$\mathbb{R}^n$ 中的向量表示法

Vector representation 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, random_good_matrix

Main idea

Recall that $\mathcal{E}_n = \{ \be_1, \ldots, \be_n \}$ is the standard basis of $\mathbb{R}^n$.
For any vector $\bv = (c_1, \ldots, c_n)\in\mathbb{R}^n$, it can be written as

$$ \bv = c_1\be_1 + \cdots + c_n\be_n. $$

Similarly, let $\beta = \{ \bu_1, \ldots, \bu_n \}$ be a basis of $\mathbb{R}^n$.
Every vector $\bv\in\mathbb{R}^n$ has a unique way to be written as a linear combination

$$ \bv = c_1\bu_1 + \cdots + c_n\bu_n. $$

We call the vector $(c_1,\ldots, c_n)\in\mathbb{R}^n$ the vector representation of $\bv$ with respect to the basis $\beta$, denoted as $[\bv]_\beta$.
Since every vector in $\mathbb{R}^n$ can be written as a linear combinatoin of $\beta$ and the way of writing it is unique, it is a one-to-one correspondence between $\bv$ and $[\bv]_\beta$.

Let let $\beta = \{ \bu_1, \ldots, \bu_n \}$ be a basis of $\mathbb{R}^n$ and
$A$ the $n\times n$ matrix whose columns are vectors in $\beta$.
Since $\beta$ is a basis, $A$ is invertible.
By definition,

$$ A[\bv]_\beta = \bv \text{ and } A^{-1}\bv = [\bv]_\beta. $$

When $\beta$ is the standard basis of $\mathbb{R}^n$, $A = I_n$ and $[\bv]_\beta = \bv$.
Therefore, our usual way of writting a vector is the vector representation with respect to the standard basis.

In the case when $\beta$ is an orthonormal basis, $A$ is an orthogonal matrix and $A^{-1} = A\trans$.
Therefore,

$$ A[\bv]_\beta = \bv \text{ and } A\trans\bv = [\bv]_\beta. $$

Side stories

• vector representation algebra
• define new inner product

Experiments

Exercise 1

執行以下程式碼。
令 $\beta = \{ \bu_1, \ldots, \bu_3 \}$ 為 $A$ 的行向量且
已知其為 $\mathbb{R}^3$ 的基底。

Run the code below. Let $\beta = \{ \bu_1, \ldots, \bu_3 \}$ be the columns of $A$. Suppose $\beta$ is a basis of $\mathbb{R}^3$.

### code
set_random_seed(0)
print_ans = False
m,n,r = 3,3,3
A = random_good_matrix(m,n,r, bound=3)
x1 = vector(random_int_list(n, 3))
x2 = vector(random_int_list(n, 3))
v1,v2 = A*x1, A*x2
k = choice([3,4,5])

print("A =")
show(A)
print("v1 =", v1)
print("v2 =", v2)
print("k =", k)

if print_ans:
    Ainv = A.inverse()
    print("[v1]_beta =", Ainv * v1)
    print("[v2]_beta =", Ainv * v2)
    print("[v1 + v2]_beta =", Ainv * (v1 + v2))
    print("[v1]_beta + [v2]_beta =", Ainv * v1 + Ainv * v2)
    print("[k * v1]_beta =", Ainv * (k*v1))
    print("k * [v1]_beta =", k * Ainv * v1)
    print("< [v1]_beta, [v2]_beta > =", (Ainv * v1).inner_product(Ainv * v2))
    print("< v1, v2 > =", (v1).inner_product(v2))

Exercise 1(a)

求出 $[\bv_1]_\beta$ 及 $[\bv_2]_\beta$。

Find $[\bv_1]_\beta$ and $[\bv_2]_\beta$.

Exercise 1(b)

判斷是否 $[\bv_1 + \bv_2]_\beta = [\bv_1]_\beta + [\bv_2]_\beta$。

Check if $[\bv_1 + \bv_2]_\beta = [\bv_1]_\beta + [\bv_2]_\beta$.

Exercise 1(c)

判斷是否 $[k\bv_1]_\beta = k[\bv_1]_\beta$。

Check if $[k\bv_1]_\beta = k[\bv_1]_\beta$.

Exercise 1(d)

判斷是否 $\inp{\bv_1}{\bv_2} = \inp{[\bv_1]_\beta}{[\bv_2]_\beta}$。

Check if $\inp{\bv_1}{\bv_2} = \inp{[\bv_1]_\beta}{[\bv_2]_\beta}$.

