# 矩陣的列空間

Row space of a matrix

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

$\newcommand{\trans}{^\top}
\newcommand{\adj}{^{\rm adj}}
\newcommand{\cof}{^{\rm cof}}
\newcommand{\inp}[2]{\left\langle#1,#2\right\rangle}
\newcommand{\dunion}{\mathbin{\dot\cup}}
\newcommand{\bzero}{\mathbf{0}}
\newcommand{\bone}{\mathbf{1}}
\newcommand{\ba}{\mathbf{a}}
\newcommand{\bb}{\mathbf{b}}
\newcommand{\bc}{\mathbf{c}}
\newcommand{\bd}{\mathbf{d}}
\newcommand{\be}{\mathbf{e}}
\newcommand{\bh}{\mathbf{h}}
\newcommand{\bp}{\mathbf{p}}
\newcommand{\bq}{\mathbf{q}}
\newcommand{\br}{\mathbf{r}}
\newcommand{\bx}{\mathbf{x}}
\newcommand{\by}{\mathbf{y}}
\newcommand{\bz}{\mathbf{z}}
\newcommand{\bu}{\mathbf{u}}
\newcommand{\bv}{\mathbf{v}}
\newcommand{\bw}{\mathbf{w}}
\newcommand{\tr}{\operatorname{tr}}
\newcommand{\nul}{\operatorname{null}}
\newcommand{\rank}{\operatorname{rank}}
%\newcommand{\ker}{\operatorname{ker}}
\newcommand{\range}{\operatorname{range}}
\newcommand{\Col}{\operatorname{Col}}
\newcommand{\Row}{\operatorname{Row}}
\newcommand{\spec}{\operatorname{spec}}
\newcommand{\vspan}{\operatorname{span}}
\newcommand{\Vol}{\operatorname{Vol}}
\newcommand{\sgn}{\operatorname{sgn}}
\newcommand{\idmap}{\operatorname{id}}
\newcommand{\am}{\operatorname{am}}
\newcommand{\gm}{\operatorname{gm}}
\newcommand{\mult}{\operatorname{mult}}
\newcommand{\iner}{\operatorname{iner}}$

## Main idea

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

$$
A = \begin{bmatrix}
- & \br_1 & - \\
~ & \vdots & ~ \\
- & \br_m & - \\
\end{bmatrix}
$$

be an $m\times n$ matrix and $\bv$ a vector in $\mathbb{R}^n$.
Then the $i$-th entry of $A\bv$ is

$$
(A\bv)_i = \inp{\br_i}{\bv}.
$$

A set in $\mathbb{R}^n$ of the form

$$
\{ \bv\in\mathbb{R}^n : \inp{\br}{\bv} = b \}
$$

for some vector $\br$ and scalar $b$
is called a **hyperplane**,
where $\br$ is its **normal vector**.

Therefore, if

$$
\bb = \begin{bmatrix} b_1 \\ \vdots \\ b_m \end{bmatrix},
$$

then the solution set of $A\bv = \bb$ is
the intersections of the hyperplanes given by $\inp{\br_i}{\bv} = b_i$ for $i = 1,\ldots m$.

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

The **row space** of $A$ is defined as

$$
\Row(A) = \vspan(\{\br_1, \ldots, \br_m\}).
$$

Let $V$ be a subspace in $\mathbb{R}^n$.
The **orthogonal complement** of $V$ is defined as

$$
V^\perp = \{\bw\in\mathbb{R}^n : \inp{\bw}{\bv} = 0 \text{ for all }\bv\in V\}.
$$

Thus, $\ker(A) = \Row(A)^\perp$ 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 $\inp{\br_i}{\bv} = b_i$ 畫出來的超平面。

<!-- eng start -->
Run the code below. Let the hyperplanes of $\inp{\br_i}{\bv} = b_i$ be the red, blue, and green plane.
<!-- eng end -->

### 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。

<!-- eng start -->
Choose some appropriate `r1, r2, r3` and `b1, b2, b3` so that the intersection of the three hyperplanes is a straight line.
<!-- eng end -->

r1 = vector([1,1,0])
r2 = vector([0,1,0])
r3 = vector([3,2,0])
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")

