# 建構新的向量空間

Constructing new vector spaces

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

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from lingeo import random_good_matrix
from linspace import vtop, vtom

## Main idea

Let $S_1$ and $S_2$ be two set.
The **Cartesian product** of $S_1$ and $S_2$ is

$$
S_1 \times S_2 = \{ (s_1,s_2) : s_1\in S_1, s_2\in S_2 \}.
$$

If $S_1$ and $S_2$ are finite sets, then $|S_1\times S_2| = |S_1|\times |S_2|$.

Let $U$ and $V$ be two vector spaces.
The **Cartesian product** of $U$ and $V$ is the set

$$
U\times V = \{ (\bu, \bv) : \bu\in U, \bv\in V \}
$$

along with the vector addition

$$
(\bu_1, \bv_1) + (\bu_1, \bv_1) = (\bu_1 + \bu_2, \bv_1 + \bv_2)
$$
and the scalr multiplication

$$
k(\bu, \bv) = (k\bu, k\bv).
$$

The Cartesian product of two vector spaces is again a vector space.
For example, $\mathbb{R}^2 = \mathbb{R} \times \mathbb{R}$.

Suppose $\beta_U$ and $\beta_V$ are bases of $U$ and $V$, respectively.
Then

$$
\{ (\bu, \bzero_V) : \bu\in \beta_U \} \cup \{ (\bzero_U, \bv) : \bv\in \beta_V \}
$$

is a basis of $U \times V$, where $\bzero_U$ and $\bzero_V$ are the zero vectors in $U$ and $V$, respectively.
Therefore, $\dim(U \times V) = \dim(U) + \dim(V)$ if both of $U$ and $V$ are finite-dimensional.

Let $U$ be a vector space and $V$ a subspace of $U$.
Recall that an affine subspace is of the form $\bu + V$ for some vector $\bu$.
Thus, the **quotient space** of $U$ by $V$ is the set of all affine subspaces

$$
U / V = \{ \bu + V : \bu \in U\}
$$

(here each affine subspace $\bu + V$ is treated as a vector)
along with the vector addition

$$
(\bu_1 + V) + (\bu_2 + V) = (\bu_1 + \bu_2) + V
$$

and the scalar multiplication

$$
k(\bu + V) = (k\bu) + V.
$$

For example, when $V$ is the $x,y$-plane, then the structure of $\mathbb{R}^2 / V$ is similar to $\mathbb{R}^1$, since each $z$ value decides an affine plane.

By the expanding lemma, one may obtain a basis $\beta_V$ of $V$ and expand it to a basis $\beta_U$.
Thus,

$$
\{ \bu + V : \bu \in \beta_U \setminus \beta_V \}
$$

is a basis of $U / V$.
(Note that $/$ is the quotient while $\setminus$ is the setminus.)
Therefore, $\dim(U / V) = \dim(U) - \dim(V)$ if both $U$ and $V$ are finite-dimensional.

## Side stories
- well-defined

## Experiments

##### Exercise 1

執行以下程式碼。
考慮向量空間 $\mathbb{R}^3\times \mathcal{P}_1$。

<!-- eng start -->
Run the code below. Consider the vector space $\mathbb{R}^3 \times \mathcal{P}_1$.
<!-- eng end -->

### code
set_random_seed(0)
print_ans = False
m,n,r = 3,5,choice([2,3])
A = random_good_matrix(m,n,r)
v1,v2,v3 = A[:,:3]
p1,p2,p3 = [vtop(v) for v in A[:,3:]]

print("u1 = (v1, p1) =", (v1, p1))
print("u2 = (v2, p2) =", (v2, p2))
print("u3 = (v3, p3) =", (v3, p3))


if print_ans:
    print("u1 + u2 =", (v1 + v2, p1 + p2))
    print("Linear independent?", r == 3)

