Row space of a matrix
This work by Jephian Lin is licensed under a Creative Commons Attribution 4.0 International License.
$\newcommand{\trans}{^\top} \newcommand{\adj}{^{\rm adj}} \newcommand{\cof}{^{\rm cof}} \newcommand{\inp}[2]{\left\langle#1,#2\right\rangle} \newcommand{\dunion}{\mathbin{\dot\cup}} \newcommand{\bzero}{\mathbf{0}} \newcommand{\bone}{\mathbf{1}} \newcommand{\ba}{\mathbf{a}} \newcommand{\bb}{\mathbf{b}} \newcommand{\bc}{\mathbf{c}} \newcommand{\bd}{\mathbf{d}} \newcommand{\be}{\mathbf{e}} \newcommand{\bh}{\mathbf{h}} \newcommand{\bp}{\mathbf{p}} \newcommand{\bq}{\mathbf{q}} \newcommand{\br}{\mathbf{r}} \newcommand{\bx}{\mathbf{x}} \newcommand{\by}{\mathbf{y}} \newcommand{\bz}{\mathbf{z}} \newcommand{\bu}{\mathbf{u}} \newcommand{\bv}{\mathbf{v}} \newcommand{\bw}{\mathbf{w}} \newcommand{\tr}{\operatorname{tr}} \newcommand{\nul}{\operatorname{null}} \newcommand{\rank}{\operatorname{rank}} %\newcommand{\ker}{\operatorname{ker}} \newcommand{\range}{\operatorname{range}} \newcommand{\Col}{\operatorname{Col}} \newcommand{\Row}{\operatorname{Row}} \newcommand{\spec}{\operatorname{spec}} \newcommand{\vspan}{\operatorname{span}} \newcommand{\Vol}{\operatorname{Vol}} \newcommand{\sgn}{\operatorname{sgn}} \newcommand{\idmap}{\operatorname{id}} \newcommand{\am}{\operatorname{am}} \newcommand{\gm}{\operatorname{gm}} \newcommand{\mult}{\operatorname{mult}} \newcommand{\iner}{\operatorname{iner}}$
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 $A$.
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。
Run the code below. Let the hyperplanes of $\inp{\br_i}{\bv} = b_i$ be the red, blue, and green plane.
### 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")
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。
Choose some appropriate r1, r2, r3
and b1, b2, b3
so that the intersection of the three hyperplanes is a straight line.
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")
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。
Choose some appropriate r1, r2, r3
and b1, b2, b3
so that the intersection of the three hyperplanes is a plane.
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")
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。
Choose some appropriate r1, r2, r3
and b1, b2, b3
so that the intersection of the three hyperplanes is the emptyset.
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")
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。
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.
說明如果一個超平面沒有通過原點則不是一個子空間。
Show that a hyperplane without passing through the origin is not a subspace.
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。)
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.)
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, $\bv\in\ker(A)$.
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
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
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?
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?
給定以下點
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。
Consider the following points
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.