:::warning

- [x] $r1$, $r2$, $r3$, $b1$, $b2$, $b3$ --> $\br_1$, $\br_2$, $\br_3$, $\bb_1$, $\bb_2$, $\bb_3$
- [x] While ..., we ...
- [x] Use the following typesetting:
$$
\begin{aligned}
h_1 &: x + y = 0 \text{ (red)} \\
h_2 &: y = 0 \text{ (blue)} \\
h_3 &: 3x + 2y = 0\text{ (green)}
\end{aligned}
$$
- [x] As the graphic --> As shown by the picture,
- [x] Use the following typesetting:

we can find that the three hyperplanes ~~may~~ intersact at

$$
\left\{t\begin{bmatrix} 0 \\ 0 \\1 \end{bmatrix}: t\in\mathbb{R}\right\},
$$

which can also be described by the parametrization

$$
\begin{cases}x=0\\ y=0\\ z=t\end{cases},\text{ where $t$ is a real number}.
$$
:::

#### <font color="#f00">Answer:</font>

While
$$\br_1=\begin{bmatrix}1 \\
1\\
0
\end{bmatrix},\br_2=\begin{bmatrix}0 \\
1\\
0
\end{bmatrix},\br_3=\begin{bmatrix}3 \\
2\\
0
\end{bmatrix} \text{and} ~(\bb_1,\bb_2,\bb_3)=(0,0,0),
$$

we have three different hyperplanes in three-dimensional space respectively:

$$
\begin{aligned}
h_1 &: x + y = 0 \text{ (red)} \\
h_2 &: y = 0 \text{ (blue)} \\
h_3 &: 3x + 2y = 0\text{ (green)}
\end{aligned}
$$

And there is the result below:


As shown by the picture, we can find that the three hyperplanes intersact at :

$$
\left\{t\begin{bmatrix} 0 \\ 0 \\1 \end{bmatrix}: t\in\mathbb{R}\right\}.
$$

---

##### 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。

<!-- eng start -->
Choose some appropriate `r1, r2, r3` and `b1, b2, b3` so that the intersection of the three hyperplanes is a plane.
<!-- eng end -->

r1 = vector([1,1,1])
r2 = vector([2,2,2])
r3 = vector([6,6,6])
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")

:::warning

- [x] Revise as the previous problem.
- [x] In the picture, we can only see one hyperplane (the green one) because the three hyperplanes are ~~basically congruence~~ the same.
:::

While
$$\br_1=\begin{bmatrix}1 \\
1\\
1
\end{bmatrix},\br_2=\begin{bmatrix}2 \\
2\\
2
\end{bmatrix},\br_3=\begin{bmatrix}6 \\
6\\
6
\end{bmatrix} \text{and} ~(\bb_1,\bb_2,\bb_3)=(0,0,0)
,$$

we have three different hyperplanes in three-dimensional space respectively:

$$
\begin{aligned}
h_1 &: x + y + z = 0 \text{ (red)} \\
h_2 &: 2x + 2y + 2z = 0 \text{ (blue)} \\
h_3 &: 6x + 6y + 6z = 0\text{ (green)}
\end{aligned}
$$

And there is the result below:
#### <font color="#f00">Answer:</font>

As shown by the picture, we can only see one hyperplane(the green one) because the three hyperplanes are the same.

---

##### 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。

<!-- eng start -->
Choose some appropriate `r1, r2, r3` and `b1, b2, b3` so that the intersection of the three hyperplanes is the emptyset.
<!-- eng end -->

r1 = vector([1,1,1])
r2 = vector([2,2,2])
r3 = vector([6,6,6])
b1,b2,b3 = 141,173,200

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")

#### <font color="#f00">Answer:</font>
While
$$\br_1=\begin{bmatrix}1 \\
1\\
1
\end{bmatrix},\br_2=\begin{bmatrix} 2 \\
2\\
2
\end{bmatrix},\br_3=\begin{bmatrix}6 \\
6\\
6
\end{bmatrix} \text{and} ~(\bb_1,\bb_2,\bb_3)=(141,173,200)
,$$

:::warning

- [x] See the previous problem.
:::

we have three different hyperplanes in three-dimensional space respectively:

$$
\begin{aligned}
h_1 &: x + y + z = 141 \text{ (red)} \\
h_2 &: 2x + 2y +2z = 173 \text{ (blue)} \\
h_3 &: 6x + 6y + 6z = 200\text{ (green)}
\end{aligned}
$$

And there is the result below:


As shown by the picture, we can find that the three hyperplanes never intersact.

