基底轉換 - HackMD

# 基底轉換 Change of basis ![Creative Commons License](https://i.creativecommons.org/l/by/4.0/88x31.png) This work by Jephian Lin is licensed under a [Creative Commons Attribution 4.0 International License](http://creativecommons.org/licenses/by/4.0/). $\newcommand{\trans}{^\top} \newcommand{\adj}{^{\rm adj}} \newcommand{\cof}{^{\rm cof}} \newcommand{\inp}[2]{\left\langle#1,#2\right\rangle} \newcommand{\dunion}{\mathbin{\dot\cup}} \newcommand{\bzero}{\mathbf{0}} \newcommand{\bone}{\mathbf{1}} \newcommand{\ba}{\mathbf{a}} \newcommand{\bb}{\mathbf{b}} \newcommand{\bc}{\mathbf{c}} \newcommand{\bd}{\mathbf{d}} \newcommand{\be}{\mathbf{e}} \newcommand{\bh}{\mathbf{h}} \newcommand{\bp}{\mathbf{p}} \newcommand{\bq}{\mathbf{q}} \newcommand{\br}{\mathbf{r}} \newcommand{\bx}{\mathbf{x}} \newcommand{\by}{\mathbf{y}} \newcommand{\bz}{\mathbf{z}} \newcommand{\bu}{\mathbf{u}} \newcommand{\bv}{\mathbf{v}} \newcommand{\bw}{\mathbf{w}} \newcommand{\tr}{\operatorname{tr}} \newcommand{\nul}{\operatorname{null}} \newcommand{\rank}{\operatorname{rank}} %\newcommand{\ker}{\operatorname{ker}} \newcommand{\range}{\operatorname{range}} \newcommand{\Col}{\operatorname{Col}} \newcommand{\Row}{\operatorname{Row}} \newcommand{\spec}{\operatorname{spec}} \newcommand{\vspan}{\operatorname{span}} \newcommand{\Vol}{\operatorname{Vol}} \newcommand{\sgn}{\operatorname{sgn}} \newcommand{\idmap}{\operatorname{id}} \newcommand{\am}{\operatorname{am}} \newcommand{\gm}{\operatorname{gm}} \newcommand{\mult}{\operatorname{mult}} \newcommand{\iner}{\operatorname{iner}}$ ```python from lingeo import random_good_matrix ``` ## Main idea Let $V$ be a vector space. Let $\alpha = \{\bv_1, \ldots, \bv_n\}$ and $\beta = \{\bu_1, \ldots, \bu_n\}$ be two bases of $V$. Each vector $\bb$ can have different representations $[\bb]_\alpha$ and $[\bb]_\beta$ depending on the bases used. Let $[\idmap]_\alpha^\beta$ be the $n \times n$ matrix $$ \begin{bmatrix} | & ~ & | \\ [\bv_1]_\beta & \cdots & [\bv_n]_\beta \\ | & ~ & | \\ \end{bmatrix}. $$ Then $[\idmap]_\alpha^\beta [\bb]_\alpha = [\bb]_\beta$ for any $\bb\in V$. Therefore, we call $[\idmap]_\alpha^\beta$ the **change of basis matrix** from $\alpha$ to $\beta$. (The notation $[\idmap]_\alpha^\beta$ means it is an identity map from $V$ to $V$ but we are observing the domain and the codomain from the points of view of $\alpha$ and $\beta$. This notation will be more clear when we introduce the matrix representation of a linear function.) The change of basis matrix $[\idmap]_\alpha^\beta$ can be viewed as a function taking the representation in $\alpha$ and output the representation in $\beta$. Therefore, if we have another basis $\gamma$, then we have - $[\idmap]_\alpha^\alpha = I_n$, - $[\idmap]_\beta^\gamma[\idmap]_\alpha^\beta = [\idmap]_\alpha^\gamma$, - $[\idmap]_\beta^\alpha[\idmap]_\alpha^\beta = I_n$. Note that a matrix takes its input vector from the right, so we should read a product of matrices from right to left. ## Side stories - inverse of a change of basis matrix ## Experiments ##### Exercise 1 執行以下程式碼。  