矩陣指數 - HackMD

# 矩陣指數 Matrix exponential ![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_int_list ``` ## Main idea We know the exponential function can be written as $$ e^x = \frac{1}{0!} + \frac{1}{1!}x + \frac{1}{2!}x^2 + \cdots. $$ In a similar way, the **matrix exponential** of a square matrix $A$ is defined as $$ e^A = \frac{1}{0!}I + \frac{1}{1!}A + \frac{1}{2!}A^2 + \cdots. $$ The computation of the matrix exponential can be a tedious work, at least by hand. However, when $D$ is a diagonal matrix, its matrix exponential can be easily obtained by $$ \begin{aligned} e^D &= \frac{1}{0!}I + \frac{1}{1!}D + \frac{1}{2!}D^2 + \cdots \\ &= \frac{1}{0!} \begin{bmatrix} 1 & ~ & ~ \\ ~ & \ddots & ~ \\ ~ & ~ & 1 \end{bmatrix} + \frac{1}{1!} \begin{bmatrix} \lambda_1 & ~ & ~ \\ ~ & \ddots & ~ \\ ~ & ~ & \lambda_n \end{bmatrix} + \frac{1}{2!} \begin{bmatrix} \lambda_1^2 & ~ & ~ \\ ~ & \ddots & ~ \\ ~ & ~ & \lambda_n^2 \end{bmatrix} + \cdots \\ &= \begin{bmatrix} \frac{1}{0!} + \frac{1}{1!}\lambda_1 + \frac{1}{2!}\lambda_1^2 + \cdots & ~ & ~ \\ ~ & \ddots & ~ \\ ~ & ~ & \frac{1}{0!} + \frac{1}{1!}\lambda_n + \frac{1}{2!}\lambda_n^2 + \cdots \end{bmatrix} \\ &= \begin{bmatrix} e^{\lambda_1} & ~ & ~ \\ ~ & \ddots & ~ \\ ~ & ~ & e^{\lambda_n} \end{bmatrix}. \end{aligned} $$ Moreover, suppose $A$ is diagonalizable as $D = Q^{-1}AQ$ and $A = QDQ^{-1}$. Then the matrix exponential can be computed as $$ \begin{aligned} e^A &= \frac{1}{0!}I + \frac{1}{1!}A + \frac{1}{2!}A^2 + \cdots \\ &= \frac{1}{0!}I + \frac{1}{1!}QAQ^{-1} + \frac{1}{2!}QA^2Q^{-1} + \cdots \\ &= Q\left(\frac{1}{0!}I + \frac{1}{1!}D + \frac{1}{2!}D^2 + \cdots \right)Q^{-1} \\ &= Qe^DQ^{-1}. \end{aligned} $$ ## Side stories - Taylor expansion - Cayley transform - solution to $\dot{\bx} = A\bx$ ## Experiments ##### Exercise 1 執行以下程式碼。令 $s_k = \sum_{r = 0}^k \frac{1}{r!}A^r$ 為計算 $e^A$ 時的部份和。定義 $\|B - C\|^2$ 為 $B - C$ 矩陣各項的平方和。  Run the code below. Let $s_k = \sum_{r = 0}^k \frac{1}{r!}A^r$ be the partial sum of $e^A$. Define $\|B - C\|^2$ as the square sum of entries of $B - C$.  ```python ### code set_random_seed(0) print_ans = False n = 2 A = matrix(n, random_int_list(n^2, 3)) pretty_print(LatexExpr("A ="), A) tk = [identity_matrix(n), A] for k in range(2,16): tk.append(tk[-1] * A / k) sk = tk[:1] for k in range(1,16): sk.append(sk[-1] + tk[k]) pretty_print(LatexExpr("s_{5} ="), N(sk[5])) pretty_print(LatexExpr("s_{10} ="), N(sk[10])) pretty_print(LatexExpr("s_{15} ="), N(sk[15])) if print_ans: print("| s5 - s10 |^2 =", (sk[5] - sk[10]).norm("frob")) print("| s10 - s15 |^2 =", (sk[10] - sk[15]).norm("frob")) ``` By setting `seed=0`. $A=\begin{bmatrix}-3&3\\1&2\end{bmatrix}.