##### Exercise 1(a)

計算 $\bu_1 + \bu_2$。

<!-- eng start -->
Calculate $\bu_1 + \bu_2$.
<!-- eng end -->

##### Exercise 1(b)

判斷 $\{ \bu_1, \bu_2, \bu_3 \}$ 是否線性獨立。

<!-- eng start -->
Is $\{ \bu_1, \bu_2, \bu_3 \}$ linearly independent?
<!-- eng end -->

## Exercises

##### Exercise 2

考慮 $V = \mathbb{R}^3 \times \mathbb{R}^2$。

<!-- eng start -->
Consider $V = \mathbb{R}^3 \times \mathbb{R}^2$.
<!-- eng end -->

##### Exercise 2(a)

求 $V$ 中的零向量。

<!-- eng start -->
What is the zero vector in $V$?
<!-- eng end -->

##### Exercise 2(b)

令

$$
\begin{aligned}
\bv_1 &= ((1,1,1), (1,1)) \\
\bv_2 &= ((0,1,1), (1,1)) \\
\bv_3 &= ((0,0,1), (1,1)) \\
\end{aligned}
$$

且 $S = \{ \bv_1, \bv_2, \bv_3 \}$。
判斷 $\vspan(S)$ 是否可以生成全空間 $V$。

<!-- eng start -->
Let

$$
\begin{aligned}
\bv_1 &= ((1,1,1), (1,1)) \\
\bv_2 &= ((0,1,1), (1,1)) \\
\bv_3 &= ((0,0,1), (1,1)) \\
\end{aligned}
$$

and $S = \{ \bv_1, \bv_2, \bv_3 \}$. Determine if $\vspan(S)$ equal the whole space $V$.
<!-- eng end -->

##### Exercise 2(c)

判斷 $S$ 是否線性獨立。

<!-- eng start -->
Is $S$ linearly independent?
<!-- eng end -->

##### Exercise 3

考慮 $V = \mathcal{P}_2 \times \mathcal{P}_1$。
令

$$
\begin{aligned}
p_1 &= (x+1)(x+2), \\
p_2 &= (x+1)(x^2 + x + 1). \\
\end{aligned}
$$

<!-- eng start -->
Consider the vector space $V = \mathcal{P}_2 \times \mathcal{P}_1$. Let

$$
\begin{aligned}
p_1 &= (x+1)(x+2), \\
p_2 &= (x+1)(x^2 + x + 1). \\
\end{aligned}
$$
<!-- eng end -->

##### Exercise 3(a)

令 $\operatorname{ptov}_d$ 為把 $\mathcal{P}_d$ 中的多項式寫為 $\mathbb{R}^{d+1}$ 中向量的函數。
建一個矩陣 $A$ 其行向量分別為
$\operatorname{ptov}_4(p_1),
\operatorname{ptov}_4(xp_1),
\operatorname{ptov}_4(x^2p_1),
\operatorname{ptov}_4(p_2),
\operatorname{ptov}_4(xp_2)$。
寫出 $A$。

<!-- eng start -->
Let $\operatorname{ptov}_d$ be the function sending a polynomial in $\mathcal{P}_d$ to a vector in $\mathbb{R}^{d+1}$ by recording its coefficients from lower degree to higher dergee. Construct a matrix $A$ whose columns are $\operatorname{ptov}_4(p_1)$, $\operatorname{ptov}_4(xp_1)$, $\operatorname{ptov}_4(x^2p_1)$, $\operatorname{ptov}_4(p_2)$, and $\operatorname{ptov}_4(xp_2)$. Calculate $A$.
<!-- eng end -->

##### Exercise 3(b)

驗證對任何 $a\in\mathcal{P}_2$ 及 $b\in\mathcal{P}_1$

$$
A \begin{bmatrix}
\operatorname{ptov}_2(a) \\
\operatorname{ptov}_1(b)
\end{bmatrix} =
\operatorname{ptov}_4(ap_1 + bp_2)
$$

