# 矩陣的列空間

This work by Jephian Lin is licensed under a [Creative Commons Attribution 4.0 International License](http://creativecommons.org/licenses/by/4.0/).

{%hackmd 5xqeIJ7VRCGBfLtfMi0_IQ %}

## Main idea

##### Matrix-vector multiplication (by row)

$$
A = \begin{bmatrix} - & {\bf r}_1 & - \\
~ & \vdots & ~ \\ - & {\bf r}_m & - \\
\end{bmatrix}
$$
be an $m\times n$ matrix and ${\bf v}$ a vector in $\mathbb{R}^n$.
Then the $i$-th entry of $A{\bf v}$ is
$$(A{\bf v})_i = \langle{\bf r}_i, {\bf v}\rangle.$$

A set in $\mathbb{R}^n$ of the form
$$\{ {\bf v}\in\mathbb{R}^n : \langle{\bf r}, {\bf v}\rangle = b \}$$
for some vector ${\bf r}$ and scalar $b$
is called a **hyperplane**,
where ${\bf r}$ is its **normal vector**.

Therefore, if
$${\bf b} = \begin{bmatrix} b_1 \\ \vdots \\ b_m \end{bmatrix},$$
then the solution set of $A{\bf v} = {\bf b}$ is
the intersections of the hyperplanes given by $\langle{\bf r}_i, {\bf v}\rangle = b_i$ for $i = 1,\ldots m$.

A **hyperplane** is a subspace if and only if it contains the origin ${\bf 0}$,
which is equivalent to the corresponding $b$ is $0$.

The **row space** of $A$ is defined as
$$\operatorname{Row}(A) = \operatorname{span}(\{{\bf r}_1, \ldots, {\bf r}_m\}).$$

Let $V$ be a subspace in $\mathbb{R}^n$.
The **orthogonal complement** of $V$ is defined as
$$V^\perp = \{{\bf w}\in\mathbb{R}^n : \langle{\bf w},{\bf v}\rangle = 0 \text{ for all }{\bf v}\in V\}.
$$
Thus, $\operatorname{ker}(A) = \operatorname{Row}(A)^\perp$ for any matrix $A$.

## Side stories
- paramertrization
- partition of space

## Experiments

##### Exercise 1
執行下方程式碼。
紅色、藍色、綠色的平面分別為 $\langle{\bf r}_i,{\bf v}\rangle = b_i$ 畫出來的超平面。

### code
set_random_seed(0)
print_ans =False
r1 = vector([1,0,0])
r2 = vector([0,1,0])
r3 = vector([0,0,1])
b1,b2,b3 = 0,0,0

H.<x,y,z> = HyperplaneArrangements(QQ)
h1 = r1[0]*x + r1[1]*y + r1[2]*z - b1
h2 = r2[0]*x + r2[1]*y + r2[2]*z - b2
h3 = r3[0]*x + r3[1]*y + r3[2]*z - b3

h1.plot(color="red") + h2.plot(color="blue") + h3.plot(color="green")

##### Exercise 1(a)
設定一些 `r1, r2, r3` 及 `b1, b2, b3` 使得三個超平面的交集為一直線。

r1 = vector([1,0,0])
r2 = vector([0,1,0])
r3 = vector([1,1,0])
b1,b2,b3 = 0,0,0

H.<x,y,z> = HyperplaneArrangements(QQ)
h1 = r1[0]*x + r1[1]*y + r1[2]*z - b1
h2 = r2[0]*x + r2[1]*y + r2[2]*z - b2
h3 = r3[0]*x + r3[1]*y + r3[2]*z - b3

h1.plot(color="red") + h2.plot(color="blue") + h3.plot(color="green")

##### Exercise 1(b)
設定一些 `r1, r2, r3` 及 `b1, b2, b3` 使得三個超平面的交集為一平面。

r1 = vector([1,0,0])
r2 = vector([1,0,0])
r3 = vector([1,0,0])
b1,b2,b3 = 0,0,0

H.<x,y,z> = HyperplaneArrangements(QQ)
h1 = r1[0]*x + r1[1]*y + r1[2]*z - b1
h2 = r2[0]*x + r2[1]*y + r2[2]*z - b2
h3 = r3[0]*x + r3[1]*y + r3[2]*z - b3

h1.plot(color="red") + h2.plot(color="blue") + h3.plot(color="green")

##### Exercise 1(c)
設定一些 `r1, r2, r3` 及 `b1, b2, b3` 使得三個超平面的交集為空集合。

r1 = vector([1,0,0])
r2 = vector([1,0,0])
r3 = vector([1,0,0])
b1,b2,b3 = 1,2,0

H.<x,y,z> = HyperplaneArrangements(QQ)
h1 = r1[0]*x + r1[1]*y + r1[2]*z - b1
h2 = r2[0]*x + r2[1]*y + r2[2]*z - b2
h3 = r3[0]*x + r3[1]*y + r3[2]*z - b3

h1.plot(color="red") + h2.plot(color="blue") + h3.plot(color="green")

