$ \vec{v} = c_1 \vec{u}_1 + c_2 \vec{v}_2 + c_3 \vec{u}_3 … c_n \vec{v}_n $
For example
$$ \vec{v} = \left(\begin{array}{r} 2& \\ -2& \\ 1 \end{array}\right), \vec{e}_1 = \left(\begin{array}{r} 0& \\ 2& \\ 0& \end{array}\right), \vec{e}_2 = \left(\begin{array}{r} 1& \\ -2& \\ 1& \end{array}\right), \vec{e}_3 = \left(\begin{array}{r} -1& \\ -1& \\ -2& \end{array}\right), $$
Prove that $\vec{v}$ can be expressed as a linear combination of $e_n$:
$$ \vec{v} = c_1 \left(\begin{array}{r} 0& \\ 2& \\ 0& \end{array}\right) + \ c_2 \left(\begin{array}{r} 1& \\ -2& \\ 1& \end{array}\right) + \ c_3 \left(\begin{array}{r} -1& \\ -1& \\ -2& \end{array}\right) $$
$$ \Leftrightarrow \left(\begin{array}{rrrrrrrrrrr} c_1 \cdot 0 + c_2 \cdot 1 + c_3 \cdot (-1) \\ c_1 \cdot 2 + c_2 \cdot (-2) + c_3 \cdot (-1) \\ c_1 \cdot 0 + c_2 \cdot 1 + c_3 \cdot (-2) \end{array}\right) $$
$$ \Leftrightarrow \left(\begin{array}{rrrrrrrrrrr} 2 =& c_2 - c_3 \\ -2 =& 2c_1 - 2c_2 - c_3 \\ 1 =& c_2 - 2c_3 \end{array}\right) $$
$$ \Leftrightarrow \left(\begin{array}{rrrrrrrrrrr} c_3 =& c_2 - 2 \\ c_1 =& c_2 + \frac{c_3}{2} - 1 \\ c_2 =& 1 + 2c_3 \end{array}\right) $$
$c_3 = c_2 - 2$
$\Leftrightarrow c_3 = (1 + 2c_3) - 2$
$\Leftrightarrow c_3 = 1$
$c_2 = 1 + 2c_3$
$\Leftrightarrow c_2 = 1 + 2(1)$
$\Leftrightarrow c_2 = 3$
$c_1 = c_2 + \frac{c_3}{2} - 1$
$\Leftrightarrow c_1 = (3) + \frac{(1)}{2} - 1$
$\Leftrightarrow c_1 = \frac{(5)}{2} $ \
$ \therefore \vec{v}$ can be expressed as $\frac{5}{2} \vec{e_1} + 3\vec{e_2} + \vec{e_3}$
Mod is basically division but it always results in an integer by resulting with a remainder
$$ a \mod n = r $$ $$ 10 \mod 3 = 1 $$
let $ \mathbb{Z} = \lbrace 0, \ \pm 1, \ \pm 2, \ \pm 3, \ … \ \pm \infty \rbrace $, the set of integers is infinite. Let $ P \in \mathbb{Z} $ be a prime number. $P \in \lbrace 2, 3, 5, 7, 11 \rbrace$. There are infinite prime numbers.
Define a fixed Prime P:
$$ \mathbb{Z}_P = \mathbb{Z} / P\mathbb{Z}$ \text{ := the set of all remainders } \mathbb{Z} \text{ module P}$$
$$ m = p k + r, 0 \leq r \leq p -1 $$
Write $ [r] := \lbrace p k + r \rbrace $ (aka: each [n] contains every number that leaves a remainder of n after being divided by p)
Then $ \mathbb{Z}_p = \lbrace [0], [1], [2], [3], … [p - 1] \rbrace $ Note:$\mathbb{Z}_p$ has p elements
Examples:
p = 2, $ \ \mathbb{Z}_2 = \lbrace [0], [1] \rbrace$
p = 3, $ \ \mathbb{Z}_3 = \lbrace [0], [1], [2] \rbrace$
For $ t \in \mathbb{N}$, define $ \mathbb{Z}_p^t = \mathbb{Z}_p \cdot \mathbb{Z}_p \cdot \mathbb{Z}_p … \mathbb{Z}_p \rbrace $ t times or $ \lbrace (x_1), (x_2), … (x_t) | x_i \in \mathbb{p} \text{ for each } i, 1 \leq i \leq t \rbrace $
let $ \vec{v} = ( v_1, v_2, v_3. … v_n) \in \mathbb{R}^n, \vec{u} = ( u_1, u_2, u_3. … u_n) \in \mathbb{R}^n $
Dot product produces a real number:
$ \vec{u} \cdot \vec{v} = (u_1 \cdot v_1 + u_2 \cdot v_2 + u_3 \cdot v_3 … u_n \cdot v_n ) $
Algebra of Dot Product:
Length: $|| \vec{u} || := \sqrt{\vec{u} \cdot \vec{u}} $
$ ||\vec{u}||$ is the length/magnitude of the vector.
Also: $||\vec{u}|| \leq 0 \text{ unless } \vec{u} = 0$
Also also: $ \vec{u}, \vec{v}$ are orthogonal when $\vec{u}, \vec{v} = 0, \Leftrightarrow \vec{u} \perp \vec{v}$
Angle:
The Angle $\theta$ between $\vec{u}$ and $\vec{v}$ is measured in the smallest distance from $\vec{u}$ to $\vec{v}$ counter-clockwise.
The Angle Theorem: for any two vectors:
$ \vec{u} \cdot \vec{v} = ||\vec{u}|| \cdot ||\vec{v}|| \cos \theta, ||\vec{u}|| \ \neq \ 0, ||\vec{v} || \ \neq \ 0$,
when 0 is between $\vec{u}$ and $\vec{v}$.