\input ../common.inc
\begin{document}
My laptop has 6\% power, so i'm writing this from TTY and it doesn't have russian keyboard

we're gonna be using:\\
$\wrapa{\alpha f, x} = \alpha\wrapa{f, x} = \wrapa{f, \alpha x},\quad f \in \c L^*$

Let $\ss{e_1, \dots, e_n}$ eb a basis in $\c L \implies \all x \in \c L\quad x^je_j;\\
\wrapa{e^j, x} := x^j,\;j \in \overline{1, n}.$

We must prove: $\wrapa{e^j, \alpha^1x_1 + \alpha^2x_2} = \alpha_1\wrapa{e^j, x_1} + \alpha_2\wrapa{e^j, x_2}\\
x_1 = \alpha^1x_1 + \alpha^2x_2 = \wrapb{\alpha^1x_1 + \alpha^2x_2}^je_j = \alpha^1x_1^je_j + \alpha^1x_2^je_j = 
\wraps{\wrapb{\alpha^1x_1 + \alpha^2x_2}^j - \alpha^1x_1^j - \alpha^1x_1^j}\cdot e_j = \theta \implies [] = 0 \implies$

\[
\wrapb{\alpha^1x_1 + \alpha^2x_2}^j = \alpha^1x_1^j + \alpha^2x_2^j
\]

$\wrapb{\alpha^1x_2 + \alpha^2x_2}^j = \wrapa{e^j, \alpha^1x_1 + \alpha^2x_2};\;x_1^j = \wrapa{e^j, x_1};\;x_2^j = \wrapa{e^j, x_2} \implies
\wrapa{e^j, \alpha^1x_2 + \alpha^2x_2} = \alpha^1\wrapa{e^j, x_1} + \alpha^1\wrapa{e^j, x_2}$

\textbf{Lemm 1.}: $\ss{e^1, \dots, e^n}$ forms a basis in $\c L^*$
\\
Prove: 
\ntab \textit{fullness}: $\wrapa{f, x} = \wrapa{f, x^je_j} = x^j\wrapa{f, e_j} = \wrapa{e^j, x}\underbrace{\wrapa{f, e_j}}_{=\alpha}\\
\wrapa{e^j, x}\alpha = \wrapa{\alpha e^j, x} = \wrapa{\wrapa{f, e_j}\cdot e^j, x} \implies \all x \in \c L : f = \wrapa{f, e_j}e^j$
- we can 
\ntab \textit{Linear independence}: $\wrapa{e^j, e_k} = \wraps{e_k = 0e_1 + \dots + 1e_k + \dots + 0e_n = \delta^j_ke_j} = \delta_{jk}$\\
 $\wrapa{\alpha_je^j, e_k} = \wrapa{\theta^*, e_k} = 0 = \alpha_j\delta^j_k = \alpha_k, k \in \overline{1, k} \implies$ Linearly independant

Kinda important example: $\ss{P^n}$ - set of polynoms with power $\leq n, p^(t) = \alpha_0 + \alpha_1t + \dots + \alpha_nt^n\\
\text{basis: } \ss{1, t, t^2, \dots, t^n}$ (to proove: fullness is trivial, linear independence proff goes like this: let $t=0 \implies \alpha_0 = 0$, then differentiate $n$ times and $\all \alpha_i = 0$). \\
Taylor expansion of $p(t) \in \s P^n = \dst\sum_{k=0}^n = \frac{P^{(k)}(0)}{k!}t^k$ - which is same as basis expansion!

meanwhile, Lets try to find $\wrapb{P^{(*)}}^*$\\
The operation: $\wrapa{id\cdot|_{t=0}, P(t)} := P(0), \wrapa{\at{\frac{d}{dt}\cdot}{t=0}, p(t)} := P'(0), \dots, \wrapa{\at{\frac{d^n}{dt^n}}{t=0}, P(t)} := 
P^{(n)}(0)$\\
so, basis in covector space $\wrapb{P^{(*)}}^*: \ss{\at{id\cdot}{t=0}, \at{\frac{d}{dt}\cdot}{t=0}, \frac{d^2}{dt^2}_{\dots}, \dots, \frac{d^n}{dt^n}}_{\dots}$