## Exercises

:::success

Exercise 2 寫得很好！
:::

##### Exercise 2
超平面的基本性質。

##### Exercise 2(a)
找兩個向量 ${\bf u}_1, {\bf u}_2$
使得 $\left\{\begin{bmatrix}x\\y\\z\end{bmatrix} : x + y + z = 0\right\} = \operatorname{span}(\{{\bf u}_1, {\bf u}_2\})$。
（可以令 $y = c_1$ 及 $z = c_2$ 來算出解的參數式。）
這讓我們更確定一個通過原點的超平面是一個子空間。

答:
令 $y = c_1$ 及 $z = c_2$，則 $x = - c_1 - c_2$，
此時
$$\begin{bmatrix}x\\y\\z\end{bmatrix} = \begin{bmatrix}- c_1 - c_2\\c_1\\c_2\end{bmatrix},
$$
可改寫成
$$\begin{bmatrix}x\\y\\z\end{bmatrix} = c_1\begin{bmatrix}- 1\\1\\0\end{bmatrix} + c_2\begin{bmatrix}- 1\\0\\1\end{bmatrix}.
$$
其中
$${\bf u}_1 = \begin{bmatrix}- 1\\1\\0\end{bmatrix} , {\bf u}_2 = \begin{bmatrix}- 1\\0\\1\end{bmatrix}.
$$
即為所求。

##### Exercise 2(b)
說明如果一個超平面沒有通過原點則不是一個子空間。

答:
子空間必通過原點，若一個超平面未通過原點，即一子空間加上一平移向量，稱為仿射子空間。

##### Exercise 2(c)
實際上﹐每個齊次線性方程組（$A{\bf v} = {\bf 0}$）
的解都可以用參數式表達。
找兩個向量 ${\bf u}_1, {\bf u}_2$
使得 $\left\{\begin{bmatrix}x\\y\\z\\w\end{bmatrix} :
\begin{array}{ccccc}
x & +y & & +w & =0 \\
& & z & +w &= 0 \\
\end{array}\right\} = \operatorname{span}(\{{\bf u}_1, {\bf u}_2\})$。
（可以令 $y = c_1$ 及 $w = c_2$ 來算出解的參數式。）

答:
令 $y = c_1$ 及 $w = c_2$，則 $x = - c_1 - c_2$，$z = - c_2$，
此時
$$\begin{bmatrix}x\\y\\z\\w\end{bmatrix} = \begin{bmatrix}- c_1 - c_2\\c_1\\- c_2\\c_2\end{bmatrix},
$$
可改寫成
$$\begin{bmatrix}x\\y\\z\\w\end{bmatrix} = c_1\begin{bmatrix}- 1\\1\\0\\0\end{bmatrix} + c_2\begin{bmatrix}-1\\0\\- 1\\1\end{bmatrix}.
$$
其中
$${\bf u}_1 = \begin{bmatrix}- 1\\1\\0\\0\end{bmatrix} , {\bf u}_2 = \begin{bmatrix}-1\\0\\- 1\\1\end{bmatrix}.
$$
即為所求。

##### Exercise 3
證明 $\operatorname{ker}(A) = \operatorname{Row}(A)^\perp$。

Sample:
Let ${\bf r}_1, \ldots, {\bf r}_m$ be the rows of $A$.

**"$\subseteq$"**
If ${\bf v}\in\operatorname{ker}(A)$, then ...
...
Therefore, ${\bf v}\in\operatorname{Row}(A)^\perp$.

**"$\supseteq$"**
If ${\bf v}\in\operatorname{Row}(A)^\perp$, then ...
...
Therefore, ${\bf v}\in\operatorname{ker}(A)$.

:::warning
檢討前請先回顧一下 $\ker(A)$、$\Row(A)$、還有 $V^\perp$ 的定義。檢討的時候可以再討論更正。

用中英文寫都可以，重點是邏輯正確且該解釋到的有解釋到。
:::

證明: **[由鄭子安同學提供]**
Let $A$ be an $m\times n$ matrix, and let ${\bf r}_1, \ldots, {\bf r}_m$ be the rows of $A$. Then $\ker(A)$ is the same as
$$
\{\bv\in\mathbb{R}^n: \inp{\bv}{{\bf r}_i} \text{ for } i = 1, \ldots, m\}
$$
by definition.

