This work by Jephian Lin is licensed under a Creative Commons Attribution 4.0 International License.
Subspaces in $\mathbb{R}^n$
This work by Jephian Lin is licensed under a Creative Commons Attribution 4.0 International License.
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from lingeo import random_int_list, draw_span
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
$$c_1\bu_1 + \cdots + c_k\bu_k,$$
for some vectors $\bu_1,\ldots, \bu_k\in S$ 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 $\vspan(S)$.
(We vacuously define $\vspan(\emptyset) = \{\bzero\}$.)
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.
Let $A = \begin{bmatrix} a_{ij}\end{bmatrix}$ be an $m\times n$ matrix and $\bv = (v_1,\ldots,v_n)$ a vector in $\mathbb{R}^n$.
Then $A\bv$ 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 $A\bx = \bb$, while
it is homogeneous when $\bb = \bzero$.
The solution set of $A\bv = \bzero$ is called the kernel of $A$, denoted as $\ker(A)$.
That is, $\ker(A) = \{\bv\in\mathbb{R}^n : A\bv = \bzero\}$.
The kernel of an $m\times n$ matrix is a subspace in $\mathbb{R}^n$.
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 $\bu_1$ 和 $\bu_2$。
黑色向量為 $\bb$。
問 $\bb$ 是否是 $\{\bu_1, \bu_2\}$ 的線性組合？
若是，求 $c_1,c_2$ 使得 $\bb = c_1\bu_1 + c_2\bu_2$。
Run the code below. Let the origin be the orange point. Let $\bu_1$ and $\bu_2$ be the red and blue vectors with its tails at the origin. Let $\bb$ be the black vector. Is $\bb$ a linear combination of $\{\bu_1, \bu_2\}$? If yes, find $c_1$ and $c_2$ so that $\bb = c_1\bu_1 + c_2\bu_2$.
### code
set_random_seed(0)
print_ans = False
while True:
l = random_int_list(9)
A = matrix(3, l)
if A.det() != 0:
break
u1 = vector(A[0])
u2 = vector(A[1])
u3 = vector(A[2])
inside = choice([0,1,1])
coefs = random_int_list(2, 2)
if inside:
b = coefs[0]*u1 + coefs[1]*u2
else:
b = coefs[0]*u1 + coefs[1]*u2 + 3*u3
print("u1 =", u1)
print("u2 =", u2)
print("b =", b)
pic = draw_span([u1,u2])
pic += arrow((0,0,0), b, width=5, color="black")
show(pic)
if print_ans:
if inside:
print("b is a linear combination of { u1, u2 } since b = %s u1 + %s u2."%(coefs[0], coefs[1]))
else:
print("b is not a linear combination of { u1, u2 }.")
令
$$ \be_1 = \begin{bmatrix}1\\0\\0\end{bmatrix}, \be_2 = \begin{bmatrix}0\\1\\0\end{bmatrix}, \be_3 = \begin{bmatrix}0\\0\\1\end{bmatrix}. $$說明 $\mathbb{R}^3 = \vspan(\{\be_1, \be_2, \be_3\})$，
因此它是一個子空間。
（要說明每一個 $\mathbb{R}^3$ 中的向量都可以寫成 $c_1\be_1 + c_2\be_2 + c_3\be_3$ 的形式。）
Let
$$ \be_1 = \begin{bmatrix}1\\0\\0\end{bmatrix}, \be_2 = \begin{bmatrix}0\\1\\0\end{bmatrix}, \be_3 = \begin{bmatrix}0\\0\\1\end{bmatrix}. $$Explain why $\mathbb{R}^3 = \vspan(\{\be_1, \be_2, \be_3\})$. As a consequence, it is a subspace. (You have to show that every vector in $\mathbb{R}^3$ can be written as $c_1\be_1 + c_2\be_2 + c_3\be_3$.)
令
$$ \bb = \begin{bmatrix}1\\2\\-3\end{bmatrix}, \bu_1 = \begin{bmatrix}-1\\1\\0\end{bmatrix}, \bu_2 = \begin{bmatrix}0\\-1\\1\end{bmatrix}. $$說明 $\bb\in\vspan(\{\bu_1, \bu_2\})$。
Let
$$ \bb = \begin{bmatrix}1\\2\\-3\end{bmatrix}, \bu_1 = \begin{bmatrix}-1\\1\\0\end{bmatrix}, \bu_2 = \begin{bmatrix}0\\-1\\1\end{bmatrix}. $$Explain why $\bb\in\vspan(\{\bu_1, \bu_2\})$.
令
$$ \bb = \begin{bmatrix}1\\1\\1\end{bmatrix}, \bu_1 = \begin{bmatrix}-1\\1\\0\end{bmatrix}, \bu_2 = \begin{bmatrix}0\\-1\\1\end{bmatrix}. $$說明 $\bb\notin\vspan(\{\bu_1, \bu_2\})$。
Let
$$ \bb = \begin{bmatrix}1\\1\\1\end{bmatrix}, \bu_1 = \begin{bmatrix}-1\\1\\0\end{bmatrix}, \bu_2 = \begin{bmatrix}0\\-1\\1\end{bmatrix}. $$Explain why $\bb\notin\vspan(\{\bu_1, \bu_2\})$.