---

## Exercises

##### Exercise 2

超平面的基本性質。

<!-- eng start -->
Basic properties of a hyperplane.
<!-- eng end -->

##### Exercise 2(a)

找兩個向量 $\bu_1, \bu_2$
使得 $\left\{\begin{bmatrix}x\\y\\z\end{bmatrix} : x + y + z = 0\right\} = \vspan(\{\bu_1, \bu_2\})$。
（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。

<!-- eng start -->
Find vectors $\bu_1$ and $\bu_2$ so that $\left\{\begin{bmatrix}x\\y\\z\end{bmatrix} : x + y + z = 0\right\} = \vspan(\{\bu_1, \bu_2\})$. (You may assume $y = c_1$ and $z = c_2$ to parametrize the equation.) This indicates that a hyperplane passing through the origin is likely to be a subspace.
<!-- eng end -->

:::warning

- [x] Assume that ... . Then ... , so we ... .
- [x] and rewrite as --> This can be rewritten as ...
- [x] ~~the above is the answer.~~
:::

#### <font color="#f00">Answer:</font>

Assume that $y = c_1$ and $z = c_2.$
Then $x = - c_1 - c_2,$
so we can get
$$\begin{bmatrix}x\\y\\z\\\end{bmatrix}=\begin{bmatrix}-c_1-c_2\\c_1\\c_2\end{bmatrix}.$$
This can be rewritten as
$$\begin{bmatrix}x\\y\\z\end{bmatrix}= c_1\begin{bmatrix}- 1\\1\\0\end{bmatrix} + c_2\begin{bmatrix}-1\\0\\1\end{bmatrix},$$
so we can get
$$ {\bu}_1=\begin{bmatrix}-1\\1\\0\end{bmatrix}, {\bu}_2 = \begin{bmatrix}-1\\0\\1\end{bmatrix}.
$$

---

##### 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。

<!-- eng start -->
Show that a hyperplane without passing through the origin is not a subspace.
<!-- eng end -->

:::warning

Well... yes, so the emphasise is why a subspace must pass through the orign?

- [x] $u$ --> $\bu$
- [x] 0$u$ --> $0\bu$; put the zero in the math mode
:::

#### <font color="#f00">Answer:</font>

If $V$ is a subspace, then $V=\vspan(S)$ for some $S$.
If $S=\emptyset,$ then $V=\vspan(\emptyset)={0}.$
If $\bu\in S,$ then $\bzero = 0\bu$ is a linear combination.
Hence a subspace must pass through the origin, so if a hyperplane does not pass through the origin, it is not a subspace.

---

##### Exercise 2(c)

實際上，每個齊次線性方程組（$A\bv = \bzero$）
的解都可以用參數式表達。
找兩個向量 $\bu_1, \bu_2$
使得 $\left\{\begin{bmatrix}x\\y\\z\\w\end{bmatrix} :
\begin{array}{ccccc}
x & +y & & +w & =0 \\
& & z & +w &= 0 \\
\end{array}\right\} = \vspan(\{\bu_1, \bu_2\})$。
（可以令 $y = c_1$ 及 $w = c_2$ 來算出解的參數式。）

<!-- eng start -->
Indeed, the solution set of a homogeneous system of linear equations (i.e., $A\bv = \bzero$) can be parametrized as the span of some vectors.

