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\(\mathbb{R}^n\) 中的子空間
This work by Jephian Lin is licensed under a Creative Commons Attribution 4.0 International License.
\(\newcommand{\trans}{^\top} \newcommand{\adj}{^{\rm adj}} \newcommand{\cof}{^{\rm cof}} \newcommand{\inp}[2]{\left\langle#1,#2\right\rangle} \newcommand{\dunion}{\mathbin{\dot\cup}} \newcommand{\bzero}{\mathbf{0}} \newcommand{\bone}{\mathbf{1}} \newcommand{\ba}{\mathbf{a}} \newcommand{\bb}{\mathbf{b}} \newcommand{\bc}{\mathbf{c}} \newcommand{\bd}{\mathbf{d}} \newcommand{\be}{\mathbf{e}} \newcommand{\bh}{\mathbf{h}} \newcommand{\bp}{\mathbf{p}} \newcommand{\bq}{\mathbf{q}} \newcommand{\br}{\mathbf{r}} \newcommand{\bx}{\mathbf{x}} \newcommand{\by}{\mathbf{y}} \newcommand{\bz}{\mathbf{z}} \newcommand{\bu}{\mathbf{u}} \newcommand{\bv}{\mathbf{v}} \newcommand{\bw}{\mathbf{w}} \newcommand{\tr}{\operatorname{tr}} \newcommand{\nul}{\operatorname{null}} \newcommand{\rank}{\operatorname{rank}} %\newcommand{\ker}{\operatorname{ker}} \newcommand{\range}{\operatorname{range}} \newcommand{\Col}{\operatorname{Col}} \newcommand{\Row}{\operatorname{Row}} \newcommand{\spec}{\operatorname{spec}} \newcommand{\vspan}{\operatorname{span}} \newcommand{\Vol}{\operatorname{Vol}} \newcommand{\sgn}{\operatorname{sgn}} \newcommand{\idmap}{\operatorname{id}} \newcommand{\am}{\operatorname{am}} \newcommand{\gm}{\operatorname{gm}} \newcommand{\mult}{\operatorname{mult}} \newcommand{\iner}{\operatorname{iner}}\)
Main idea
Let \(S\) be a set of (possibily infinitely many) vectors in \(\mathbb{R}^n\).
A linear combination of \(S\) is a vector of the form \[c_1{\bf u}_1 + \cdots + c_k{\bf u}_k,\] for some vectors \({\bf u}_1,\ldots, {\bf u}_k\in S\) and
some scalars \(c_1,\ldots,c_k\in\mathbb{R}\).
Although \(S\) can have infinitely many vectors, a linear combination only uses finitely many vectors in \(S\).
The span of \(S\) is the set of all linear combination of \(S\),
denoted by \(\operatorname{span}(S)\).
(We vacuously define \(\operatorname{span}(\emptyset) = \{{\bf 0}\}\).)
Let \(V\) be a subset of \(\mathbb{R}^n\). Then the following two conditions are equivalent.
If either one of the two conditions holds, then \(V\) is called a subspace of \(\mathbb{R}^n\).
A system of linear equations has the form
\[\left[\begin{array}{ccccccc} a_{11}x_1 & + & \cdots & + & a_{1n}x_n & = & b_1 \\ \vdots & ~ & ~ & ~ & \vdots & = & \vdots \\ a_{m1}x_1 & + & \cdots & + & a_{mn}x_n & = & b_m \\ \end{array}\right]\] for some variables \(x_1,\ldots,x_n\), and some numbers \(a_{ij}\)'s and \(b_1,\ldots,b_m\).
When \(b_1 = \cdots = b_m = 0\), it is a homogeneous system of linear equations.
An \(m\times n\) matrix \(A\) over \(\mathbb{R}\) is array
\[\begin{bmatrix} a_{11} & \cdots & a_{1n} \\ \vdots & ~ & \vdots \\ a_{m1} & \cdots & a_{mn} \\ \end{bmatrix}\]
for some numbers \(a_{ij}\)'s.
Matrix-vector multiplication (by entry)
Let \(A = \begin{bmatrix} a_{ij}\end{bmatrix}\) be an \(m\times n\) matrix and \({\bf v} = (v_1,\ldots,v_n)\) a vector in \(\mathbb{R}^n\).
