}(\mathbf{s}) = \{p: \textbf{dist}(s, p) \leq \textbf{dist}(s', p), \forall s' \in S \} \] Where $\mathbf{Vor}(\mathbf{s})$ is the Voronoi region for a point $s$

Two points Voronoi Diagram

Three points Voronoi Diagram

Differential, Forwad Differentiation, can be implemented in Haskell \begin{align*} (x + \varepsilon x') + (y + \varepsilon y') &= (x + y) + \varepsilon(x' + y') \\ (x + \varepsilon x')(y + \varepsilon y') &= xy + \varepsilon(x'y + y'x) \\ f(x + \varepsilon x') &= f(x) + \varepsilon f'(x)x' \\ f(g(x + \varepsilon x')) &= f(g(x + \varepsilon x')) + \varepsilon f'(g(x + \varepsilon x')) \\ f(g(x + \varepsilon x')) &= f( g(x) + \varepsilon g'(x) x') \\ f(g(x + \varepsilon x')) &= f(g(x)) + \varepsilon f'(g(x)) g'(x)x' \\ \end{align*}

If $f$ is continuous on closed interval $[a, b]$ and differential on open interval $(a, b)$, then there exists point $c$ such that \begin{align*} f'(c) = \frac{f(b) - f(a)}{b - a} \end{align*}

We all talk about Algebra from high school to University, but we hardly were given any defintion of Algebra.
This is the defintion from Hopf Lecture Notes

An algebra $A$ can be defined as triple $A=(V, *, I)$ where $V$ is Vector Space, $*$ is multiplication and $I$ is identity
Given Vector Space $V$ over the field $K$, e.g. $K = \mathbb{C}$, Algebra satisfies following properties:
\begin{align*} &a, b, c \in V \\ &*: V \times V \rightarrow V \quad \text{Bilinearity} \tag{1}\\ &a * b * c = a * (b * c) \quad \text{Associativity} \\ &1*a = a*1 \quad \text{Unitality} \\ \end{align*}
Example: n dimension matrix over the field $K = \mathbb{C}$
It is easy to show all the matrices form a Vector Space since it satisfied all the Vector Space properties
\begin{align*} &\forall m_1, m_2, m_3 \in V \\ &m_1 + m_2 + m_3 = m_1 + (m_2 + m_3) \\ &m_1 + 0 = 0 + m_1 \\ &m_1 + m_2 = 0 \\ &... \\ &\text{ What is the basics of a Vector Space in two dimensions, e.g.} \\ &A= \begin{bmatrix} 1 & 2 \\ 3 & 4 \\ \end{bmatrix} \\ &\text{ The following is the basics of for two dimensions matrix } \\ &\left\{ e_1= \begin{bmatrix} 1 & 0 \\ 0 & 0 \\ \end{bmatrix}, e_2= \begin{bmatrix} 0 & 1 \\ 0 & 0 \\ \end{bmatrix}, e_3= \begin{bmatrix} 0 & 0 \\ 1 & 0 \\ \end{bmatrix}, e_4= \begin{bmatrix} 0 & 0 \\ 0 & 1 \\ \end{bmatrix} \right\} \\ &\text{Any 2x2 matrix can be written in linear combination of } \left\{e_1, e_2, e_3, e_4 \right\} \\ &A = 1e_1 + 2e_2 + 3e_3 + 4e_4 \\ \end{align*} Once we have a Vector Space, We can define the multiplication$(*)$ from $*: A \times A \rightarrow A$

We know Inner Product is Bilinearity, let's see the meaning of Bilinearity for Inner Product
\begin{align*} &\forall u, v, w \in V, \alpha \in F, \text{ e.g. } F = \mathbb{R} \\ &\left< , \right> : V \times V \rightarrow \mathbb{R} \quad \text{ definition of Inner Product} \\ &\left< u, v \right> = u^{T}v \\ &\left< u + v, w \right> = \left< u, w \right> + \left< v, w \right> \quad \text{ linear on the left} \\ &\left< u, v + w \right> = \left< u, v \right> + \left< u, w \right> \quad \text{ linear on the right} \\ \end{align*} It is very similar to distributive law, e.g.
\begin{align*} a(b + c) &= ab + ac = (b + c)a \quad \text{Multiplication is bilinear}\\ \left< a, b + c \right> &= \left< a, b \right> + \left< a, c \right> \\ \end{align*} The multiplication can be defined as matrix multiplication
\begin{align*} &*: V \times V \rightarrow V \\ &*: M \times M \rightarrow M \quad \text{ where } V = M\\ &*:(M, M) \rightarrow M \quad \text{Prefix notation}\\ &M*M \rightarrow M \quad \text{Infix notation} \\ &\forall A, B, C \in M, \alpha \in \mathbb{F} \\ &\text{Check the Associativity} \\ &A * B * C = A * (B * C) \\ &\text{Check the Bilinearity} \\ &A(B + C) = A B + A C = (B + C)A \\ &\text{Check Unitarity} \\ &I A = A I \quad \text{ where $I$ is identity matrix}\\ & \alpha A B = \alpha (AB) = A (\alpha B) = \end{align*} We did not need to check the multiplication of scalar and M since it is the property of Vector Space, not the Bilinear property

