向量、長度、角度

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

from lingeo import random_int_vec, draw_two_vec

Main idea

A vector over \(\mathbb{R}\) of dimension \(n\) is a sequence \({\bf x} = (x_1, \ldots, x_n)\) of numbers in \(\mathbb{R}\).
(In contrast, we often call a number in \(\mathbb{R}\) a scalar.)

The collection of all vectors over \(\mathbb{R}\) of dimension \(n\) is denoted by \(\mathbb{R}^n\).
We often use \({\bf 0}\) and \({\bf 1}\) to refer to \((0,\ldots,0)\) and \((1,\ldots,1)\) of appropriate dimensions, respectively.

The length of a vector \({\bf x} = (x_1,\ldots, x_n)\) is defined as
\[\|{\bf x}\| = \sqrt{x_1^2 + \cdots + x_n^2}. \]

The inner product of two vectors \({\bf x} = (x_1, \ldots, x_n)\) and \({\bf y} = (y_1, \ldots, y_n)\) is defined as
\[\langle{\bf x},{\bf y}\rangle = x_1y_1 + \cdots + x_ny_n. \]

We have the following properties:

1. \(\|{\bf x}\| \geq 0\) and the equality holds if and only if \({\bf x} = {\bf 0}\).
2. \(\|k{\bf x}\| = |k|\cdot\|{\bf x}\|\).
3. \(\|{\bf x}\| + \|{\bf y}\| \geq \|{\bf x} - {\bf y}\|\) (triangle inequality).
4. \(\|{\bf x}\|^2 = \langle{\bf x},{\bf x}\rangle\).
5. \(\langle{\bf x}_1 + {\bf x}_2,{\bf y}\rangle = \langle{\bf x}_1,{\bf y}\rangle + \langle{\bf x}_2,{\bf y}\rangle\).
6. \(\langle k{\bf x},{\bf y}\rangle = k\langle{\bf x},{\bf y}\rangle\).
7. \(\langle {\bf x},{\bf y}\rangle = \langle {\bf y},{\bf x}\rangle\).
8. \(|\langle {\bf x},{\bf y}\rangle| \leq \|{\bf x}\|\cdot \|{\bf y}\|\) (Cauchy–Schwarz inequality).

Thanks to the Cauchy–Schwarz inequality, we define the angle between two vectors \({\bf x}\) and \({\bf y}\) as the angle \(\theta\) such that \(\cos\theta = \frac{\langle{\bf x},{\bf y}\rangle}{\|{\bf x}\|\|{\bf y}\|}\).
We \({\bf x}\) and \({\bf y}\) are orthogonal if \(\langle{\bf x},{\bf y}\rangle = 0\).

Side stories

• cosine law
• orthogonal
• projection

Experiments

Exercise 1

執行下方程式碼。
令 \(O\) 為原點、
\(A\) 和　\(B\)　分別為 x, y 的向量終點。
（可以選用自己喜好的 ramdom_seed）

### code
set_random_seed(0)
print_ans = False
x = random_int_vec(6)
y = random_int_vec(6)
print("x =", x)
print("y =", y)
pic = draw_two_vec(x, y)
pic.axes(False)
show(pic)

if print_ans:
    print("OA =", x.norm())
    print("OB =", y.norm())
    print("AB =", (x-y).norm())
    cos_cos = (x.norm()**2 + y.norm()**2 - (x-y).norm()**2) / 2 / x.norm() / y.norm()
    cos_inner = x.inner_product(y) / x.norm() / y.norm()
    print("cos by cos law =", 
          cos_cos, "=", 
          N(cos_cos))
    print("cos by inner product =", 
          cos_inner, "=", 
          N(cos_inner))

Exercise 1(a)

計算 \(\overline{OA}\)、\(\overline{OB}\)、和 \(\overline{AB}\) 的長度，
並利用餘絃定理來計算 \(\cos\angle AOB\)。