Then \(A{\bf v}\) is a vector in \(\mathbb{R}^m\) whose \(i\)-th entry is
\[\sum_{k=1}^n a_{ik}v_k. \]
Thus, every system of linear equation can be written as \(A{\bf x} = {\bf b}\), while
it is homogeneous when \({\bf b} = {\bf 0}\).
The solution set of \(A{\bf v} = {\bf 0}\) is called the kernel of \(A\), denoted as \(\operatorname{ker}(A)\).
That is, \(\ker(A) = \{{\bf v}\in\mathbb{R}^n : A{\bf v} = {\bf 0}\}\).
The kernel of an \(m\times n\) matrix is a subspace in \(\mathbb{R}^n\).
Side stories
Experiments
Exercise 1
執行下方程式碼。
原點為橘色點、 從原點延伸出去的紅色向量和淡藍色向量分別為 \({\bf u}_1\) 和 \({\bf u}_2\)。
黑色向量為 \({\bf b}\)。
問 \({\bf b}\) 是否是 \(\{{\bf u}_1, {\bf u}_2\}\) 的線性組合?
若是﹐求 \(c_1,c_2\) 使得 \({\bf b} = c_1{\bf u}_1 + c_2{\bf u}_2\)。
答:[由林子翔同學提供]
當
seed = 0
時,題目給的數字為\(\bb = (2,16,10)\),
\({\bf u}_1= (-4,3,5)\),
\({\bf u}_2=(-5,-5,0)\)。
再把 \({\bf b} = c_1{\bf u}_1 + c_2{\bf u}_2\) 寫成 \[ \begin{bmatrix}2\\16\\10\end{bmatrix}= c_1\begin{bmatrix}-4\\3\\5\end{bmatrix}+ c_2\begin{bmatrix}-5\\-5\\0\end{bmatrix}, \] 可推得
\[ \left\{\begin{aligned} -4c_1-5c_2 &= 2, \\ 3c_1-5c_2 &= 16, \\ 5c_1+0c_2 &= 10. \end{aligned}\right. \] 因此 \(c_1 = 2\), \(c_2 = -2\)。
帶回原式 \({\bf b} = c_1{\bf u}_1 + c_2{\bf u}_2\),
可推得 \({\bf b} = 2{\bf u}_1 + -2{\bf u}_2\),
故 \({\bf b}\) 是 \(\{{\bf u}_1, {\bf u}_2\}\) 的線性組合。
Exercises
Exercise 2(a)
令
\[{\bf e}_1 = \begin{bmatrix}1\\0\\0\end{bmatrix}, {\bf e}_2 = \begin{bmatrix}0\\1\\0\end{bmatrix}, {\bf e}_3 = \begin{bmatrix}0\\0\\1\end{bmatrix}. \]
說明 \(\mathbb{R}^3 = \operatorname{span}(\{{\bf e}_1, {\bf e}_2, {\bf e}_3\})\)﹐
因此它是 個子空間。
(要說明每一個 \(\mathbb{R}^3\) 中的向量都可以寫成 \(c_1{\bf e}_1 + c_2{\bf e}_2 + c_3{\bf e}_3\) 的形式。)
Let \[\begin{bmatrix}a_1\\a_2\\a_3\end{bmatrix}\in \mathbb{R}^3.\] We may solve\[c_1 \begin{bmatrix}1\\0\\0\end{bmatrix}+ c_2 \begin{bmatrix}0\\1\\0\end{bmatrix}+ c_3 \begin{bmatrix}0\\0\\1\end{bmatrix}= \begin{bmatrix}a_1\\a_2\\a_3\end{bmatrix}\] and get \[ c_1=a_1 ,c_2=a_2,c_3=a_3. \] Therefore, every vector in \(\mathbb{R}^3\) can be written as a linear combination of \(\{\be_1,\be_2,\be_3\}\).