Find vectors $\bu_1$ and $\bu_2$ so that $\left\{\begin{bmatrix}x\\y\\z\\w\end{bmatrix} :
\begin{array}{ccccc}
x & +y & & +w & =0 \\
& & z & +w &= 0 \\
\end{array}\right\} = \vspan(\{\bu_1, \bu_2\})$. (You may assume $y = c_1$ and $w = c_2$ to parametrize the equations.)
<!-- eng end -->

:::warning

Revise it according to 2(a).
:::

#### <font color="#f00">Answer:</font>

Assume that $y=c_1$ and $w=c_2.$
Then we can get $x=-c_1-c_2$ and $z=-c_2,$
so we can get
$$\begin{bmatrix}x\\y\\z\\w\end{bmatrix}=\begin{bmatrix}-c_1-c_2\\c_1\\-c_2\\c_2\end{bmatrix}.$$
This can be rewritten as
$$\begin{bmatrix}x\\y\\z\\w\end{bmatrix}=c_1\begin{bmatrix}-1\\1\\0\\0\end{bmatrix}+c_2\begin{bmatrix}-1\\0\\-1\\1\end{bmatrix},$$
then we get that
$${\bu}_1=\begin{bmatrix}-1\\1\\0\\0\end{bmatrix},{\bu}_2=\begin{bmatrix}-1\\0\\-1\\1\end{bmatrix}
.$$

---

##### Exercise 3

證明 $\ker(A) = \Row(A)^\perp$。

<!-- eng start -->
Prove that $\ker(A) = \Row(A)^\perp$.
<!-- eng end -->

Sample:
Let $\br_1, \ldots, \br_m$ be the rows of $A$.

**"$\subseteq$"**
If $\bv\in\ker(A)$, then ...
...
Therefore, $\bv\in\Row(A)^\perp$.

**"$\supseteq$"**
If $\bv\in\Row(A)^\perp$, then ...
...
Therefore, $\bv\in\ker(A)$.

:::warning

- [x] Suppose ... . Then ... .
- [x] Assume ... .
- [x] According to matrix ..., we ~~can~~ know

$$
\left\{\begin{aligned}
\bv_1\cdot\bx &= 0, \\
&\vdots \\
\bv_n\cdot\bx &= 0.
\end{aligned}\right.
$$

- [x] The goal is **proving** ... . Therefore, it is enough to show that $\inp{\bx}{\br} = 0$ for any $\bv\in\Row(A)$. Since any $\bv\in\Row(A)$ can be written as $\br = ...$, we have

$$
\inp{\bx}{\br} = \inp{\bx}{r_1\bv_1 + \cdots + r_n\bv_n} = ... = 0.
$$

- [x] In $\{ ... | ... \}$, change $|$ to $\mid$ or $:$.
- [x] I think "Then $\ker(A)= \{\bx\in\mathbb{R}^n \vert\ A\bx=0\}.$" is not necessary.
- [x] $\Row(A)=\{\br_1\bv_1+⋯+\br_n\bv_n\vert\br_1,⋯,\br_n\in\mathbb{R}\},$ --> $\Row(A)=\{r_1\bv_1+⋯+r_n\bv_n \vert r_1,⋯,r_n\in\mathbb{R}\},$
:::

#### <font color="#f00">Answer:</font>

Suppose $\bx\in\ker(A).$
Then $A\bx=0.$
Assume $$A = \begin{bmatrix} - & \bv_1 & - \\~ & \vdots & ~ \\- & \bv_n & - \\\end{bmatrix}.$$
According matrix, we know:$\begin{cases}\bv_1\cdot\bx=0,\\ \vdots\\ \bv_n\cdot\bx=0.\end{cases}$
The goal is proving $\bx\in\Row(A)^\perp.$
Therefore, it is enough to show that $\inp{\bx}{\br} = 0$ for any $\bv\in\Row(A)$.
$\Row(A)=\{r_1\bv_1+⋯+r_n\bv_n\mid r_1,⋯, r_n\in\mathbb{R}\},$
and $(\Row(A))^\perp=\{y\in\mathbb{R}^n\mid r_1\bv_1\cdot\by+,⋯,+ r_n\bv_n\cdot\by=0,\forall\br_1,⋯,\br_n\in\mathbb{R}\},$
means that $\by$ is orthogonal to any vector in $\Row(A).$
To prove that $\bx$ can play the role of $\by$,
that is, to prove $\br_1\bv_1\cdot\bx+,⋯,+\br_n\bv_n\cdot\bx=0.$
From the combination of the aforementioned inner products,
multiply the $k-th$ formula by $\br_k$ and add it to get the above formula.
By the definition of $(\Row(A))^\perp,$ we can know $\bx\in(\Row(A))^\perp.$
And suppose $\bx\in(\Row(A))^\perp,$ according the definition,
for any $r_1,⋯, r_n\in\mathbb{R},$ $r_1\bv_1\cdot\bx+,⋯,+ r_n\bv_n\cdot\bx=0$ would be true.