**1. Claim:** $\ker(A) \subseteq \Row(A)^\perp$。

Let ${\bf v}\in\operatorname{ker}(A)$.
Then $\langle{\bv},{\bf r}_i\rangle = 0$ for all $i = 1 , 2 , \ldots , m$.
This implies that ${\bf v} \perp {\bf r}_1 , \ldots , {\bf r}_m$.
Then, we have
$$\begin{aligned}
\langle{\bv},c_1{\bf r}_1 + \cdots + c_m{\bf r}_m\rangle &= \langle{\bv},c_1{\br}_1\rangle + \cdots + \langle{\bv},c_m{\br}_m\rangle \\
&= c_1\langle{\bv},{\bf r}_1\rangle + \cdots + c_m\langle{\bv},{\bf r}_m\rangle \\
&= 0
\end{aligned}
$$
for all constants $c_1 , \cdots , c_m \in\mathbb{R}$.
In other words, $\langle{\bv},{\bf r}\rangle = 0$ for all ${\bf r} \in \operatorname{Row}(A)$.
Therefore, ${\bf v}\in\operatorname{Row}(A)^\perp$.

**2. Claim:** $\ker(A) \supseteq \Row(A)^\perp$。

Let ${\bf v}\in\operatorname{Row}(A)^\perp$.
Then $\langle{\bv},c_1{\bf r}_1 + \cdots + c_m{\bf r}_m\rangle = 0$ for all $c_1 , \ldots , c_m$ $\in$ $\mathbb{R}$.
This implies that ${\bf v}$ $\perp$ $c_1{\bf r}_1 + \ldots + c_m{\bf r}_m$.
By taking $c_1 = 1 , c_2 = \cdots = c_m = 0$,
we have ${\bf v}$ $\perp$ ${\bf r}_1$.
Similarly, ${\bf v}$ $\perp$ ${\bf r}_i$ for $i = 1 , \ldots , m$.
Therefore, ${\bf v}\in\operatorname{ker}(A)$.

**3.**
In summary,
by **Claims 1.** and **2.**
we conclude that $\operatorname{ker}(A) = \operatorname{Row}(A)^\perp$.
**Q.E.D.**

##### Exercise 4

一個超平面會把 $\mathbb{R}^n$ 分割成兩部份。
更精確來說﹐
給定法量向 ${\bf r}$ 和偏移量 $b$﹐
整個 空間會被分成三部份
正部︰$\{{\bf v}: \langle{\bf r},{\bf v}\rangle > b\}$、
負部︰$\{{\bf v}: \langle{\bf r},{\bf v}\rangle < b\}$、
邊界︰$\{{\bf v}: \langle{\bf r},{\bf v}\rangle = b\}$（超平面本身）。

##### Exercise 4(a)
考慮法向量 ${\bf r} = (1,1,1)$ 和偏移量 $b = 5$ 所定義出來的超平面。
問點
${\bf v}_1 = (0,2,3)$、
${\bf v}_2 = (1,0,1)$、
${\bf v}_3 = (3,2,1)$
分別落在超平面的正部、負部、或是邊界？

:::warning

- [x] 因為答案已經不是在描述集合，所以不用使用 $\{?: ...\}$ 的集合符號（實際上你們的用法也不正確）。直接把 "$\{{\bf v}_1 = (0,2,3): \langle{\bf r},{\bf v}_1\rangle = b\}$ 落在邊界，" 改成 "$\inp{{\bf r}}{\bv_1} = 0 + 2 + 3 = 5 = b$，所以 $\bv_1$ 落在邊界"。其它部份用同樣方式修改。
:::

答:
將 ${\bf v}_1$ 、 ${\bf v}_2$ 和 ${\bf v}_3$ 帶入，
可知
$\langle{\bf r},{\bf v}_1\rangle = 0 + 2 + 3 = b，$ 所以 ${\bf v}_1$ 落在邊界，
$\langle{\bf r},{\bf v}_2\rangle = 1 + 0 + 1 = 2 < b，$ 所以 ${\bf v}_2$ 落在負部，
$\langle{\bf r},{\bf v}_3\rangle = 3 + 2 + 1 = 6 > b，$ 所以 ${\bf v}_3$ 落在正部。

##### Exercise 4(b)

給定以下點
${\bf v}_1 = (3,4,5,6)$、
${\bf v}_2 = (2,3,6,7)$、
${\bf v}_3 = (3,1,8,6)$、
${\bf v}_4 = (5,5,5,3)$、
${\bf v}_5 = (0,0,0,0)$、
${\bf v}_6 = (1,1,2,2)$、
${\bf v}_7 = (1,3,-1,-2)$、
${\bf v}_8 = (1,2,3,4)$、
找一組法向量以及偏移量使得
其定義出來的超平面讓
${\bf v}_1, {\bf v}_2, {\bf v}_3, {\bf v}_4$ 落在正部、
${\bf v}_5, {\bf v}_6, {\bf v}_7, {\bf v}_8$ 落在負部。

:::success
灣得否！
:::

答:
取法向量 ${\bf r} = (1,1,1,1)$ 和偏移量 $b = 11$ 所定義出來的超平面，
即可滿足上述條件。

:::info
除了 Exercise 3 以外數學全部正確，格式也沒太多要改的。

目前得分 4/5
:::