\section{Basis transition}

we will mark \textbf{old basis} as $\ss{e_1, \dots, e_n}$\\
and \textbf{new basis} is $\ss{e_{1'}, e_{2'}, \dots, e_{n'}}$\\
and the \textbf{"new new" basis}: $\ss{e_{1''}, \dots, e_{n''}}$\\
as you could guess, the "new new new" basis would be with three ticks


$e_{j'} = e^i_{j'}e_i \ioi \mrx{e_{1'} & \dots &  e_{n'}} = \mrx{e_1 & \dots & e_n}\mrx{
	e^1_{1'} & \dots & e^1_{n'}\\
	\vdots & \ddots & \vdots\\
	e^n_{1'} & \dots & e^n_{n'}
} \ioi E' = E\cdot C$, where $C$ is transition matrix, while $\det C \neq 0$. Let's prove that: \\
Let $\det C = 0. C = \mrxddet{C_{1'},\dots, C_{n'}} \implies \e \alpha^{i'} : \alpha^{1'}C_{1'} + \dots + \alpha^{n'}C_{n'} = 0\\
e_{i'} = c^i_{i'}e_i = \mrx{e_1, \dots, e_n}\mrx{c^1_{i'} \\ \vdots \\ c^n_{i'}} = EC_{i'}, e_{i'} = EC_{i'}\\
\alpha^{1'}\cdot e_{1'} ; E\alpha^{1'}\cdot C_{i'} = E0 = \theta$, that means that basis $e'$ is linearly dependant $\implies \det C \neq 0$

$e_{i'} = c^i_{i'}e_i \ioi E' = EC \implies \e C^{-1}: E = E'C^{-1}\hspace{1cm}C=\wrapb{c^i_{i'}}^n_{n'};\; C^{-1} = \wrapb{c^{i'}_i}^{n'}_n$\\
This makes an easy way to denote whether the transition is from old ot new or from new to old basis

$(e_1, \dots, e_n) = (e_{1'}, \dots, e_{n'})\mrx{
	b^{1'}_1 & \dots & b^{1'}_n \\
	\vdots & \ddots & \vdots \\
	b^{n'}_1 & \dots & b^{n'}_n \\
} = E'B \implies \system{
	e_i = c^{i'}_ie_{i'}\\
	e_{i'} = c^i_{i'}e_i
}$ - \textbf{covariant law}

\subsection{Basis coordinates transistion}
$x = x^{j'}e_{j'};\; x = x^je_j \implies ?x^{j'} = c^{j'}_jx^j?$
\ntab Proof: $e_j = c^{j'}_je_{j'} \implies x^{j'}e_{j'} = x^jc^{j'}_je_{j'} \ioi \wraps{x^{j'} - c^{j'}_jx^j}e_{j'} = \theta \implies
x^{j'} = c^{j'}_jx^j$ - \textbf{countervariant law}

Im matrix form: \\$\system{
	X_e' = C^{-1}X_e\\
	E' = EC
}$

\section{Linear operator}

Let there be $\c L_1, \c L_2$

\textbf{Linear operator} on $\c L_1, \c L_2$: $\c A : \c L_1 \to \c L_2$ is a linear operator IF $\c A(\alpha^1x_1 + \alpha^2x_2) = \alpha^1\c A(x_1) + \alpha^2\c A(x_2)$

Example: take $\s K^{m\times n}, \c L_1 = X \in \s K^{n\times 1}, Y \in L_2 = \s K^{m\times 1}\\
AX \in \c L_2$, this is linear operator

\subsection{kernel and image}

$\ker A := \ss{X \in \c L_1 : AX_1 = \theta_2 \in \c L_1}$ - this is subset (or equal to) $L_1$ that gets you $\theta$ in $\c L_2$ after applying A\\
$\im A := \ss{Y \in \c L_2 : \e k \in L_1, y = A(X)}$ - this is subset (or equal to) $L_2$ that you can get from $A(\c L_1)$, 

Lets prove that $\ker$ is subspace of $\c L_1$, $\im$ is subspace of $\c L_2$:\\
Let $X_1, X_2 \in \ker A: AX_1 = AX_2 = \theta_2 \implies A(\alpha^1 X_1 +\alpha^2 X_2) = \alpha^1 A(X_1) + \alpha^2 A(X_2) = \theta_2 \in \ker A$\\
and I CANT FUCKING SEE THE PROOF ON THE OTHER PART OF BLACKBOARD FUUUUCK: but the idea is (talke 2 vectors from image, take their linear combination and it 
appears to be also in that image

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