$ $s_5=\begin{bmatrix} -0.525000000000000 & 6.85000000000000 \\ 2.28333333333333 & 10.8916666666667\end{bmatrix}.$ $s_{10}=\begin{bmatrix} 1.17567708333333 & 6.23704613095238 \\ 2.07901537698413 & 11.5707539682540\end{bmatrix}.$ $s_{15}=\begin{bmatrix} 1.15645523757869 & 6.24794131648453 \\ 2.08264710549484 & 11.5696907650529\end{bmatrix}.$ --- ##### Exercise 1(a) 計算 $\|s_5 - s_{10}\|^2$。  Find 計算 $\|s_5 - s_{10}\|^2$.  ##### **Answer:** $$ \|s_5 -s_{10}\|^2= \left\|\begin{bmatrix} -1.70067708333333 &0.612953869047620 \\ 0.204317956349200 &-0.679087301587300 \end{bmatrix}\right\|^2=3.77092037781944 $$ --- ##### Exercise 1(b) 計算 $\|s_{10} - s_{15}\|^2$。  Find 計算 $\|s_{10} - s_{15}\|^2$.  ##### **Answer:** $$ \|s_{10} -s_{15}\|^2= \left\|\begin{bmatrix} 0.0192218457546400 & -0.0108951855321502 \\ -0.00363172851070992 & 0.00106320320109887 \end{bmatrix}\right\|^2=0.000502504275017477 $$ --- ##### Remark 在高等微積分會學到，在一個完備的空間裡（像是 $\mathbb{R}$ 或是 $\mathbb{R}^n$ 或是跟矩陣長很像的 $\mathbb{R}^{n^2}$），如果一個數列有 $\| s_n - s_m \|$ 隨著 $n,m$ 變大而距離愈來愈小，則 $\{s_n\}$ 數列會收斂。  In the future analysis courses, we will learn that in a complete space (such as $\mathbb{R}$, $\mathbb{R}^n$, or $\mathbb{R}^{n^2}$, which has a similar stucture as matrices), any sequence $\{s_n\}$ with $\| s_n - s_m \|$ going to zero as $n,m$ goes to infinity will converge.  :::info What do the experiments try to tell you? (open answer) ... ::: ##### **Answer:** How to compute $e^A$ and as the remark mentioned. --- ## Exercises ##### Exercise 2 令 $O_n$ 為 $n\times n$ 的零矩陣，用定義計算 $e^{O_n}$。  Let $O_n$ be the $n\times n$ zero matrix. Find $e^{O_n}$ by definition.  ##### **Answer:** $e^{O_n}= \frac{1}{0!}\mathit{I}_n + \frac{1}{1!}O_n + \frac{1}{2!} {O_n}^2+ \cdots = \mathit{I}_n.$ Thus, $e^{O_n}=\mathit{I}_n$, no matter what $n$ is. --- ##### Exercise 3 令 $$ A = \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix}. $$ 依照以下步驟求出 $e^{tA}$，其中 $t$ 是變數。  Let $$ A = \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix}. $$ Use the given instructions to find $e^{tA}$, where $t$ is a variable.  ##### Exercise 3(a) 計算 $A, A^2, A^3, A^4$ 並找出 $A^n$ 的規律。  Find $A, A^2, A^3, A^4$. Then find a pattern of $A^n$.  ##### **Answer :** ##### **part(a) :** By calculation, $A = \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix},\space A^2 = \begin{bmatrix} -1 & 0 \\ 0 & -1 \end{bmatrix},\space A^3 = \begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix},\space A^4 = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = \mathit{I}_2.