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向量空間中的向量表示法

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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
from linspace import ptov, vtop

Main idea

Let \(V\) be a vector space.
Let \(\beta = \{ {\bf u}_1, \ldots, {\bf u}_n \}\) be a basis of \(V\).
Every vector \({\bf v}\in V\) has a unique way to be written as a linear combination
\[{\bf v} = c_1{\bf u}_1 + \cdots + c_n{\bf u}_n. \]
We call the vector \((c_1,\ldots, c_n)\in\mathbb{R}^n\) the vector representation of \({\bf v}\) with respect to the basis \(\beta\), denoted as \([{\bf v}]_\beta\).

Let \(\mathcal{P}_d\) be the vector space of all polynomials of degree at most \(d\).
Let \(\beta = \{1, x, \ldots, x^d\}\) be the standard basis of \(\mathcal{P}_d\).
Then \([p]_\beta = \operatorname{ptov}(p)\) is simply writing down the coefficients of \(p\) into a vector in \(\mathbb{R}^{d+1}\).

Let \(\mathcal{M}_{m.n}\) be the vector space of all \(m\times n\) matrices.
Let \(\beta = \{E_{11}, \ldots, E_{1n}, \ldots, E_{m1}, \ldots, E_{mn}\}\) be the standard basis of \(\mathcal{M}_{m,n}\).
Then \([A]_\beta = \operatorname{mtov}(A)\) is simply writing down the entries of \(A\) into a vector in \(\mathbb{R}^{mn}\) in the row-major order.

Side stories

  • use the standard basis to solve the vector representation
  • new inner product

Experiments

Exercise 1

執行以下程式碼。
\(\mathcal{P}_2\) 為所有次數小於等於 \(2\) 的多項式所形成的向量空間。
\(\alpha = \{1, x, x^2\}\)\(\mathcal{P}_2\) 的標準基底。
\(\beta = \{1, (1+x), (1+x)^2\}\)\(\mathcal{P}_2\) 的另一組基底。

seed \(=69\),得 \(p = 2 + 3x + 2x^2\)

### code
set_random_seed(0)
print_ans = False
d = 2
p1,p2,p3 = 1, 1+x, (1+x)^2
p = vtop(vector(random_int_list(d+1)))

print("p =", p)

if print_ans:
    A = matrix([ptov(p1, d), ptov(p2, d), ptov(p3, d)]).transpose()
    p_alpha = ptov(p, d)
    p_beta = A.inverse() * p_alpha
    print("[p]_alpha =", ptov(p, d))
    print("[p]_beta =", p_beta)
    print("A =")
    show(A)
Exercise 1(a)

求出 \([p]_\alpha\)

\(Ans:\)
根據觀察,\(p = 2 \cdot 1 + 3 \cdot x + 2 \cdot x^2\),所以 \([p]_\alpha = \begin{bmatrix} 2\\3\\2 \end{bmatrix}\)

Exercise 1(b)

求出 \([p]_\beta\)

\(Ans:\)
\(p = c_1 + c_2(1+x) + c_3(1+x)^2\),其中 \(c_1,c_2,c_3 \in \mathbb{R}\)

\[ \begin{aligned} p &= c_1 + c_2(1+x) + c_3(1+x)^2\\ &= (c_1 + c_2 + c_3) + (c_2 + 2c_3)x + c_3x^2\\ &= 2 + 3x + 2x^2 \end{aligned} \] 觀察 \(x^2\) 項得 \(c_3 = 2\),再看 \(x\) 項得 \(c_2 + 4 = 3\)\(c_2 = -1\),最後常數項 \(c_1 + 1 = 2\)\(c_1 = 1\)
所以 \(p = 1 \cdot 1 + (-1) \cdot (1+x) + 2 \cdot x^2\)
\([p]_\beta = \begin{bmatrix} 1\\-1\\2 \end{bmatrix}\)

Exercise 1©

\(p_1, \ldots, p_3\)\(\beta\) 中的各多項式。
寫出 \(3\times 3\) 矩陣 \(A\)
其各行向量為 \([p_1]_\alpha, \ldots, [p_3]_\alpha\)

\(Ans:\)
根據題目 \(p_1 = 1\)\(p_2 = 1 + x\)\(p_3 = 1 + 2x + x^2\)
稍微改寫一下,可以得
\[ p_1 = 1 \cdot 1 + 0 \cdot x + 0 \cdot x^2\\ p_2 = 1 \cdot 1 + 1 \cdot x + 0 \cdot x^2\\ p_3 = 1 \cdot 1 + 2 \cdot x + 1 \cdot x^2\\ A = \begin{bmatrix} 1 & 1 & 1\\ 0 & 1 & 2\\ 0 & 0 & 1 \end{bmatrix} \]