都成立。

<!-- eng start -->
Show that

$$
A \begin{bmatrix}
\operatorname{ptov}_2(a) \\
\operatorname{ptov}_1(b)
\end{bmatrix} =
\operatorname{ptov}_4(ap_1 + bp_2)
$$

for any $a\in\mathcal{P}_2$ and $b\in\mathcal{P}_1$.
<!-- eng end -->

##### Exercise 3(c)

求出所有可以讓 $ap_1 + bp_2 = 0$ 的 $(a,b)\in V$。

<!-- eng start -->
Find all $(a,b)\in V$ such that $ap_1 + bp_2 = 0$.
<!-- eng end -->

##### Exercise 4

令 $U$ 為一向量空間而 $V$ 為其一子空間。

<!-- eng start -->
Let $U$ be a vector space and $V$ its subspace.
<!-- eng end -->

##### Exercise 4(a)

證明以下敘述等價：

1. $\bu_1 + V = \bu_2 + V$.
2. $\bu_1 - \bu_2 \in V$.

因此另外一個定義商空間的方法是定義向量之間的關係：

$$
\bu_1 \sim \bu_2 \iff \bu_1 - \bu_2 \in V.
$$

可以證明這樣的關係是一個等價關係。
如此一來 $U / \sim$ 和 $U / V$ 的概念是一樣的。

<!-- eng start -->
Show that the following are equivalent:

1. $\bu_1 + V = \bu_2 + V$.
2. $\bu_1 - \bu_2 \in V$.

Therefore, the quotient space can be defined by the relation of vectors in $V$:

$$
\bu_1 \sim \bu_2 \iff \bu_1 - \bu_2 \in V.
$$

One may verify that this is an equivalence relation. Thus, $U / \sim$ and $U / V$ are equivalent.
<!-- eng end -->

##### Exercise 4(b)

我們可以不管直觀上的任何意義來定義加法：

$$
(\bu_1 + V) + (\bu_2 + V) = (\bu_1 + \bu_2) + V.
$$

然而要小心的是
如果 $\bu_1 + V$ 和 $\bu'_1 + V$ 一樣、
同時 $\bu_2 + V$ 和 $\bu'_2 + V$ 一樣﹐
那麼加出來的 $(\bu_1 + \bu_2) + V$ 和 $(\bu'_1 + \bu'_2) + V$ 也會一樣嗎？
符合這樣性質的定義我們稱為是**定義完善的**（well-defined）。

證明商空間上定義的向量加法是定義完善的。

<!-- eng start -->
We may naïvely define the addition of two elements in the quotient space:

$$
(\bu_1 + V) + (\bu_2 + V) = (\bu_1 + \bu_2) + V.
$$

However, we need to be careful about the potential contradiction: Suppose $\bu_1 + V = \bu'_1 + V$ and $\bu_2 + V = \bu'_2 + V$. Is it true that $(\bu_1 + \bu_2) + V = (\bu'_1 + \bu'_2) + V$? If this is true, then we say our definition of the addition is **well-defined** .

Show that the addition on the elements in the quotient space is well-defined.
<!-- eng end -->

##### Exercise 4(c)

證明商空間上的純量乘法

$$
k(\bu + V) = (k\bu) + V
$$

是定義完善的。

<!-- eng start -->
Show that the definition of the scalar multiplication

$$
k(\bu + V) = (k\bu) + V
$$

is well-defined.
<!-- eng end -->

##### Exercise 5

證明笛卡爾積做出來的新結構是一個向量空間。
找出一組基底並證明其正確性。

<!-- eng start -->
Show that the Cartesian product of two vector spaces is again a vector space. Find a basis of the product and show that it is indeed a basis.
<!-- eng end -->

##### Exercise 6

證明商空間做出來的新結構是一個向量空間。
找出一組基底並證明其正確性。

<!-- eng start -->
Show that the quotient space of two vector spaces is again a vector space. Find a basis of the product and show that it is indeed a basis.
<!-- eng end -->