Definition of Bilinear form on a Vector Space $V$ over the field $\mathbb{F}$ is a map \[ H: V \times V \rightarrow \mathbb{F} \\ \] The map satifieds following properties:
\begin{align*} &\forall u, v, w \in V, \alpha \in \mathbb{F} \\ &H(u + v, w) = H(u, w) + H(v, w) \\ &H(u, v + w) = H(u, v) + H(u, w) \\ &H( \alpha u, v) = \alpha H(u, v) \\ &H( u, \alpha v) = \alpha H(u, v) \\ \end{align*} The $H$ is similar like $\times$ multiplication that we have known
\begin{align*} H(w, u + v) &= H(w, u) + H(w, v) \\ \times(w, u + v) &= \times(w, u) + \times(w, v) \\ w \times (u + v) &= w \times u + w \times v \\ \end{align*} What we have familiarized is Inner Product, but Inner Product is more specific
\begin{align*} \left< u, u \right > \geq 0 \Leftrightarrow u = 0 \\ \end{align*} Definition: Let $g$ be a bilinear form on space $V$, and let $\mathcal{\beta} = \{b_1, b_2, ... , b_n \}$ be a basis of $V$. Then the Matrix $G = (g_{i,j}) = g(b_i, b_j)$ is called the matrix of bilinear form $g$ with respect to the basis $\beta$, we will call matrix $G$ a Gram matrix of $g$

let $v, w \in V$ are represented in $\beta$ as \begin{align*} \left[ v \right]_{\beta} &= \left[ \begin{array}{c} \alpha_1 \\ \alpha_2 \\ \vdots \\ \alpha_3 \end{array} \right] \quad \left[ w \right]_{\beta} = \left[ \begin{array}{c} \sigma_1 \\ \sigma_2 \\ \vdots \\ \sigma_3 \end{array} \right] \\ v &= \left[ \alpha_1, \alpha_2, \dots \alpha_n \right] \left[ b_1, b_2, \dots b_n \right] \\ v &= \alpha_1 b_1 + \alpha_2 b_2 + \dots + \alpha_n b_n \tag{1}\\ w &= \sigma_1 b_1 + \sigma_2 b_2 + \dots + \sigma_n b_n \tag{2} \end{align*} then bilinearity By Linearity \begin{align*} g(v, w) &= g(\sum_{i=1}^{n} \alpha_i b_i, \sum_{j=1}^{n} \sigma_j b_j) \\ g(v, w) &= g(\alpha_1 b_1 + \dots + \alpha_n b_n, \sum_{j=1}^{n} \sigma_j b_j) \\ g(v, w) &= g(\alpha_1 b_1, \sum_{j=1}^{n} \sigma_j b_j) + \dots + g(\alpha_n b_n, \sum_{j=1}^{n} \sigma_j b_j)\\ g(v, w) &= \sum_{i=1}^{n} g(\alpha_i b_i, \sum_{j=1}^{n} \sigma_j b_j)\\ g(v, w) &= \sum_{i=1}^{n} g(\alpha_i b_i, \sigma_1 b_1 + \dots + \sigma_n b_n)\\ g(v, w) &= \sum_{i=1}^{n} g(\alpha_i b_i, \sigma_1 b_1) + \dots + \sum_{i=1}^{n} g(\alpha_i b_i, \sigma_n b_n)\\ g(v, w) &= \sum_{i=1}^{n} \sum_{j=1}^{n} g(\alpha_i b_i, \sigma_j b_j) \\ g(v, w) &= \sum_{i=1}^{n} \sum_{j=1}^{n} \alpha_i \sigma_j g(b_i, b_j) \\ \text{From} (1) (2) \\ g(v, w) &= \left[ v \right]_{\beta}^{T} G \left[ w \right]_{\beta} \\ \end{align*} Concrete example:
\begin{align*} G &= \begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix} \\ g(v, w) &= \left[ x_1, y_1 \right] G \left[ \begin{array}{c} x_2 \\ y_2 \end{array} \right] \\ g(v, w) &= \left< v, w \right> = x_1 x_2 + y_1 y_2 \\ \\ G &= \begin{bmatrix} 0 & 1\\ -1 & 0 \end{bmatrix} \\ g(v, w) &= \det \left( \begin{bmatrix} x_1 & x_2 \\ y_1 & y_2 \end{bmatrix} \right) = x_1 y_2 - x_2 y_1 \\ \end{align*} A bilinear form is call symmetric form if ...
A bilinear form is call skew symmetric form if ...
matrix bilinear form ...
quadratic form
inner product
cross product
determinant function
http://localhost/pdf/bilinearforms1.pdf