:::warning

- [x] Conversely, suppose $\bx \in \Row(A)^\perp$. We claim that $\bx\in\ker(A)$. Recall that $r_1\bv_1 + \cdots + r_n\bv_n$ is in $\Row(A)$ for any $r_1, \ldots, r_n\in\mathbb{R}$. In particular,

$$
\bv_1 = 1\bv_1 + 0\bv_2 + \cdots + 0\bv_n \in \Row(A)
$$

and similarly $\bv_k\in\Row(A)$ for each $k = 1, \ldots, n$. Therefore, $\inp{\bv_1}{\bx} = \cdots = \inp{\bv_n}{\bx} = 0$. By the definition of $\ker(A)$ ...

:::

The next goal is proving $\bx\in\ker(A),$ as proving $A\bx=0.$

And $\begin{cases}\bv_1\cdot\bx=0\\ \vdots\\ \bv_n\cdot\bx=0\end{cases},$these n equations need to be proved.

Conversely, suppose $\bx \in \Row(A)^\perp$.
We claim that $\bx\in\ker(A)$. Recall that $r_1\bv_1 + \cdots + r_n\bv_n$ is in $\Row(A)$ for any $r_1, \ldots, r_n\in\mathbb{R}$.
In particular,

$$
\bv_1 = 1\bv_1 + 0\bv_2 + \cdots + 0\bv_n \in \Row(A)
$$

and similarly $\bv_k\in\Row(A)$ for each $k = 1, \ldots, n$.
Therefore, $\inp{\bv_1}{\bx} = \cdots = \inp{\bv_n}{\bx} = 0$.
By the definition of $\ker(A),$ we can know
$\ker(A) = \Row(A)^\perp.$

---

##### 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 $\br$ 和偏移量 $b$，
整個 $\mathbb{R}^n$ 空間會被分成三部份

- 正部︰$\{\bv: \inp{\br}{\bv} > b\}$、
- 負部︰$\{\bv: \inp{\br}{\bv} < b\}$、
- 邊界︰$\{\bv: \inp{\br}{\bv} = b\}$（超平面本身）。

<!-- eng start -->
A hyperplane cuts $\mathbb{R}^n$ into two regions. To be more precise, given a normal vector $\br$ and a bias $b$, the whole space $\mathbb{R}^n$ is composed of three parts

- positive side: $\{\bv: \inp{\br}{\bv} > b\}$,
- negative side: $\{\bv: \inp{\br}{\bv} < b\}$,
- boundary $\{\bv: \inp{\br}{\bv} = b\}$ (the hyperplane itself).
<!-- eng end -->

##### Exercise 4(a)

考慮法向量 $\br = (1,1,1)$ 和偏移量 $b = 5$ 所定義出來的超平面。
問點
$\bv_1 = (0,2,3)$、
$\bv_2 = (1,0,1)$、
$\bv_3 = (3,2,1)$
分別落在超平面的正部、負部、或是邊界？

<!-- eng start -->
Let $\br = (1,1,1)$ and $b = 5$ be the normal vector and the bias of a normal plane. For each of $\bv_1 = (0,2,3)$, $\bv_2 = (1,0,1)$, and $\bv_3 = (3,2,1)$, determine whether it belongs to the positive side, the negative side, or the boundary?
<!-- eng end -->