Exercise 2(b)
令
\[{\bf b} = \begin{bmatrix}1\\2\\-3\end{bmatrix}, {\bf u}_1 = \begin{bmatrix}-1\\1\\0\end{bmatrix}, {\bf u}_2 = \begin{bmatrix}0\\-1\\1\end{bmatrix}. \]
說明 \({\bf b}\in\operatorname{span}(\{{\bf u}_1, {\bf u}_2\})\)。
Consider the equation \[ \begin{bmatrix}1\\2\\-3\end{bmatrix}= c_1\begin{bmatrix}-1\\1\\0\end{bmatrix}+ c_2\begin{bmatrix}0\\-1\\1\end{bmatrix} \] , which is equivalent to \[ \left\{\begin{aligned} -c_1+0 &= 1, \\ c_1-c_2 &= 2, \\ 0+c_2 &= -3. \end{aligned}\right. \] Then \[ c_1=-1,c_2=-3. \]
Therefore, \[ {\bf b}=(-1)u_1+(-3)u_2 \] and \[ {\bf b}\in\operatorname{span}(\{{\bf u}_1,{\bf u}_2\}) .\]
Exercise 2©
令 \[{\bf b} = \begin{bmatrix}1\\1\\1\end{bmatrix}, {\bf u}_1 = \begin{bmatrix}-1\\1\\0\end{bmatrix}, {\bf u}_2 = \begin{bmatrix}0\\-1\\1\end{bmatrix}. \]
說明 \({\bf b}\notin\operatorname{span}(\{{\bf u}_1, {\bf u}_2\})\)。
Consider the equation \[ \begin{bmatrix}1\\1\\1\end{bmatrix}= c_1\begin{bmatrix}-1\\1\\0\end{bmatrix}+ c_2\begin{bmatrix}0\\-1\\1\end{bmatrix}, \] which is equivalent to \[ \left\{\begin{aligned} -c_1+0 &= 1, \\ c_1 - c_2 &= 1, \\ 0+c_2 &= 1, \end{aligned}\right. \] which implies \[ \left\{\begin{aligned} c_1 &= -1, \\ c_2 &= -2, \\ c_2 &= 1. \end{aligned}\right. \] However, \(c_2\) cannot be both \(-2\) and \(1\).
Since \(\bb\) cannot be expressed as \(c_1 \bu_1 +c_2 \bu_2\) , \({\bf b}\notin\operatorname{span}(\{{\bf u}_1,{\bf u}_2\})\).
Exercise 2(d)
令
\[{\bf b} = \begin{bmatrix}1\\2\\-3\end{bmatrix}, {\bf u}_1 = \begin{bmatrix}-1\\1\\0\end{bmatrix}, {\bf u}_2 = \begin{bmatrix}0\\-1\\1\end{bmatrix}, {\bf u}_3 = \begin{bmatrix}-1\\0\\1\end{bmatrix}. \]
找出至少兩組 \(c_1, c_2, c_3\) 使得 \({\bf b} = c_1{\bf u}_1 + c_2{\bf u}_2 + c_3{\bf u}_3\)。
這說明了線性組合的表示法不見得唯一。
We may solve the equation \[ \left\{\begin{aligned} -c_1-c_3 &=1,\\ c_1-c_2 &=2,\\ c_2+c_3 &=3. \end{aligned}\right. \] And get two solutions \[ c_1=1,c_2=-1,c_3=-2,\\ c_1=-1,c_2=-3,c_3=0. \]
Exercise 3
依照各小題的步驟來證明子空間的兩個條件等價。
並用這些條件來判斷一個集合是否為子空間。
Exercise 3(a)
證明若條件 1 成立則條件 2 成立。
Suppose \(V = \operatorname{span}(S)\) for some \(S\subseteq\mathbb{R}^n\).
We verify each of the requirements in condition 2.
nonempty
Since \(\operatorname{span}(S)\) always contains zero vector, \(V\) is nonempty.
closed under scalar multiplication
Suppose \({\bf v}\in V\) and \(k\) a scalar.
Then \({\bf v}\) can be written as \(c_1{\bf u}_1 + \cdots + c_k{\bf u}_k\) for some vectors \({\bf u}_i\in S\) and scalars \(c_i\).
Then \(k{\bf v}=kc_1{\bf u}_1 + \cdots + kc_k{\bf u}_k=d_1{\bf u}_1 + \cdots + d_k{\bf u}_k\) for scalars \(kc_i=d_i\).
