Linear Algebra
# V is Linear Space over K
V is Linear Space over $ \mathbb{K} $
$ \Leftrightarrow $ `(def)` V is Module on $ \mathbb{K} $
$ \Leftrightarrow $ `(def)`
- V is an additive group (+)
- V is given as;
- $ \mathbb{K} \times V \rightarrow V $
- $ (k,v) \mapsto kv $
where
1. $ \quad 1v = v $
2. $ \quad (a+b)v = av+bv $
3. $ \quad a(bv) = (ab)v $
4. $ \quad a(v + v') = av + av' $
$ \Leftrightarrow $
Frankly,
- "add" and "scholor times" is defined in the space
- 「和」と「スカラー倍」が定義されている空間
# Subspace W of "V over K"
Subspace $ W $ of $ V / \mathbb{k} $
$ \Leftrightarrow $ `(def)`
$$
\begin{cases}
\forall a, \forall b \in W.\quad a+b \in W \\
\forall k \in \mathbb{R}. \quad \quad ka \in W
\end{cases}
$$
**************************************
* *
* ^ *
* | / {y = 2x} *
* | / *
* | / *
* | / *
* | / *
* |/ *
* -----------+--------------> *
* /| *
* / | *
* / | *
* / | *
* / | *
* *
* *
* *
**************************************
# Quotient Space by Subspace W of V
define isomorphism on $ V $ as;
$ v, v' \in V $
$ v \simeq v' \iff $ `(def)` $ \quad v - v' \in W $
**********************************************************
* *
* ^ *
* | *
* | *
* - - - - - + - - - - - - *
* ^ | ^ *
* v' \ | / v *
* \|/ *
* -----------+-----------> *
* | *
* | *
* -----------+------------ W = {y = c} *
* | *
* | *
* *
* *
* *
**********************************************************
Quotient Space
$ V / W := V / \sim := \{ [v] \mid v \in V \} $
where
$$
\begin{cases}
[v] + [v'] = [v + v'] \\
k [v] = [kv]
\end{cases}
$$
prove well difined !
# Linear Map
$ V $ , $ W $ : vector space over $ \mathbb{K} $
## Linear Map
$ f : V \rightarrow W $ is Linear Map
$ \iff $ `(def)`
1. $ \quad \forall v , \forall v' \in V . \quad f(v + v') = f(v) + f(v') $
2. $ \quad \forall k \in \mathbb{K}, \forall v \in V . \quad f(kv) = k f(v) $
## Kernel
kernel of $ \ f \quad $ ( $ f $ : $ V \rightarrow W $ : linear map)
- $ Ker \ f := \{ v \in V \mid f(v) = 0 \} $
$ \quad \quad = f^{-1}(\{0\}) $
## Image
image of $ \ f \quad $ ( $ f $ : $ V \rightarrow W $ : linear map)
- $ Im f := \{ w \in W \mid w \in f(v) \} $
$ \quad \quad = f(V) $
## Isomorphism Theorem
$ f : V \rightarrow W $ : linear map
$ \bar{f} : V / Ker f \quad \simeq \quad Im f $
$ \quad\quad [v] \quad \mapsto \quad f(v) $
- isomorphism : linear and bijective
- 同型 : 線形写像かつ全単射
e.g.
*************************************************
* y y *
* ^/ / / / ^ / *
* /\/ / / | / *
* / /\/ / / f | / *
* / /\/ / / ---> |/ *
* +-+-+++-+>x ----o--->x *
* / / /\/ / /| *
* / / / /\/ / | *
* / / / /\ / | *
*************************************************
$$ f(\nwarrow) = 0 $$
$$ \bar{f} : \underbrace{\mathbb{R}^2 / Ker f }_{(\nwarrow)} = \underbrace{Im\ f}_{\text{( / )}} $$
# basis 基底
$ V : \text{ linear space over } \mathbb{K} $
## $ v _ 1 , ... , v _ n \in V $ is linear independent
$ \Leftrightarrow $ `(def)`
$$ a _ 1 v _ 1 + ... + a _ n v _ n = 0 \Rightarrow a _ 1 = ... = a _ n = 0 $$
## Subspace spanned by $ v _ 1 , ... , v _ n \in V $
$ \Leftrightarrow $ `(def)`
$$ Span \{ v _ 1 , ... , v _ n \} := \{ a _ 1 v _ 1 + ... + a _ n v _ n \mid a _ i \in \mathbb{K} \} $$
or we write $ < v _ 1 , ... , v _ n > $
## $ v _ 1 , ... , v _ n \in V $ is basis of $ V $
$ \Leftrightarrow $ `(def)`
1. $ v _ 1 , ... , v _ n \in V $ is linear independent
2. $ V = Span \{ v _ 1 , ... , v _ n \} $
# dimension
Def .
if basis of $ V $ has n elements $ e _ 1 , ... , e _ n $,
$$ dim V := n $$
# prove that dimension is well-defined
suppose 2 basises of $ V $
- $ e _ 1 , ... , e _ n $
- $ d _ 1 , ... , d _ m $
then $ e _ i $ can be described as
$$ e _ i = \sum^{n} _ {j=1} a _ {ij} d _ j $$
this is
$$
(e _ 1 , ... , e _ n ) = ( d _ 1 , ... , d _ m)
\left(
\begin{array}{ccc}
a _ {11} & \cdots & a _ {1n} \\
\vdots & & \vdots \\
a _ {m1} & \cdots & a _ {nm}
\end{array}
\right)
$$
similarly
$$
(d _ 1 , ... , d _ m ) = ( e _ 1 , ... , e _ n)
\left(
\begin{array}{ccc}
b _ {11} & \cdots & b _ {1m} \\
\vdots & & \vdots \\
b _ {n1} & \cdots & b _ {mn}
\end{array}
\right)
$$
so we name these matricses $ A $ and $ B $,
$$ AB = E _ n \quad\wedge\quad BA = E _ m $$
$$ \therefore A \text{ is bijective } $$
$ A : \mathbb{R}^m \rightarrow \mathbb{R}^n $
suppose $ m \gt n $
$ A $ is injective (単射) $ \iff $ $ ( A v = 0 \Rightarrow v = 0 ) $
$$ A v = 0 $$
$ \hspace{7cm}\text{------------------------} \Downarrow $ operate with [Elementary matrix](../2016-03-02-elementary-matrix.md)
$$ P A v = 0 $$
$$ \text{------------------------} \Downarrow $$
$$
\left(
\begin{array}{ccc | c}
1 & & & \\
& \ddots & & 0 \\
& & 1 &
\end{array}
\right)
\left(
\begin{array}{c}
v _ 1 \\
\vdots \\
v _ m
\end{array}
\right)
= 0
$$
but , non 0 solution found
$$
\left(
\begin{array}{c}
v _ 1 \\
\vdots \\
\vdots \\
v _ m
\end{array}
\right)
=
\left(
\begin{array}{c}
0 \\
\vdots \\
0 \\
1
\end{array}
\right)
$$
this is contradiction
$$ \therefore m \le n $$
similary,
$$ n \le m $$
$ \quad \hspace{6cm} \therefore m = n \hspace{4cm} $ $ \Box \quad \text{well-defined} $