Run the code below.  ```python ### code set_random_seed(0) print_ans = False A = random_good_matrix(3,3,3, bound=3) B = random_good_matrix(3,3,3, bound=3) e1,e2,e3 = identity_matrix(3).transpose() u1,u2,u3 = A.transpose() v1,v2,v3 = B.transpose() print("alpha has three vectors:") print("e1 =", e1) print("e2 =", e2) print("e3 =", e3) print("beta has three vectors:") print("u1 =", u1) print("u2 =", u2) print("u3 =", u3) print("gamma has three vectors:") print("v1 =", v1) print("v2 =", v2) print("v3 =", v3) if print_ans: print("[id]_alpha^beta =") show(A.inverse()) print("[id]_beta^alpha =") show(A) print("[id]_beta^gamma =") show(B.inverse() * A) print("[id]_alpha^gamma =") show(B.inverse()) ``` :::warning - [x] Missing period. ::: By setting `seed = 0` , we get $$\alpha = \{ \begin{bmatrix} 1 \\ 0 \\ 0 \\\end{bmatrix}, \begin{bmatrix} 0 \\ 1 \\ 0 \\\end{bmatrix}, \begin{bmatrix} 0 \\ 0 \\ 1 \\\end{bmatrix} \} $$ , $$\beta = \{ \begin{bmatrix} 1 \\-3 \\ 0 \\\end{bmatrix}, \begin{bmatrix} 3 \\-8 \\-1 \\\end{bmatrix}, \begin{bmatrix} 1 \\-1 \\-1 \\\end{bmatrix} \} $$ and $$\gamma = \{ \begin{bmatrix} 1 \\-3 \\-2 \\\end{bmatrix}, \begin{bmatrix} 1 \\-2 \\-2 \\\end{bmatrix}, \begin{bmatrix} 2 \\-5 \\-3 \\\end{bmatrix} \}. $$ ##### Exercise 1(a) 計算 $[\idmap]_\alpha^\beta$ 及 $[\idmap]_\beta^\alpha$，並確認 $[\idmap]_\beta^\alpha [\idmap]_\alpha^\beta = I_3$。  Find $[\idmap]_\alpha^\beta$ and $[\idmap]_\beta^\alpha$. Then verify $[\idmap]_\beta^\alpha [\idmap]_\alpha^\beta = I_3$.  :::warning - [x] Make your notation consistent: $\bc_i$ --> $c_i$ ::: Answer: $$ [\idmap]_\beta^\alpha = \left[ \begin{array}{ccc} 1 & 3 & 1 \\ -3 & -8 & -1 \\ 0 & -1 & -1 \\ \end{array} \right]. $$ Let $\alpha = \{\be_1, \be_2, \be_3\}$ , $\beta = \{\bv_1, \bv_2, \bv_3\}$ and $\be_1 = c_1\bv_1 + c_2\bv_2 + c_3\bv_3$ , $c_1,c_2,c_3 \in \mathbb R$. The vector representation of $\be_1$ is $\begin{bmatrix} c_1 \\c_2 \\ c_3 \\\end{bmatrix}$ with respect to $\beta$, so we know $\begin{bmatrix} c_1 \\c_2 \\ c_3 \\\end{bmatrix}$ is the first column of $[\idmap]_\alpha^\beta$. By solving $$ \left[ \begin{array}{ccc} 1 & 3 & 1 \\ -3 & -8 & -1 \\ 0 & -1 & -1 \\ \end{array} \right] \begin{bmatrix} c_1 \\c_2 \\ c_3 \\ \end{bmatrix} = \begin{bmatrix} 1 \\0 \\ 0 \\ \end{bmatrix},$$ we get $$\begin{bmatrix} c_1 \\c_2 \\ c_3 \\ \end{bmatrix} = \begin{bmatrix} 7 \\-3 \\ 3 \\ \end{bmatrix}.$$ Repeat the above steps and replace $\be_1$ with $\be_2$ and $\be_3$, we know $$ [\idmap]_\alpha^\beta = \left[ \begin{array}{ccc} 7 & 2 & 5 \\ -3 & -1 & -2 \\ 3 & 1 & 1 \\ \end{array} \right].