令
$$ \bb = \begin{bmatrix}1\\2\\-3\end{bmatrix}, \bu_1 = \begin{bmatrix}-1\\1\\0\end{bmatrix}, \bu_2 = \begin{bmatrix}0\\-1\\1\end{bmatrix}, \bu_3 = \begin{bmatrix}-1\\0\\1\end{bmatrix}. $$找出至少兩組 $c_1, c_2, c_3$ 使得 $\bb = c_1\bu_1 + c_2\bu_2 + c_3\bu_3$。
這說明了線性組合的表示法不見得唯一。
Let
$$ \bb = \begin{bmatrix}1\\2\\-3\end{bmatrix}, \bu_1 = \begin{bmatrix}-1\\1\\0\end{bmatrix}, \bu_2 = \begin{bmatrix}0\\-1\\1\end{bmatrix}, \bu_3 = \begin{bmatrix}-1\\0\\1\end{bmatrix}. $$Find at least two combinations of $c_1, c_2, c_3$ such that $\bb = c_1\bu_1 + c_2\bu_2 + c_3\bu_3$. This shows that the forms of linear combinations might not be unique.
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。
Use the given instructions to show the following statements are equivalent.
Then you may use the equivalent conditions to check if a set is a subspace.
Sample:
Suppose $V = \vspan(S)$ for some $S\subseteq\mathbb{R}^n$.
We verify each of the requirements in condition 2.
nonempty
Since $\vspan(S)$ always contains ..., $V$ is nonempty.
closed under scalar multiplication
Suppose $\bv\in V$ and $k$ a scalar.
Then $\bv$ can be written as $c_1\bu_1 + \cdots + c_k\bu_k$ for some vectors $\bu_i\in S$ and scalars $c_i$.
Then ...
So $k\bv\in\vspan(S) = V$.
closed under vector addition
Suppose $\bv_1,\bv_2\in V$.
Then $\bv_1$ can be written as ...,
and $\bv_2$ can be written as ...
Then ...
So $\bv_1 + \bv_2 \in\vspan(S) = V$.
Sample:
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 $V = \vspan(V)$.
$V\subseteq\vspan(V)$
...
$\vspan(V)\subseteq V$
Let $\bu$ be an element of $\vspan(V)$.
Then $\bu$ can be written as ...
...
Therefore, $\bu\in V$.
判斷 $\left\{\begin{bmatrix}1\\1\\1\end{bmatrix}\right\}$ 是否為一子空間。
Check if $\left\{\begin{bmatrix}1\\1\\1\end{bmatrix}\right\}$ is a subspace.
判斷 $\left\{t\begin{bmatrix}1\\1\\1\end{bmatrix} : t\in\mathbb{R}\right\}$ 是否為一子空間。
Check if $\left\{t\begin{bmatrix}1\\1\\1\end{bmatrix} : t\in\mathbb{R}\right\}$ is a subspace.
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。
Check if $\left\{\begin{bmatrix}x\\y\\z\end{bmatrix} : x^2 + y^2 + z^2 = 1\right\}$ is a subspace.
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。
Check if $\left\{\begin{bmatrix}x\\y\\z\end{bmatrix} : x^2 + y^2 + z^2 \geq 1\right\}$ is a subspace.
判斷 $\left\{\begin{bmatrix}x\\y\\z\end{bmatrix} : x \geq 0\right\}$ 是否為一子空間。
Check if $\left\{\begin{bmatrix}x\\y\\z\end{bmatrix} : x \geq 0\right\}$ is a subspace.
判斷 $\left\{\begin{bmatrix}x\\y\\z\end{bmatrix} : x + y + z = 0\right\}$ 是否為一子空間。
Check if $\left\{\begin{bmatrix}x\\y\\z\end{bmatrix} : x + y + z = 0\right\}$ is a subspace.
判斷 $\left\{\begin{bmatrix}x\\y\\z\end{bmatrix} : x + y + z = 1\right\}$ 是否為一子空間。
Check if $\left\{\begin{bmatrix}x\\y\\z\end{bmatrix} : x + y + z = 1\right\}$ is a subspace.
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。
Check if $\left\{\begin{bmatrix}x\\y\\z\end{bmatrix} : \begin{aligned} x + y + z &= 0 \\ x + 2y + 3z &= 0 \end{aligned} \right\}$ is a subspace.
證明對任何 $S\subseteq\mathbb{R}^n$ 都有 $\vspan(\vspan(S)) = \vspan(S)$。
Show that $\vspan(\vspan(S)) = \vspan(S)$ for any subset $S\subseteq\mathbb{R}^n$.
令 $S\subseteq\mathbb{R}^n$。
證明以下敘述等價：
Let $S\subseteq\mathbb{R}^n$. Show the following statements are equivalent.