\[ \textbf{Inner Product} \] \[ \text{Positivity} \] \[ \langle\mathbf{v}, \mathbf{v}\rangle \geq 0 \] \[ \langle \mathbf{v} , \mathbf{v} \rangle = \mathbf{0} \iff \mathbf{v} = \mathbf{0}\] \[ \text{Linearity in the first component} \] \[ \langle c_{1}\mathbf{v_1} \,, \mathbf{v_2}\rangle = c_{1}\langle \mathbf{v_1}, \mathbf{v_2}\rangle \] \[ \langle \mathbf{v_1} + \mathbf{v_2} \,, \mathbf{v_3} \rangle = \langle \mathbf{v_1} \,, \mathbf{v_3}\rangle + \langle \mathbf{v_2} \,, \mathbf{v_3}\rangle\] \[ \langle c_{1}\mathbf{v_1} + c_{2}\mathbf{v_2}, \mathbf{v_3}\rangle = c_{1}\langle \mathbf{v_1}, \mathbf{v_3}\rangle + c_{2}\langle\mathbf{v_2}, \mathbf{v_3} \rangle \] \[ \text{Conjugate Symmetic}\] \[ \langle \mathbf{v_1}, \mathbf{v_2} \rangle = \overline{\langle \mathbf{v_2}, \mathbf{v_1} \rangle}\] \[ \text{Properties of Inner product}\] \[ \langle \mathbf{v_1} \,, \lambda \mathbf{v_2} \rangle = \overline{\langle \lambda\mathbf{v_2} \,, \mathbf{v_1} \rangle} = \overline{\lambda} \overline{\langle \mathbf{v_2} \,, \mathbf{v_1} \rangle} = \overline{\lambda} \langle \mathbf{v_1} \,, \mathbf{v_2} \rangle \]
\[ \textbf{Outer Product} \] The outer product $\vec{u} \times \vec{v}$ is equivalent to $u v^{T}$, for instance \[ u \otimes v = u v^{T} = \left[ \begin{array}{c} u_1 \\ u_2 \\ u_3 \end{array} \right] \left[ \begin{array}{cccc} v_1 & v_2 & v_3 & v_4 \end{array} \right] = \begin{bmatrix} u_0 v_1 & u_0 v_2 & u_0 v_3 & u_0 v_4 \\ u_2 v_1 & u_2 v_2 & u_2 v_3 & u_2 v_4 \\ u_3 v_1 & u_3 v_2 & u_3 v_3 & u_3 v_4 \\ \end{bmatrix} \] \[ \textbf{Matrix Multiplication definited as Outer Product} \] \[ \begin{bmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \\ \end{bmatrix} \begin{bmatrix} b_{11} & b_{12} \\ b_{21} & b_{22} \\ \end{bmatrix} = \begin{bmatrix} a_{11} \\ a_{21} \end{bmatrix} \otimes \begin{bmatrix} b_{11} \\ b_{12} \end{bmatrix} + \begin{bmatrix} a_{12} \\ a_{22} \end{bmatrix} \otimes \begin{bmatrix} b_{21} \\ b_{22} \end{bmatrix} = \begin{bmatrix} a_{11} \\ a_{21} \end{bmatrix} \begin{bmatrix} b_{11} & b_{12} \end{bmatrix} + \begin{bmatrix} a_{12} \\ a_{22} \end{bmatrix} \begin{bmatrix} b_{21} & b_{22} \end{bmatrix} \]

