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矩陣的列空間

Creative Commons License
This work by Jephian Lin is licensed under a Creative Commons Attribution 4.0 International License.

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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, r3b1, 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, r3b1, 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©

設定一些 r1, r2, r3b1, 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

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 \(y = c_1\)\(z = c_2\) 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:
\(y = c_1\)\(z = c_2\),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©

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 \(y = c_1\)\(w = c_2\) 來算出解的參數式。)

答:
\(y = c_1\)\(w = c_2\),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, \({\bf v}\in\operatorname{ker}(A)\).

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?

  • 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:
\({\bf v}_1\)\({\bf v}_2\)\({\bf v}_3\) 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。

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。

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

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