令 \(\bx = (-4,3,5,-5,-5,0) = \overrightarrow{OA}\) 及
\({\bf y} = (3,-3,3,4,-4,-3) = \overrightarrow{OB}\)。
則 \(\|{\overrightarrow{OA}}\| = \sqrt{100} = 10\), \(\|{\overrightarrow{OB}}\| = \sqrt{68} = 2\sqrt{17}\)。
另一方面，\(\overrightarrow{AB} = \by - \bx = (7,-6,-2,9,1,-3)\)
所以 \(\|\overrightarrow{AB}\| = \sqrt{180} = 6\sqrt{5}\)。
最後我們得到 \(\cos\angle AOB = \frac{\|{\overrightarrow{OA}}\|^2 + \|{\overrightarrow{OB}}\|^2 - \|{\overrightarrow{AB}}\|^2}{2\|{\overrightarrow{OA}}\|\|{\overrightarrow{OB}}\|} = \frac{-3} {10\sqrt{17}}\)。

Exercise 1(b)

令 \(\theta\) 為 x, y 之間用內積定出來的夾角﹐
求 \(\cos\theta\)。

沿用 1(a) 之 \(\overrightarrow{OA}\)、\(\overrightarrow{OB}\)： \[ \overrightarrow{OA} = (-4,3,5,-5,-5,0)\\ \overrightarrow{OB} = (3,-3,3,4,-4,-3) \] 由內積定義我們得到 \(\langle{\overrightarrow{OA}}, {\overrightarrow{OB}}\rangle = -6\)。

我們由內積的性質得知
\[-6 = \|{\overrightarrow{OA}}\|\|{\overrightarrow{OB}}\|\cos\angle AOB.\]
最後得到 \[\cos\angle AOB = \frac{-3}{10\sqrt{17}}.\]

Exercise 1©

前兩題算出來總是一樣嗎？
請用代數的方法證明。

Sample:
It is known that \(\overline{OA} = \|{\bf x}\|\), \(\overline{OB} = \|{\bf y}\|\), and \(\overline{AB} = \|{\bf x} - {\bf y}\|\).
Thus,
\[ \frac{\overline{OA}^2 + \overline{OB}^2 - \overline{AB}^2}{2\overline{OA}\cdot \overline{OB}} = \frac{\|{\bf x}\|^2 + \|{\bf y}\|^2 - \|{\bf x} - {\bf y}\|^2}{2\|{\bf x}\|\|{\bf y}\|}=\frac{2\langle {\bf x},{\bf y}\rangle}{2\|{\bf x}\|\|{\bf y}\|}=\frac{\langle {\bf x},{\bf y}\rangle}{\|{\bf x}\|\|{\bf y}\|}. \]

Exercises

Exercise 2

將每個向量都展開
（像是 \({\bf x} = (x_1,\ldots, x_n)\)）﹐
把每個長度和內積的定義也都展開。
證明 Main idea 中最後列出來的那些性質
（除了三角不等式和柯西不等式以外）。

(1)
Recall the length of a vector \({\bf x} = (x_1,\ldots, x_n)\) is defined as
\[\|{\bf x}\| = \sqrt{x_1^2 + \cdots + x_n^2}. \]

Also, we know \(x^2\geq 0\) for any \(x\in\mathbb{R}\) and the equality holds if and only if \(x = 0\).

Since \(x_1^2 , \cdots, x_n^2 \geq{\bf 0}\), we have \(\|\bx\| \geq 0\).
Moveover, when \(\|{\bf x}\| = \sqrt{x_1^2 + \cdots + x_n^2} = 0\), \({x_1^2 , \cdots, x_n^2}\) must all be \(0\), which made \({\bf x} = {\bf 0}\).