An binary relation $\leq$ defined on Set $A$ is called $\textbf{partial order}$ on set $A$ if following conditions are hold identifically on set $A$ \begin{align*} &a \leq a \\ &a \leq b \text{ and } b \leq a \Rightarrow a = b \\ &a \lt b \text{ and } b \lt c \Rightarrow a \lt c \\ &\text{If in additional, for every } a, b \in A \\ &a \leq b \text{ or } b \leq a \\ &\text{ then } \leq \text{ is total order on } A \\ \end{align*}

\begin{equation} \begin{aligned} \left< u, v \right> &= \|\vec{u}\| \| \vec{v}\| \cos{\phi} \\ \cos \phi &= \frac{\left < u , v \right >}{\|u\|\|v\| } \\ \vec{u} \text{ project on } \vec{v} \\ \|u\| \cos \phi \frac{v}{\|v\|} &= \|u\| \frac{ \left < u, v \right>}{\|u\|\|v\|} \frac{v}{\|v\|}\\ \|u\| \frac{v}{\|v\|} \cos \phi &= \frac{ \left < u, v \right>}{ \left < v, v \right> } v \\ proj_v &= \frac{\left < u, v \right>}{ \left < v, v \right> } v = v \frac{\left< v, u \right>}{\left< v, v \right>} = \frac{v (v^{T} u)}{v^T v} = \frac{(v v^{T}) u}{v^{T}v} \\ \\ \implies proj_v &= \left < u, v \right> v \quad \text{ where } \|v\| = 1\\ \implies proj_v &= v v^{T} u \quad \text{ where } \|v\| = 1\\ v v^{T} &\text{ is projection matrix from u onto v} \text{ where } \|v\| = 1\\ \end{aligned} \end{equation}

The above theorem is so obvious.

1. A straigh line segment can be drawn joining any two points
2. A straigh line segment can be extended indefinitely to a straigh line
3. Given any straigh line segment, a straigh line segment can be drawn having the segment as radius and one endpoint as center

Given vectors $u = (b_1, b_2, b_3), v = (c_1, c_2, c_3)$, the cross product of $u, v$ is the following:
\begin{equation} \begin{aligned} u \times v &= w = \begin{bmatrix} i & j & k \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3 \end{bmatrix} \\ \det{A}&=i (-1)^{1 + 1} \begin{vmatrix} b_2 & b_3 \\ c_2 & c_3 \end{vmatrix} + j (-1)^{2 + 1} \begin{vmatrix} a_2 & a_3 \\ c_2 & c_3 \end{vmatrix} + k (-1)^{3 + 1} \begin{vmatrix} a_2 & a_3 \\ b_2 & b_3 \\ \end{vmatrix} \end{aligned} \end{equation} The direction of cross product can be determinated by the Right Hand Rule
The magnitude can be computed as following:
\[ u \times v = \|u\| \|v\| \sin{\alpha} \vec{n} \quad \text{ where } \vec{n} \text{ is the unit vector that is perpendicular to the plane which contains } \vec{u} \vec{v} \]
The magnitude of cross product is just the volumn of parallelepiped $\vec{u}, \vec{v}, \vec{w}$

Draw line parallel to $\overline{P_1 P_2}$ and pass $O$ \[ P_1 \oplus P_2 = P_3 \]

\[ \begin{aligned} r^2 = \overline{OP'} \oplus \overline{OP} \\ \frac{r}{\overline{OP'}} = \frac{\overline{OP}}{r} \end{aligned} \] if $\overline{OP'} = 0$, then it maps to $\infty$

If $f(x_1) = f(x_2) \in Y$ then $x_1 = x_2 \in X$ $ \Rightarrow f(x)$ is injective
Injective means there is unique value $x \in X$ for each $y \in Y$

If $\forall y \in Y \quad \exists x \in X $ such as $f(x) = y \Rightarrow f(x)$ is subjective.
Subjective means each $y \in Y$ there always exists $x \in X$ such that $f(x) = y$

Function $f : X \rightarrow Y$ is continuous if followings are True
$\exists \, \epsilon > 0 \in X$ such that $\exists \, \delta > 0 \in Y$ and $|\, f(x + \epsilon)\,| < \delta$