Exercise 1(d)

驗證並說明為什麼 \(A[p]_\beta = [p]_\alpha\)

  • 以下證明所有的 \(p \in \mathcal{P}_2\)\([p]_\beta = \begin{bmatrix} c_1 \\ c_2 \\ c_3 \end{bmatrix}\) >
    以下證明所有的 \(p \in \mathcal{P}_2\) 等式都成立。
    假設 \([p]_\beta = \begin{bmatrix} c_1 \\ c_2 \\ c_3 \end{bmatrix}\),則 \(p = c_1p_1 + c_2p_2 + c_3p_3\)

\(Ans:\)
先驗證例子
\[ \begin{aligned} A[p]_\beta &= \begin{bmatrix} 1 & 1 & 1\\ 0 & 1 & 2\\ 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} 1\\ -1\\ 2\\ \end{bmatrix}\\&= \begin{bmatrix} 1 \cdot 1 + 1 \cdot (-1) + 1 \cdot 2\\ 0 \cdot 1 + 1\cdot (-1) + 2 \cdot 2\\ 0 \cdot 1 + 0 \cdot (-1) + 1 \cdot 2 \end{bmatrix}\\ &=\begin{bmatrix} 2\\ 3\\ 2 \end{bmatrix}\\ &= [p]_\alpha \end{aligned} \]

以下證明所有的 \(p \in \mathcal{P}_2\) 等式都成立。

假設 \([p]_\beta = \begin{bmatrix} c_1 \\ c_2 \\ c_3 \end{bmatrix}\),則 \(p = c_1p_1 + c_2p_2 + c_3p_3\) \[ \begin{aligned} A[p]_\beta &= A\begin{bmatrix}c_1 \\ c_2 \\ c_3\end{bmatrix}\\ &=A\begin{bmatrix}c_1 \\ 0 \\ 0\end{bmatrix} + A\begin{bmatrix}0 \\ c_2 \\ 0\end{bmatrix} + A\begin{bmatrix}0 \\ 0 \\ c_3\end{bmatrix}\\ &= c_1[p_1]_\alpha + c_2[p_2]_\alpha + c_3[p_3]_\alpha\\ &= [p]_\alpha \end{aligned} \]

Exercises

Exercise 2

\(\mathcal{P}_2\) 為所有次數小於等於 \(2\) 的多項式所形成的向量空間。
\(p = 2 + 3x + 4x^2\)

Exercise 2(a)

\(\beta = \{1,x,x^2\}\)\(\mathcal{P}_2\) 的一組基底。
\([p]_\beta\)

\(Ans:\)
根據觀察 \(p = 2 \cdot 1 + 3 \cdot x + 4 \cdot x^2\),所以 \([p]_\beta = \begin{bmatrix} 2\\3\\4 \end{bmatrix}\)

Exercise 2(b)

\(\beta = \{1,(1-x),(1-x)^2\}\)\(\mathcal{P}_2\) 的一組基底。
\([p]_\beta\)

\(Ans:\)
應用1(d),取 \(A = \begin{bmatrix} 1 & 1 & 1\\ 0 & -1 & -2\\ 0 & 0 & 1 \end{bmatrix}\)\(A[p]_\beta = [p]_\alpha\)
先求出 \(A^{-1} = \begin{bmatrix} 1 & 1 & 1\\ 0 & -1 & -2\\ 0 & 0 & 1 \end{bmatrix}\),所以
\[ [p]_\beta = A^{-1}[p]_\alpha = \begin{bmatrix} 9\\-11\\4 \end{bmatrix}. \]

Exercise 2©

\(\beta = \{1,x,x(x-1)\}\)\(\mathcal{P}_2\) 的一組基底。
\([p]_\beta\)

  • \(\begin{bmatrix} 1 & x & x(x-1) \end{bmatrix} \cdot \begin{bmatrix} C_1 \\ C_2 \\ C_3 \end{bmatrix} =\) 這部份好像不用?

\(Ans:\)

\([p]_\beta = \begin{bmatrix} C_1 \\ C_2 \\ C_3 \end{bmatrix}\)

滿足
\[ C_1 \cdot 1 + C_2 \cdot x + C_3 \cdot x(x-1) = 2 + 3x + 4x^2, \]

解得 \(C_1 = 2,C_2 = 7,C_3 = 4\) ,

\([p]_\beta = \begin{bmatrix} 2 \\ 7 \\ 4 \end{bmatrix}\)