$$ Finally, $$ \left[ \begin{array}{ccc} 1 & 3 & 1 \\ -3 & -8 & -1 \\ 0 & -1 & -1 \\ \end{array} \right] \left[ \begin{array}{ccc} 7 & 2 & 5 \\ -3 & -1 & -2 \\ 3 & 1 & 1 \\ \end{array} \right] = I_3. $$ ##### Exercise 1(b) 計算 $[\idmap]_\beta^\gamma$ 及 $[\idmap]_\alpha^\gamma$，並確認 $[\idmap]_\beta^\gamma [\idmap]_\alpha^\beta = [\idmap]_\alpha^\gamma$。  Find $[\idmap]_\beta^\gamma$ and $[\idmap]_\alpha^\gamma$. Then verify $[\idmap]_\beta^\gamma [\idmap]_\alpha^\beta = [\idmap]_\alpha^\gamma$.  :::warning - [x] Make your notation consistent: $\bc_i$ --> $c_i$ ::: Answer: Let $\beta = \{\bv_1, \bv_2, \bv_3\}$ , $\gamma = \{\bu_1, \bu_2, \bu_3\}$ and $\bv_1 = c_1\bu_1 + c_2\bu_2 + c_3\bu_3$ , $c_1,c_2,c_3 \in \mathbb R$. The vector representation of $\bv_1$ is $\begin{bmatrix} c_1 \\c_2 \\ c_3 \\\end{bmatrix}$ with respect to $\gamma$, so we know $\begin{bmatrix} c_1 \\c_2 \\ c_3 \\\end{bmatrix}$ is the first column of $[\idmap]_\beta^\gamma$. By solving $$ \left[ \begin{array}{ccc} 1 & 1 & 2 \\ -3 & -2 & -5 \\ -2 & -2 & -3 \\ \end{array} \right] \begin{bmatrix} c_1 \\c_2 \\ c_3 \\ \end{bmatrix} = \begin{bmatrix} 1 \\-3 \\ 0 \\ \end{bmatrix},$$ we get $$\begin{bmatrix} c_1 \\c_2 \\ c_3 \\ \end{bmatrix} = \begin{bmatrix} -1 \\-2 \\ 2 \\ \end{bmatrix}.$$ Repeat the above steps and replace $\bv_1$ with $\bv_2$ and $\bv_3$, we know $$ [\idmap]_\beta^\gamma = \left[ \begin{array}{ccc} -1 & -3 & -2 \\ -2 & -4 & 1 \\ 2 & 5 & 1 \\ \end{array} \right].$$ Using the same method, we get $$ [\idmap]_\alpha^\gamma = \left[ \begin{array}{ccc} -4 & -1 & -1 \\ 1 & 1 & -1 \\ 2 & 0 & 1 \\ \end{array} \right].$$ Finally, $$ [\idmap]_\beta^\gamma [\idmap]_\alpha^\beta = \left[ \begin{array}{ccc} -1 & -3 & -2 \\ -2 & -4 & 1 \\ 2 & 5 & 1 \\ \end{array} \right] \left[ \begin{array}{ccc} 7 & 2 & 5 \\ -3 & -1 & -2 \\ 3 & 1 & 1 \\ \end{array} \right] = \left[ \begin{array}{ccc} -4 & -1 & -1 \\ 1 & 1 & -1 \\ 2 & 0 & 1 \\ \end{array} \right] = [\idmap]_\alpha^\gamma. $$ ## Exercises ##### Exercise 2 對於以下的各向量空間 $V$、以及兩個基底 $\alpha$ 和 $\beta$﹐ 求出 $[\idmap]_\alpha^\beta$ 和 $[\idmap]_\beta^\alpha$。  For each of the following vector space $V$ and two bases $\alpha$ and $\beta$, find $[\idmap]_\alpha^\beta$ and $[\idmap]_\beta^\alpha$.  ##### Exercise 2(a) 令 $V = \mathbb{R}^3$、 $\alpha = \{\be_1, \be_2, \be_3\}$ 為標準基底、 $$ \beta = \left\{\begin{aligned} \bu_1 = (0,0,1), \\ \bu_2 = (0,1,1), \\ \bu_3 = (1,1,1) \end{aligned}\right\}. $$  Let $V = \mathbb{R}^3$, $\alpha = \{\be_1, \be_2, \be_3\}$ be the standard basis, and $$ \beta = \left\{\begin{aligned} \bu_1 = (0,0,1), \\ \bu_2 = (0,1,1), \\ \bu_3 = (1,1,1) \end{aligned}\right\}. $$  :::warning - [x] Make your notation consistent: $\bc_i$ --> $c_i$ ::: $Ans:$ $$[\operatorname{id}]_\beta^\alpha =\left[ \begin{array}{ccc} 0 & 0 & 1 \\ 0 & 1 & 1 \\ 1 & 1 & 1 \\ \end{array} \right]. $$ Let $\alpha = \{\be_1, \be_2, \be_3\}$ , $\beta = \{\bu_1, \bu_2, \bu_3\}$ and $\be_1 = c_1\bu_1 + c_2\bu_2 + c_3\bu_3$ , $c_1,c_2,c_3 \in \mathbb R$. The vector representation of $\be_1$ is $\begin{bmatrix} c_1 \\c_2 \\ c_3 \\\end{bmatrix}$ with respect to $\beta$, so we know $\begin{bmatrix} c_1 \\c_2 \\ c_3 \\\end{bmatrix}$ is the first column of $[\idmap]_\alpha^\beta$. By solving $$ \left[ \begin{array}{ccc} 0 & 0 & 1 \\ 0 & 1 & 1 \\ 1 & 1 & 1 \\ \end{array} \right] \begin{bmatrix} c_1 \\c_2 \\ c_3 \\ \end{bmatrix} = \begin{bmatrix} 1 \\0 \\ 0 \\ \end{bmatrix},$$ we get $$\begin{bmatrix} c_1 \\c_2 \\ c_3 \\ \end{bmatrix} = \begin{bmatrix} 0 \\-1 \\ 1 \\ \end{bmatrix}.$$ Repeat the above steps and replace $\be_1$ with $\be_2$ and $\be_3$, we know $$ [\idmap]_\alpha^\beta = \left[ \begin{array}{ccc} 0 & -1 & 1 \\ -1 & 1 & 0 \\ 1 & 0 & 0 \\ \end{array} \right]. $$ ##### Exercise 2(b) 令 $V = \mathbb{R}^3$、 $\alpha = \{\be_1, \be_2, \be_3\}$ 為標準基底、 $$ \beta = \left\{\begin{aligned} \bu_1 = (\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}), \\ \bu_2 = (\frac{1}{\sqrt{2}},-\frac{1}{\sqrt{2}},0), \\ \bu_3 = (\frac{1}{\sqrt{6}},\frac{1}{\sqrt{6}},-\frac{2}{\sqrt{6}}) \end{aligned}\right\}. $$  Let $V = \mathbb{R}^3$, $\alpha = \{\be_1, \be_2, \be_3\}$ be the standard basis, and $$ \beta = \left\{\begin{aligned} \bu_1 = (\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}), \\ \bu_2 = (\frac{1}{\sqrt{2}},-\frac{1}{\sqrt{2}},0), \\ \bu_3 = (\frac{1}{\sqrt{6}},\frac{1}{\sqrt{6}},-\frac{2}{\sqrt{6}}) \end{aligned}\right\}. $$  :::warning - [x] Make your notation consistent: $\bc_i$ --> $c_i$ Note that the matrix in your answer is an orthogonal matrix, so its inverse is the same as its transpose. ::: $Ans:$ $$[\operatorname{id}]_\beta^\alpha =\left[ \begin{array}{ccc} \frac{1}{\sqrt{3}} & \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{6}} \\ \frac{1}{\sqrt{3}} & -\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{6}} \\ \frac{1}{\sqrt{3}} & 0 & -\frac{2}{\sqrt{6}} \\ \end{array} \right].$$ Let $\alpha = \{\be_1, \be_2, \be_3\}$ , $\beta = \{\bu_1, \bu_2, \bu_3\}$ and $\be_1 = c_1\bu_1 + c_2\bu_2 + c_3\bu_3$ , $c_1,c_2,c_3 \in \mathbb R$. The vector representation of $\be_1$ is $\begin{bmatrix} c_1 \\c_2 \\ c_3 \\\end{bmatrix}$ with respect to $\beta$, so we know $\begin{bmatrix} c_1 \\c_2 \\ c_3 \\\end{bmatrix}$ is the first column of $[\idmap]_\alpha^\beta$. By solving $$ \left[ \begin{array}{ccc} \frac{1}{\sqrt{3}} & \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{6}} \\ \frac{1}{\sqrt{3}} & -\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{6}} \\ \frac{1}{\sqrt{3}} & 0 & -\frac{2}{\sqrt{6}} \\ \end{array} \right] \begin{bmatrix} c_1 \\c_2 \\ c_3 \\ \end{bmatrix} = \begin{bmatrix} 1 \\0 \\ 0 \\ \end{bmatrix},$$ we get $$\begin{bmatrix} c_1 \\c_2 \\ c_3 \\ \end{bmatrix} = \begin{bmatrix} \frac{1}{\sqrt{3}} \\\frac{1}{\sqrt{2}} \\ -\frac{1}{\sqrt{6}} \\ \end{bmatrix}.