So \(k{\bf v}\in\operatorname{span}(S) = V\).
closed under vector addition
Suppose \({\bf v}_1,{\bf v}_2\in V\).
Then \({\bf v}_1\) can be written as \(a_1{\bf u}_1 + \cdots + a_k{\bf u}_k\) for some vectors \({\bf u}_i\in S\) and scalars \(a_i\),
and \({\bf v}_2\) can be written as \(b_1{\bf u}_1 + \cdots + b_k{\bf u}_k\) for some vectors \({\bf u}_i\in S\) and scalars \(b_i\).
Thus, \({\bf v}_1 + {\bf v}_2=(a_1+b_1){\bf u}_1 + \cdots + (a_k+b_k){\bf u}_k=c_1{\bf u}_1 + \cdots + c_k{\bf u}_k\) for scalars \(a_i+b_i=c_i\).
So \({\bf v}_1 + {\bf v}_2 \in\operatorname{span}(S) = V\).
Exercise 3(b)
證明若條件 2 成立則條件 1 成立。
Suppose \(V\) is a nonempty subset of \(\mathbb{R}^n\) and is closed under scalar multiplication and vector addition.
It is enough to show that \(V = \operatorname{span}(V)\).
\(V\subseteq\operatorname{span}(V)\)
Clearly, each element in \(V\) is a linear combination of \(V\) and is in \(\vspan(V)\).
\(\operatorname{span}(V)\subseteq V\)
Let \({\bf u}\) be an element of \(\operatorname{span}(V)\).
Then \({\bf u}\) can be written as \(c_1{\bf u}_1 + \cdots + c_k{\bf u}_k\) for some vectors \({\bf u}_i\in V\) and scalars \(c_i\).
From the assumption, \(V\) is closed under scalar multiplication and vector addition, so \({\bf u}\in V\).
Exercise 3©
判斷 \(\emptyset\) 是否為一子空間。
Because \(\emptyset\) not conform with "\(V\) is a nonempty subset" in condition2, so \(\emptyset\) is not a subspace.
Exercise 3(d)
判斷 \(\{{\bf 0}\}\) 是否為一子空間。
Because we vacuously define \(\operatorname{span}(\emptyset) = \{{\bf 0}\}\), so \(\{{\bf 0}\}\) is a subspace.
Exercise 3(e)
判斷 \(\left\{\begin{bmatrix}1\\1\\1\end{bmatrix}\right\}\) 是否為一子空間。
觀察\(\begin{bmatrix}1\\1\\1\end{bmatrix} +\begin{bmatrix}1\\1\\1\end{bmatrix} = \begin{bmatrix}2\\2\\2\end{bmatrix}\),發現集合中的兩向量相加會得到一個集合外的向量。
因為加法沒有封閉性,所以 \(\left\{\begin{bmatrix}1\\1\\1\end{bmatrix}\right\}\) 不為一子空間。
Exercise 3(f)
判斷 \(\left\{t\begin{bmatrix}1\\1\\1\end{bmatrix} : t\in\mathbb{R}\right\}\) 是否為一子空間。
(1) \(V\) 通過原點。
(2) Let \({\bf v}_1,{\bf v}_2\in \left\{t\begin{bmatrix}1\\1\\1\end{bmatrix} :t\in\mathbb{R}\right\}.\)
Then \[ {\bf v}_1 = t_1\begin{bmatrix}1\\1\\1\end{bmatrix}, {\bf v}_2 = t_2\begin{bmatrix}1\\1\\1\end{bmatrix} \] for some \(t_1,t_2\in\mathbb{R}\).
Then \[ {\bf v}_1 + {\bf v}_2 = \begin{bmatrix}t_1\\t_1\\t_1\end{bmatrix} +\begin{bmatrix}t_2\\t_2\\t_2\end{bmatrix} = \begin{bmatrix}t_1 + t_2\\t_1 + t_2\\t_1 + t_2\end{bmatrix} = t_1 + t_2\begin{bmatrix}1\\1\\1\end{bmatrix}\in\left\{t\begin{bmatrix}1\\1\\1\end{bmatrix} :t\in\mathbb{R}\right\} \] so, \(\left\{t\begin{bmatrix}1\\1\\1\end{bmatrix} : t\in\mathbb{R}\right\}\) is a subspace.