Exercise 2(d)


\[\begin{aligned} p_1(x) &= \frac{(x-2)(x-3)}{(1-2)(1-3)}, \\ p_2(x) &= \frac{(x-1)(x-3)}{(2-1)(2-3)}, \\ p_3(x) &= \frac{(x-1)(x-2)}{(3-1)(3-2)}. \\ \end{aligned} \] \(\beta = \{p_1, p_2, p_3\}\)\(\mathcal{P}_2\) 的一組基底。
\([p]_\beta\)

  • 一樣,好像沒必要寫 "\(\begin{bmatrix} p_1 & p_2 & p_3 \end{bmatrix} \cdot \begin{bmatrix} C_1 \\ C_2 \\ C_3 \end{bmatrix} =\)"

\(Ans:\)
\([p]_\beta = \begin{bmatrix} C_1 \\ C_2 \\ C_3 \end{bmatrix}\)

滿足
\[ C_1 \cdot \frac{(x-2)(x-3)}{(1-2)(1-3)} + C_2 \cdot \frac{(x-1)(x-3)}{(2-1)(2-3)} + C_3 \cdot \frac{(x-1)(x-2)}{(3-1)(3-2)} = 2 + 3x + 4x^2, \]

\(x=1,x=2,x=3\) 分別帶入解得 \(C_1 = 9,C_2 = 24,C_3 = 47\)

\([p]_\beta = \begin{bmatrix} 9 \\ 24 \\ 47 \end{bmatrix}\)

good

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另解:
可以觀察到,\(\{ p_1, p_2, p_3 \}\) 是一組 Lagrange basis 對應到 \(1, 2, 3\),所以 \(p\) 可以寫成
\[ \begin{aligned} p &= 2 + 3x + 4x^2\\ &= p(1)p_1 + p(2)p_2 + p(3)p_3 \end{aligned} \] 所以
\[ \begin{aligned}{[p]}_\beta &= \begin{bmatrix} p(1)\\p(2)\\p(3) \end{bmatrix}\\ &=\begin{bmatrix}9\\24\\47\end{bmatrix}.\end{aligned} \]

Exercise 3

\(\mathcal{M}_{2,3}\) 為所有 \(2\times 3\) 矩陣所形成的向量空間。

\[A = \begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ \end{bmatrix}. \]

Exercise 3(a)

\(\beta = \{E_{11}, E_{12}, E_{13}, E_{21}, E_{22}, E_{23}\}\)\(\mathcal{M}_{2,3}\) 的標準基底。
\([A]_\beta\)

\(Ans:\)
明顯地,\(A=E_{11}+2E_{12}+3E_{13}+4E_{21}+5E_{22}+6E_{23}\)

因此,\([A]_\beta=\begin{bmatrix} 1\\2\\3\\4\\5\\6 \end{bmatrix}\)

Exercise 3(b)

\[ M_1 = \begin{bmatrix} 1 & 1 & 1 \\ 1 & 1 & 1 \\ \end{bmatrix}, M_2 = \begin{bmatrix} 0 & 1 & 1 \\ 1 & 1 & 1 \\ \end{bmatrix}, M_3 = \begin{bmatrix} 0 & 0 & 1 \\ 1 & 1 & 1 \\ \end{bmatrix}, \] \[ M_4 = \begin{bmatrix} 0 & 0 & 0 \\ 1 & 1 & 1 \\ \end{bmatrix}, M_5 = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 1 & 1 \\ \end{bmatrix}, M_6 = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 1 \\ \end{bmatrix}, \]
\([A]_\beta\)

\(Ans:\)

\([A]_\beta=\begin{bmatrix} a\\b\\c\\d\\e\\f \end{bmatrix}\) ,則 \(A=aM_1+bM_2+cM_3+dM_4+eM_5+fM_6\)

因此 \(\begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ \end{bmatrix}=\begin{bmatrix} a & a+b & a+b+c\\ a+b+c+d & a+b+c+d+e & a+b+c+d+e+f & \end{bmatrix}\)

可簡單解得 \(a=b=c=d=e=f=1\),因此, \([A]_\beta=\begin{bmatrix} 1\\1\\1\\1\\1\\1 \end{bmatrix}\)

Exercise 4

以下例子說明一些向量空間的內積
其實就是某個表示法下的 \(\mathbb{R}^n\) 標準內積。
(實際上所有有限維度空間中的內積
都可以寫成某個表示法下的 \(\mathbb{R}^n\) 標準內積。
但證明須要用到一些對角化的工具。)

Exercise 4(a)

\(\mathcal{M}_{m,n}\) 為所有 \(m\times n\) 矩陣所形成的向量空間。
我們曾經驗證過 \(\langle A, B\rangle = \operatorname{tr}(B^\top A)\) 是一種合法的內積。

說明其實 \(\operatorname{tr}(B^\top A) = \langle [A]_\beta, [B]_\beta \rangle\)
其中 \(\beta\)\(\mathcal{M}_{m,n}\) 的標準基底。