:::warning

- [x] Substitute ${\bf v}_1$ and ${\bf v}_2$ and ${\bf v}_3$ into the formula $x + y + z$.
- [x] Then we can get ... boundary, ... negative side, and ... positive side.
- [x] ~~The above is the answer.~~
:::

#### <font color="#f00">Answer:</font>

Substitute ${\bf v}_1$ and ${\bf v}_2$ and ${\bf v}_3$ into the formula $x+y+z.$
Then we can get
$\inp{\br}{\bv_1}=0+2+3=b$ , so ${\bv_1}$ is in the boundary,
$\inp{\br}{\bv_2}=1+0+1<b$ , so ${\bv_2}$ is in the negative side,
and $\inp{\br}{\bv_3}=3+2+1>b$ , so ${\bv_3}$ is in the positive side.

---

##### Exercise 4(b)

給定以下點

- $\bv_1 = (3,4,5,6)$、
- $\bv_2 = (2,3,6,7)$、
- $\bv_3 = (3,1,8,6)$、
- $\bv_4 = (5,5,5,3)$、
- $\bv_5 = (0,0,0,0)$、
- $\bv_6 = (1,1,2,2)$、
- $\bv_7 = (1,3,-1,-2)$、
- $\bv_8 = (1,2,3,4)$、

找一組法向量以及偏移量使得
其定義出來的超平面讓
$\bv_1, \bv_2, \bv_3, \bv_4$ 落在正部、
$\bv_5, \bv_6, \bv_7, \bv_8$ 落在負部。

<!-- eng start -->
Consider the following points

- $\bv_1 = (3,4,5,6)$,
- $\bv_2 = (2,3,6,7)$,
- $\bv_3 = (3,1,8,6)$,
- $\bv_4 = (5,5,5,3)$,
- $\bv_5 = (0,0,0,0)$,
- $\bv_6 = (1,1,2,2)$,
- $\bv_7 = (1,3,-1,-2)$,
- $\bv_8 = (1,2,3,4)$.

Find a normal vector and a bias so that for the corresponding hyperplane, the points $\bv_1, \bv_2, \bv_3, \bv_4$ fall in the positive side, while $\bv_5, \bv_6, \bv_7, \bv_8$ fall in the negative site.
<!-- eng end -->

:::warning

- [x] If we take ..., then $=10$.
- [x] Since we wish $\bv_1,\bv_2,\bv_3,\bv_4$ to fall in the positive side, $b$ should be ... .
- [x] Since ... negative **side**, $b$ should be ... .
- [x] In this way, ==_== we can take ... and bias as $b = 11$ to fulfill the requirement in the question.
:::

#### <font color="#f00">Answer:</font>

If we take the normal vector as ${\br}=(1,1,1,1),$
then
$\inp{\br}{\bv_1}=3+4+5+6=18$,
$\inp{\br}{\bv_2}=2+3+6+7=18$,
$\inp{\br}{\bv_3}=3+1+8+6=18$,
$\inp{\br}{\bv_4}=5+5+5+3=18$,
$\inp{\br}{\bv_5}=0+0+0+0=0$,
$\inp{\br}{\bv_6}=1+1+2+2=6$,
$\inp{\br}{\bv_7}=1+3-1-2=1$,
$\inp{\br}{\bv_8}=1+2+3+4=10$,
Since we wish $\bv_1, \bv_2, \bv_3, \bv_4$ to fall in the positive side, $b$ should be less than 18.
Since we wish $\bv_5, \bv_6, \bv_7, \bv_8$ to fall in the negative side, $b$ should be greater than 10.
In this way, we can take ${\br}=(1,1,1,1)$ and bias equals $11$ to fulfill the question.

---

:::info
There are 6 questions answered in the exercises.

2b. Need more details.
3. The idea is correct. The writting can be improved.

Overall, the mathematics aspects are great!

Collaboration: 1
4 problems: 4 (4 problems done)
1 extra problem: 1

Leftover: 3 can be improved

Moderator: 1
quality control: 1
:::