(2)
藉由代數計算，可以直接驗證 \[ \begin{aligned} \|k{\bf x}\| &= \sqrt{{(kx_1)^2} + \cdots + {(kx_n)^2}} \\ &= |k|\sqrt{x_1^2 + \cdots + x_n^2} \\ &= |k|\cdot\|{\bf x}\|, \end{aligned} \] 其中 \(\bx = (x_1,\ldots, x_n)\)。

(4)
藉由代數計算，可以直接驗證
\[ \langle{\bf x},{\bf x} \rangle = {x_1^2 + \cdots + x_n^2} = \|{\bf x}\|^2. \]

(5)
藉由代數計算，可以直接驗證 \[ \begin{aligned} \langle {\bf x}_1 + {\bf x}_2,{\bf y}\rangle &=({\bf x}_{11} + {\bf x})_{21}{\bf y} + \cdots + ({\bf x}_{1n} + {\bf x}_{2n})\by_n \\ &=({\bf x}_{11}{\bf y} + \cdots + {\bf x}_{1n}{\bf y}_n) + ({\bf x}_{21}{\bf y} + \cdots + {\bf x}_{2n}{\bf y}_{n})\\ &=\langle \bx_1 , \by \rangle + \langle \bx_2 , \by \rangle. \end{aligned} \]

(6)我們要驗證 \(\langle k{\bf x},{\bf y}\rangle = k\langle{\bf x},{\bf y}\rangle\)。
令 \({\bf x} = (x_1,\ldots, x_n)\)，\({\bf y} = (y_1,\ldots, y_n)\)。

藉由代數運算，可以直接驗證 \[ \begin{aligned} \langle k{\bf x},{\bf y}\rangle &= kx_1y_1 + \cdots + kx_ny_n\\ &= k(x_1y_1 + \cdots + x_ny_n)\\ &=k\langle{\bf x},{\bf y}\rangle. \end{aligned} \]

(7)我們要驗證 \(\langle{\bf x},{\bf y}\rangle = \langle{\bf y},{\bf x}\rangle\)。
令 \({\bf x} = (x_1,\ldots, x_n)\)，\({\bf y} = (y_1,\ldots, y_n)\)。
將 \(\langle{\bf x},{\bf y}\rangle\) 展開得到 \(x_1y_1+\cdots+x_ny_n\)，
再利用乘法交換律將式子改寫成 $y_1x_1+\cdots+y_nx_n = \langle{\bf y},{\bf x}\rangle. $

Exercise 3

證明 \(\|{\bf x} \pm {\bf y}\|^2 = \|{\bf x}\|^2 \pm 2\langle{\bf x},{\bf y}\rangle + \|{\bf y}\|^2\)。

令 \({\bf x} = (x_1,\ldots, x_n)\)，\({\bf y} = (y_1,\ldots, y_n)\)，
則\(\|{\bf x} \pm {\bf y}\|^2=\sqrt{(x_1 \pm y_1)^2 + \cdots + (x_n \pm y_n)^2}^2\)，
將平方展開後得 \((x_1^2 \pm 2x_1y_1 + y_1^2) + \cdots + (x_n^2 \pm 2x_ny_n + y_n^2)\)，
再將式子整理後得到
\[ \sqrt{x_1^2 + \cdots + x_n^2}^2 \pm 2(x_1y_1 + \cdots + x_ny_n) + \sqrt{y_1^2 + \cdots + y_n^2}^2 = \|{\bf x}\|^2 \pm 2\langle{\bf x},{\bf y}\rangle + \|{\bf y}\|^2. \]

Exercise 4

給定兩個向量 \({\bf x},{\bf y}\)。
若 \({\bf y}\) 可以寫成兩個分量相加 \({\bf y} = {\bf h} + {\bf w}\)﹐
其中 \({\bf h}\) 和 \({\bf x}\) 垂直（所以 \(\langle{\bf h},{\bf x}\rangle = 0\)）、
而 \({\bf w}\) 和 \({\bf x}\) 平行（所以 \({\bf w} = k{\bf x}\)），
求 \({\bf h}\) 和 \({\bf w}\)（用 \({\bf x}\) 和 \({\bf y}\) 表示出來）。