  • \(B\trans\) 就好,不用寫 \(B^{\trans}\)

\(Ans:\)
\(A = \begin{bmatrix} | && |\\ \ba_1 & \cdots & \ba_n\\ | && | \end{bmatrix},B \trans = \begin{bmatrix}- & \bb_1 & -\\ & \vdots &\\- & \bb_n & - \end{bmatrix},B \trans A\) 的第 \(i\) 列第 \(j\) 行元素寫作 \((B\trans A)_{ij}\)
\(A\) 的第 \(i\) 列第 \(j\) 行元素寫作 \(a_{ij}\)\(B \trans\) 的第 \(i\) 列第 \(j\) 行元素寫作 \(b_{ij}\)
可以看出 \((B \trans A)_{ij} = \langle \bb_i,\ba_j \rangle\),所以
\[ \begin{aligned} \operatorname{tr}(B \trans A) &= \langle \bb_1,\ba_1 \rangle + \cdots + \langle \bb_n,\ba_n \rangle\\ &= \sum_{j=1}^{n}\sum_{i=1}^{n}b_{ij}a_{ij}. \end{aligned} \] 因為 \([A]_\beta = (a_{11},\cdots,a_{mn}),[B]_\beta = (b_{11},\cdots,b_{mn})\)
\(\langle [A]_\beta, [B]_\beta \rangle = \sum_{j=1}^{n}\sum_{i=1}^{n}b_{ij}a_{ij}\)
所以 \(\operatorname{tr}(B^{\trans}A) = \langle [A]_\beta, [B]_\beta \rangle\)

Exercise 4(b)

\(\mathcal{P}_3\) 為所有次數小於等於 \(3\) 的多項式所形成的向量空間。
我們曾經驗證過 \(\langle p_1, p_2 \rangle = a_0b_0 + a_1b_1 + a_2b_2 + a_3b_3\) 是一種合法的內積﹐其中
\[\begin{aligned} p_1 &= a_0 + a_1x + a_2x^2 + a_3x^3, \\ p_2 &= b_0 + b_1x + b_2x^2 + b_3x^3. \\ \end{aligned} \]

說明其實 \(\langle p_1, p_2 \rangle = \langle [p_1]_\beta, [p_2]_\beta \rangle\)
其中 \(\beta\)\(\mathcal{P}_3\) 的標準基底。

  • 標點

\(Ans:\)
因為 \([p_1]_\beta = (a_0,a_1,a_2,a_3),[p_2]_\beta = (b_0,b_1,b_2,b_3)\)
所以
\[ \begin{aligned} \langle [p_1]_\beta, [p_2]_\beta \rangle &= a_0b_0 + a_1b_1 + a_2b_2 + a_3b_3\\ &= \langle p_1, p_2 \rangle. \end{aligned} \]

Exercise 4©

\(\mathcal{P}_3\) 為所有次數小於等於 \(3\) 的多項式所形成的向量空間。
我們曾經驗證過 \(\langle p_1, p_2 \rangle = p_1(1)p_2(1) + p_1(2)p_2(2) + p_1(3)p_2(3) + p_1(4)p_2(4)\) 是一種合法的內積。

\[\begin{aligned} f_1(x) &= \frac{(x-2)(x-3)(x-4)}{(1-2)(1-3)(1-4)}, \\ f_2(x) &= \frac{(x-1)(x-3)(x-4)}{(2-1)(2-3)(2-4)}, \\ f_3(x) &= \frac{(x-1)(x-2)(x-4)}{(3-1)(3-2)(3-4)}, \\ f_4(x) &= \frac{(x-1)(x-2)(x-3)}{(4-1)(4-2)(4-3)}. \\ \end{aligned} \] 說明其實 \(\langle p_1, p_2 \rangle = \langle [p_1]_\beta, [p_2]_\beta \rangle\)
其中 \(\beta = \{f_1, f_2, f_3, f_4\}\)\(\mathcal{P}_3\) 的一組基底。

\(Ans:\)
觀察可以知道 \(\beta\) 是一組 Lagrange basis 對應到 \(1,2,3,4\),因此
\[ [p]_\beta = (p(1),p(2),p(3),p(4)) \] 代入 \(\langle [p_1]_\beta, [p_2]_\beta \rangle\),得
\[ \begin{aligned} \langle [p_1]_\beta, [p_2]_\beta \rangle &= p_1(1)p_2(1) + p_1(2)p_2(2) + p_1(3)p_2(3) + p_1(4)p_2(4)\\ &= \langle p_1, p_2 \rangle. \end{aligned} \]

目前分數 6.5

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