Exercise 3(g)
判斷 \(\left\{\begin{bmatrix}x\\y\\z\end{bmatrix} : x^2 + y^2 + z^2 = 1\right\}\) 是否為一子空間。
Let \[ {\bf v}_1 = \begin{bmatrix}1\\0\\0\end{bmatrix}, {\bf v}_2 = \begin{bmatrix}0\\1\\0\end{bmatrix}. \] Then \({\bf v}_1,{\bf v}_2\in \left\{\begin{bmatrix}x\\y\\z\end{bmatrix} : x^2 + y^2 + z^2 = 1\right\}\).
But \({\bf v}_1+{\bf v}_2\notin \left\{\begin{bmatrix}x\\y\\z\end{bmatrix} : x^2 + y^2 + z^2 = 1\right\}\).
So \(\left\{\begin{bmatrix}x\\y\\z\end{bmatrix} : x^2 + y^2 + z^2 = 1\right\}\) is not a subspace.
Exercise 3(h)
判斷 \(\left\{\begin{bmatrix}x\\y\\z\end{bmatrix} : x^2 + y^2 + z^2 \geq 1\right\}\) 是否為一子空間。
這題完全沒問題喔!
\(\left\{\begin{bmatrix}x\\y\\z\end{bmatrix} : x^2 + y^2 + z^2 \geq 1\right\}\)不通過原點,因此不為子空間。
Exercise 3(i)
判斷 \(\left\{\begin{bmatrix}x\\y\\z\end{bmatrix} : x \geq 0\right\}\) 是否為一子空間。
[由林暐智同學提供]
Let \(\bv_1 = \begin{bmatrix}1\\0\\0\end{bmatrix}\) and \(k=-100\).
Then \(\bv_1 \in \left\{\begin{bmatrix}x\\y\\z\end{bmatrix} : x \geq 0\right\}\), but \(k{\bf v}_1\notin \left\{\begin{bmatrix}x\\y\\z\end{bmatrix} : x \geq 0\right\}\). So \(\left\{\begin{bmatrix}x\\y\\z\end{bmatrix} : x \geq 0\right\}\) is not a subspace.
Exercise 3(j)
判斷 \(\left\{\begin{bmatrix}x\\y\\z\end{bmatrix} : x + y + z = 0\right\}\) 是否為一子空間。
令\(\left\{\begin{bmatrix}x\\y\\z\end{bmatrix} : x + y + z = 0\right\} = V\)。
(1) \(V\) 通過原點。
(2) 令
\[{\bf v}_1 = \begin{bmatrix}a_1\\b_1\\c_1\end{bmatrix}\in\ V, {\bf v}_2 = \begin{bmatrix}a_2\\b_2\\c_2\end{bmatrix}\in\ V,p, q\in\mathbb{R}. \]
然後把兩者相加得到 \(p{\bf v}_1 + q{\bf v}_2 = \begin{bmatrix}pa_1 + qa_2\\pb_1 + qb_2 \\pc_1 + qc_1\end{bmatrix}\)。
將其帶入 \(x + y + z\) 中,發現
\[ (pa_1 + qa_2) + (pb_1 + qb_2) + (pc_1 + qc_2) = p(a_1 + b_1 + c_1) + q( a_2 + b_2 + c_2) = 0 \] 符合條件,因此 \(\left\{\begin{bmatrix}x\\y\\z\end{bmatrix} : x + y + z = 0\right\}\) 為子空間。
Exercise 3(k)
判斷 \(\left\{\begin{bmatrix}x\\y\\z\end{bmatrix} : x + y + z = 1\right\}\) 是否為一子空間。
\(\left\{\begin{bmatrix}x\\y\\z\end{bmatrix} : x + y + z = 1\right\}\) 不通過原點,因此不為子空間。
Exercise 3(l)
判斷 \(\left\{\begin{bmatrix}x\\y\\z\end{bmatrix} : \begin{aligned} x + y + z &= 0 \\ x + 2y + 3z &= 0 \end{aligned} \right\}\) 是否為一子空間。
令 \(\left\{\begin{bmatrix}x\\y\\z\end{bmatrix} : \begin{aligned} x + y + z &= 0 \\ x + 2y + 3z &= 0 \end{aligned} \right\} = V\)。
(1) \(V\) 通過原點。
(2) 令