$$ Repeat the above steps and replace $\be_1$ with $\be_2$ and $\be_3$, we know $$ [\idmap]_\alpha^\beta = \left[ \begin{array}{ccc} \frac{1}{\sqrt{3}} & \frac{1}{\sqrt{3}} & \frac{1}{\sqrt{3}} \\ \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} & 0 \\ \frac{1}{\sqrt{6}} & \frac{1}{\sqrt{6}} & -\frac{2}{\sqrt{6}} \\ \end{array} \right]. $$ ##### Exercise 2(c) 令 $V = \mathcal{P}_2$、 $\alpha = \{1, x, x^2\}$ 為標準基底、 $\beta = \{1, 1-x, (1-x)^2\}$。  Let $V = \mathcal{P}_2$, $\alpha = \{1, x, x^2\}$ be the standard basis, and $\beta = \{1, 1-x, (1-x)^2\}$.  :::warning - [x] Space after any punctuation mark. - [x] Typo: vectoers Technically, $\beta$ does not contain any vectors in $\mathbb{R}^2$. ::: **Answer** Since the vector representation of a polynomial with respect to the standard basis $\alpha$ is simply recording the coefficients, we have $$ [\operatorname{id}]_\beta^\alpha =\left[ \begin{array}{ccc} 1 & 1 & 1 \\ 0 & -1 & -2 \\ 0 & 0 & 1 \\ \end{array} \right]. $$ Then, we may compute $$ [\operatorname{id}]_\alpha^\beta = ([\idmap]_\beta^\alpha)^{-1} = \begin{bmatrix} 1 & 1 & 1 \\ 0 & -1 & -2 \\ 0 & 0 & 1 \\ \end{bmatrix}. $$  ##### Exercise 2(d) 令 $$ \begin{aligned} p_1(x) &= \frac{(x-2)(x-3)}{(1-2)(1-3)}, \\ p_2(x) &= \frac{(x-1)(x-3)}{(2-1)(2-3)}, \\ p_3(x) &= \frac{(x-1)(x-2)}{(3-1)(3-2)}. \\ \end{aligned} $$ 令 $V = \mathcal{P}_2$、 $\alpha = \{1, x, x^2\}$ 為標準基底、 $\beta = \{p_1, p_2, p_3\}$。  Let $$ \begin{aligned} p_1(x) &= \frac{(x-2)(x-3)}{(1-2)(1-3)}, \\ p_2(x) &= \frac{(x-1)(x-3)}{(2-1)(2-3)}, \\ p_3(x) &= \frac{(x-1)(x-2)}{(3-1)(3-2)}. \\ \end{aligned} $$ Let $V = \mathcal{P}_2$, $\alpha = \{1, x, x^2\}$ be the standard basis, and $\beta = \{p_1, p_2, p_3\}$.  ##### Exercise 3 說明以下三個等式的直觀解釋： - $[\idmap]_\alpha^\alpha = I_n$, - $[\idmap]_\beta^\gamma[\idmap]_\alpha^\beta = [\idmap]_\alpha^\gamma$, - $[\idmap]_\beta^\alpha[\idmap]_\alpha^\beta = I_n$.  Give some intuition to the following identities: - $[\idmap]_\alpha^\alpha = I_n$, - $[\idmap]_\beta^\gamma[\idmap]_\alpha^\beta = [\idmap]_\alpha^\gamma$, - $[\idmap]_\beta^\alpha[\idmap]_\alpha^\beta = I_n$.  ##### Exercise 4 我們已經看到每個基底轉換矩陣 $[\idmap]_\alpha^\beta$ 都是可逆的﹐而且其逆矩陣就是 $[\idmap]_\beta^\alpha$。證明每個 $n\times n$ 的可逆矩陣也可以寫成 $\mathbb{R}^n$ 中的兩組基底 $\alpha$ 和 $\beta$ 所建構出來的基底轉移矩陣。（把 $\alpha$ 取為 $A$ 的各行向量、 $\beta$ 取為標準基底。）  We have seen that every change of basis matrix $[\idmap]_\alpha^\beta$ is invertible and its inverse is $[\idmap]_\beta^\alpha$. Show that every $n\times n$ matrix can be written as the change of basis matrix in $\mathbb{R}^n$ for some basis $\alpha$ and $\beta$. Hint: You may choose $\alpha$ as the columns of $A$ and $\beta$ as the standard basis of $\mathbb{R}^n$.  :::info collaboration: 2 3 problems: 3 - done: 2a, 2b, 2c extra: 1 moderator: 1 quality control: 1 :::

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