假設 \({\bf w} = k{\bf x}\) 而 \({\bf y} - {\bf w} ={\bf h}\)。
已知 \(\langle{\bf h},{\bf x}\rangle = 0 = \langle{\bf y} - {\bf w},{\bf x}\rangle = \langle{\bf y} - k{\bf x},{\bf x}\rangle\)。
藉由代數運算得 \(0 = \langle{\bf y} - k{\bf x},{\bf x}\rangle = \langle{\bf y},{\bf x}\rangle - k\langle{\bf x},{\bf x}\rangle\)，所以 \(k = \frac{\langle{\bf y},{\bf x}\rangle}{\langle{\bf x},{\bf x}\rangle}\)。
將 \(k\) 帶入原式得 \({\bf w} = {\bf x}\)\(\frac{\langle{\bf y},{\bf x}\rangle}{\langle{\bf x},{\bf x}\rangle}\)
以及 \({\bf h} = {\bf y} - {\bf x}\)\(\frac{\langle{\bf y},{\bf x}\rangle}{\langle{\bf x},{\bf x}\rangle}\)。

Exercise 5

依照以下步驟證明三角不等式和柯西不等式等價。

三角不等式
對任意兩個向量 \({\bf x},{\bf y}\in\mathbb{R}^n\)
不等式 \(\|{\bf x}\| + \|{\bf y}\| \geq \|{\bf x} - {\bf y}\|\) 皆成立。

柯西不等式
對任意兩個向量 \({\bf x},{\bf y}\in\mathbb{R}^n\)
不等式 \(|\langle {\bf x},{\bf y}\rangle| \leq \|{\bf x}\|\cdot \|{\bf y}\|\) 皆成立。

Exercise 5(a)

證明若三角不等式成立則柯西不等式成立。

Sample:
Let \({\bf x}\) and \({\bf y}\) be any given vectors in \(\mathbb{R}^n\).
By the triangle inequality, \(\|{\bf x}\| + \|{\bf y}\| \geq \|{\bf x} - {\bf y}\|\).
因為三角不等式對所有 \(\bx\) 和 \(\by\) 都成立，我們可以把 \(\by\) 的位置代入 \(-\by\) 得到 \(\|\bx\| + \|\by\| \geq \|\bx + \by\|\)，因此 \[\|\bx\| + \|\by\| \geq \|\bx \pm \by\|.\] 把三角不等式的兩邊同時平方得到
(\(\|{\bf x}\| + \|{\bf y}\|)^2 \geq (\|{\bf x} \pm {\bf y}\|)^2\)，
這等價於
\(\|{\bf x}\|^2 + \|{\bf y}\|^2+2\|{\bf x}\|\|{\bf y}\|\geq \|{\bf x}\|^2+ \|{\bf y}\|^2\pm 2\langle {\bf x},{\bf y}\rangle\)。
因此 \(2\|{\bf x}\|\|{\bf y}\|\geq \pm2\langle {\bf x},{\bf y}\rangle\)，
也就是 \(\|{\bf x}\|\|{\bf y}\|\geq \pm\langle {\bf x},{\bf y}\rangle\)。

Therefore, \(|\langle {\bf x},{\bf y}\rangle| \leq \|{\bf x}\|\cdot \|{\bf y}\|\).
Since \({\bf x}\) and \({\bf y}\) are arbitrary, the Cauchy–Schwarz inequality holds for any vectors \({\bf x}\) and \({\bf y}\).