\[{\bf v}_1 = \begin{bmatrix}a_1\\b_1\\c_1\end{bmatrix}\in\ V, {\bf v}_2 = \begin{bmatrix}a_2\\b_2\\c_2\end{bmatrix}\in\ V,p,q \in\mathbb{R}. \] 然後把兩者相加,\(p{\bf v}_1 + q{\bf v}_2 = \begin{bmatrix}pa_1 + qa_2\\pb_1 + qb_2 \\pc_1 + qc_1\end{bmatrix}\), 帶入\(x + y + z\)中,
\((pa_1 + qa_2) + (pb_1 + qb_2) + (pc_1 + qc_2) = p(a_1 + b_1 + c_1) + q(a_2 + b_2 + c_2) = 0.\)
接著,帶入\(x + 2y + 3z\),\((pa_1 + qa_2) + 2(pb_1 + qb_2) + 3(pc_1 + qc_2) = p(a_1 + 2b_1 + 3c_1) + q(a_2 + 2b_2 + 3c_2) = 0\) 符合兩條件,因此 \(\left\{\begin{bmatrix}x\\y\\z\end{bmatrix} : \begin{aligned} x + y + z &= 0 \\ x + 2y + 3z &= 0 \end{aligned} \right\}\) 為子空間
Exercise 4(a)
證明對任何 \(S\subseteq\mathbb{R}^n\) 都有 \(\operatorname{span}(\operatorname{span}(S)) = \operatorname{span}(S)\)。
First, we clearly know that \(\operatorname{span}(S)\) can be spanned by itself.
Then, we have \(\operatorname{span}(S)\subseteq\operatorname{span}(\operatorname{span}(S))\).
Next, let \(\bv\in\operatorname{span}(\operatorname{span}(S))\).
Then, \(\bv=c_1{\bf u}_1 + \cdots + c_k{\bf u}_k\), for some vectors \({\bf u}_1,\ldots, {\bf u}_k\in \operatorname{span}(S)\) for some scalars \(c_1,\ldots, c_k\in\mathbb{R}\).
So, \({\bf u}_i\) can be written as \(d_{i1}{\bf s}_1 + \cdots + d_{ik}{\bf s}_k\) for some vectors \({\bf s}_i\in S\) and scalars \(d_{ij}\).
In other words, \(\bv=(\sum_{i=1}^k c_{i}d_{i1}){\bf s}_1 + \cdots + (\sum_{i=1}^k c_{i}d_{ik}){\bf s}_k\in\operatorname{span}(S)\).
Therefore, \(\operatorname{span}(\operatorname{span}(S))\subseteq\operatorname{span}(S)\).
Finally, we can conclude \(\operatorname{span}(\operatorname{span}(S)) = \operatorname{span}(S)\).
Exercise 4(b)
令 \(S\subseteq\mathbb{R}^n\)。
證明以下敘述等價:
From 1. to 2.
Assume \({\bf w}\in\operatorname{span}(S)\).
Clearly, \(\operatorname{span}(S)\subseteq \operatorname{span}(S\cup\{{\bf w}\})\).
Let \({\bf v}\in\operatorname{span}(S\cup\{{\bf w}\})\).
Then \({\bf v}\) can be written as \(c_1{\bf u}_1 + \cdots + c_k{\bf u}_k+c_{k+1}{\bf w}\) for some vectors \({\bf u}_i\in S\) and scalars \(c_i\).
From the assumption, \({\bf w}\in\operatorname{span}(S)\), we have \({\bf v}\in\operatorname{span}(S)\).
Then, we get \(\operatorname{span}(S\cup\{{\bf w}\}) \subseteq \operatorname{span}(S)\).
So, \(\operatorname{span}(S\cup\{{\bf w}\}) = \operatorname{span}(S)\).
From 2. to 1.
Assume \(\operatorname{span}(S\cup\{{\bf w}\}) = \operatorname{span}(S)\).
Since \(\bw\in S\cup\{\bw\}\), \(\bw\in\vspan(S\cup\{\bw\})\).
格式可改進 數學除了最後一題都正確
目前分數:4.5/5