Exercises

Exercise 2

已知

$$ A = \begin{bmatrix} 1 & 0 & 1 \\ -1 & 1 & -4 \\ 5 & -2 & 12 \end{bmatrix} $$

的反矩陣為

$$ A^{-1} = \begin{bmatrix} 4 & -2 & -1 \\ -8 & 7 & 3 \\ -3 & 2 & 1 \end{bmatrix}. $$

令 $\beta$ 為 $A$ 的行向量集合。
令 $\bv_1 = (1,1,1)$、
$\bv_2 = (1,2,3)$。
求 $[\bv_1]_\beta$ 和 $[\bv_2]_\beta$。

It is known that the inverse of $$ A = \begin{bmatrix} 1 & 0 & 1 \\ -1 & 1 & -4 \\ 5 & -2 & 12 \end{bmatrix} $$ is $$ A^{-1} = \begin{bmatrix} 4 & -2 & -1 \\ -8 & 7 & 3 \\ -3 & 2 & 1 \end{bmatrix}. $$ Let $\beta$ be the columns of $A$. Let $\bv_1 = (1,1,1)$ and $\bv_2 = (1,2,3)$. Find $[\bv_1]_\beta$ and $[\bv_2]_\beta$.

Exercise 3

已知

$$ A = \begin{bmatrix} \frac{1}{\sqrt{3}} & \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{6}} \\ \frac{1}{\sqrt{3}} & -\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{6}} \\ \frac{1}{\sqrt{3}} & 0 & -\frac{2}{\sqrt{6}} \end{bmatrix} $$

為一垂直矩陣。
令 $\beta$ 為 $A$ 的行向量集合。
令 $\bv_1 = (1,1,1)$、
$\bv_2 = (1,2,3)$。
求 $[\bv_1]_\beta$ 和 $[\bv_2]_\beta$。

It is known that $$ A = \begin{bmatrix} \frac{1}{\sqrt{3}} & \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{6}} \\ \frac{1}{\sqrt{3}} & -\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{6}} \\ \frac{1}{\sqrt{3}} & 0 & -\frac{2}{\sqrt{6}} \end{bmatrix} $$ is an orthogonal matrix. Let $\beta$ be the columns of $A$. Let $\bv_1 = (1,1,1)$ and $\bv_2 = (1,2,3)$. Find $[\bv_1]_\beta$ and $[\bv_2]_\beta$.

Exercise 4

令 $\beta$ 為 $\mathbb{R}^n$ 中的一組基底。
定義

$$ \begin{aligned} f : \mathbb{R}^n &\rightarrow \mathbb{R}^n \\ \bv &\mapsto [\bv]_\beta \\ \end{aligned} $$

為一函數。

Let $\beta$ be a basis of $\mathbb{R}^n$. Define the function $$ \begin{aligned} f : \mathbb{R}^n &\rightarrow \mathbb{R}^n \\ \bv &\mapsto [\bv]_\beta \\ \end{aligned} $$

Exercise 4(a)

驗證 $f$ 為一線性函數。

Verify that $f$ is linear.

Exercise 4(b)

判斷 $f$ 是否是嵌射。

Is $f$ injective?

Exercise 4(c)

判斷 $f$ 是否是映射。

Is $f$ surjective?

Exercise 4(d)

求出 $f$ 的矩陣表示法 $[f]$。

Find the matrix representation $[f]$ of $f$.

Exercise 5

回顧一個內積 $\inp{\cdot}{\cdot}$ 必須符合以下的條件：

1. $\inp{\bx_1 + \bx_2}{\by} = \inp{\bx_1}{\by} + \inp{\bx_2}{\by}$.
2. $\inp{k\bx}{\by} = k\inp{\bx}{\by}$.
3. $\inp{\bx}{\by} = \inp{\by}{\bx}$.
4. $\inp{\bx}{\bx} \geq 0$, and the equality holds if and only if $\bx = \bzero$.

Recall that an inner product $\inp{\cdot}{\cdot}$ has to have the following properties: 1. $\inp{\bx_1 + \bx_2}{\by} = \inp{\bx_1}{\by} + \inp{\bx_2}{\by}$. 2. $\inp{k\bx}{\by} = k\inp{\bx}{\by}$. 3. $\inp{\bx}{\by} = \inp{\by}{\bx}$. 4. $\inp{\bx}{\bx} \geq 0$, and the equality holds if and only if $\bx = \bzero$.

Exercise 5(a)

令 $\beta$ 為 $\mathbb{R}^n$ 中的一組基底。
定義一個新的雙變數函數 $\inp{\bx}{\by}_\beta = \inp{[\bx]_\beta}{[\by]_\beta}$﹐
其中 $\inp{[\bx]_\beta}{[\by]_\beta}$ 指的是 $\mathbb{R}^n$ 中的標準內積。
驗證 $\inp{\cdot}{\cdot}_\beta$ 也是 $\mathbb{R}^n$ 上的另一種內積。

Let $\beta$ be a basis of $\mathbb{R}^n$. Define a new bivariate function $\inp{\bx}{\by}_\beta = \inp{[\bx]_\beta}{[\by]_\beta}$, where $\inp{[\bx]_\beta}{[\by]_\beta}$ is the standard inner product on $\mathbb{R}^n$. Verify that $\inp{\cdot}{\cdot}_\beta$ is also an inner product on $\mathbb{R}^n$.

Exercise 5(a)

證明當 $\beta$ 是單位長垂直基底時﹐任意向量都有 $\inp{\bx}{\by}_\beta = \inp{\bx}{\by}$。

Show that if $\beta$ is orthonormal, then $\inp{\bx}{\by}_\beta = \inp{\bx}{\by}$ for any vectors.