Exercise 5(b)

證明若柯西不等式成立則三角不等式成立。

Sample:

Let \({\bf x}\) and \({\bf y}\) be any given vectors in \(\mathbb{R}^n\).
By the Cauchy–Schwarz inequality, \(|\langle {\bf x},{\bf y}\rangle| \leq \|{\bf x}\|\cdot \|{\bf y}\|\).
Thus, \(\|{\bf x}\|\|{\bf y}\|\geq \pm\langle {\bf x},{\bf y}\rangle\).
By adding \(\|{\bf x}\|^2 + \|{\bf y}\|^2\) on both sides, we get \(\|{\bf x}\|^2 + \|{\bf y}\|^2+2\|{\bf x}\|\|{\bf y}\|\geq \|{\bf x}\|^2+ \|{\bf y}\|^2 \pm\ 2\langle {\bf x},{\bf y}\rangle\).
By completing the squares, we have (\(\|{\bf x}\| + \|{\bf y}\|)^2 \geq (\|{\bf x} - {\bf y}\|)^2\).
While both \(\|{\bf x}\| + \|{\bf y}\|\) and \(\|{\bf x} - {\bf y}\|\) are positive, \(\|{\bf x}\| + \|{\bf y}\| \geq \|{\bf x} - {\bf y}\|\).

Since \({\bf x}\) and \({\bf y}\) are arbitrary, the triangle inequality holds for any vectors \({\bf x}\) and \({\bf y}\).

Exercise 6

依照以下步驟證明柯西不等式。

Exercise 6(a)

證明當 \(\|{\bf x}\| = \|{\bf y}\| = 1\) 時柯西不等式成立。

Sample:
Let \({\bf x}\) and \({\bf y}\) be two vectors of length \(1\).

Since \(\|{\bf x} - {\bf y}\|^2 \geq 0\), we have
\(\|{\bf x}\|^2+ \|{\bf y}\|^2 - 2\langle {\bf x},{\bf y}\rangle\geq 0\).
Since \(\|{\bf x} + {\bf y}\|^2 \geq 0\), then
\(\|{\bf x}\|^2+ \|{\bf y}\|^2\)+\(2\langle {\bf x},{\bf y}\rangle\geq 0\).

Therefore, \(\pm2\langle {\bf x},{\bf y}\rangle\geq -2\) and \(2 \geq\pm2\langle {\bf x},{\bf y}\rangle\).
Therefore, \(|\langle{\bf x},{\bf y}\rangle| \leq 1\) whenever \(\|{\bf x}\| = \|{\bf y}\| = 1\).

Exercise 6(b)

證明對任何非零向量 \({\bf x},{\bf y}\in\mathbb{R}^n\) 柯西不等式皆成立。

Sample:

Now suppose \({\bf x}\) and \({\bf y}\) be arbitrary nonzero vectors in \(\mathbb{R}^n\).
Let \({\bf x}' = {\bf x}/\|{\bf x}\|\) and \({\bf y}' = {\bf y}/\|{\bf y}\|\).
Thus, \(\|{\bf x}'\| = \|{\bf y}'\| = 1\).
帶回 (a) 得到 \(|\langle {\bf x}',{\bf y}'\rangle| \leq \ 1\)，
也就是\(\frac{|\langle{\bf x}, {\bf y}\rangle|}{\|{\bf x}\|\|{\bf y}\|} \leq \ 1\)。
故\(|\langle {\bf x},{\bf y}\rangle| \leq \|{\bf x}\|\cdot \|{\bf y}\|\)。
Therefore, \(|\langle {\bf x},{\bf y}\rangle| \leq \|{\bf x}\|\cdot \|{\bf y}\|\) for any nonzero vectors \({\bf x}\) and \({\bf y}\).

Exercise 6©

考慮剩餘的可能性並完成柯西不等式的證明。

當 \(\|{\bf x}\|\) 或 \(\|{\bf y}\|\) 等於 \(0\) 時，\(\|{\bf x}\|\cdot \|{\bf y}\|=0\) 且 \(\inp{\bx}{\by} = 0\)。
所以 \(|\langle {\bf x},{\bf y}\rangle| =\|{\bf x}\|\cdot \|{\bf y}\|\)。

第七題太長所以被移到別區
Exercise 7