$ ##### **part(b) :** From the pattern of $A$ ~ $A^4$, we find that 1. If $n = 4k+1$, $k \in \mathbb{N}$. Then $A^n = \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix}.$ 2. If $n = 4k+2$, $k \in \mathbb{N}$. Then $A^n = \begin{bmatrix} -1 & 0 \\ 0 & -1 \end{bmatrix}.$ 3. If $n = 4k+3$, $k \in \mathbb{N}$. Then $A^n = \begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix}.$ 4. If $n = 4k$, $k \in \mathbb{N}$. Then $A^n = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = \mathit{I}_2.$ --- ##### Exercise 3(b) 利用定義求出 $e^{tA}$，並解釋 $t$ 在旋轉矩陣中扮演的角色。  Find $e^{tA}$ by definition. Then explain what is the geometric meaning of $t$.  ##### **Answer :** From the definition of $e^{A}$ and the result of 3(a), we are able to orginize $e^{tA}$ to the folowing form, $\displaystyle e^{tA} = \frac{1}{0!}I + \frac{1}{1!}{tA} + \frac{1}{2!}t^2A^2 + \cdots$ $= \begin{bmatrix} \frac{1}{0!} - \frac{1}{2!}t^2 + \frac{1}{4!}t^4 - \cdots & -\frac{1}{1!}t + \frac{1}{3!}t^3 - \frac{1}{5!}t^5 + \cdots \\ \frac{1}{1!}t - \frac{1}{3!}t^3 + \frac{1}{5!}t^5 - \cdots & \frac{1}{0!} - \frac{1}{2!}t^2 + \frac{1}{4!}t^4 - \cdots \end{bmatrix} = \begin{bmatrix} \sum_{n = 0}^\infty \frac{{(-1)}^n}{2n!}t^{2n} & \sum_{n = 0}^\infty \frac{{(-1)}^n+1}{2n+1!}t^{2n+1} \\ \sum_{n = 0}^\infty \frac{{(-1)}^n}{2n+1!}t^{2n+1} & \sum_{n = 0}^\infty \frac{{(-1)}^n}{2n!}t^{2n} \end{bmatrix} = \begin{bmatrix} \cos t & -\sin t \\ \sin t & \cos t \end{bmatrix}.$ By observation, $e^{tA}$ is a rotation matrix, and $t$ is the the angle of rotation. --- ##### Exercise 4 對以下矩陣 $A$ 求出 $e^A$。  For the following matrices $A$, find $e^A$.  ##### Exercise 4(a) 已知 $$ \begin{bmatrix} 4 & 0 \\ 0 & 6 \end{bmatrix} = \begin{bmatrix} 1 & 1 \\ 1 & -1 \end{bmatrix}^{-1} \begin{bmatrix} 5 & -1 \\ -1 & 5 \end{bmatrix} \begin{bmatrix} 1 & 1 \\ 1 & -1 \end{bmatrix}. $$ 令 $$ A = \begin{bmatrix} 5 & -1 \\ -1 & 5 \end{bmatrix}. $$  It is known that $$ \begin{bmatrix} 4 & 0 \\ 0 & 6 \end{bmatrix} = \begin{bmatrix} 1 & 1 \\ 1 & -1 \end{bmatrix}^{-1} \begin{bmatrix} 5 & -1 \\ -1 & 5 \end{bmatrix} \begin{bmatrix} 1 & 1 \\ 1 & -1 \end{bmatrix}. $$ Let $$ A = \begin{bmatrix} 5 & -1 \\ -1 & 5 \end{bmatrix}. $$  ##### **Answer:** Known from the given diagonal matrix equation: $$ A= \begin{bmatrix} 5 & -1 \\ -1 & 5 \end{bmatrix}= \begin{bmatrix} 1 & 1 \\ 1 & -1 \end{bmatrix} \begin{bmatrix} 4 & 0 \\ 0 & 6 \end{bmatrix} \begin{bmatrix} 1 & 1 \\ 1 & -1 \end{bmatrix}^{-1}. $$ Let $Q= \begin{bmatrix} 1 & 1 \\ 1 & -1 \end{bmatrix}$ and we apply $e^A=Qe^DQ^{-1}$ $$ e^A= \begin{bmatrix} 1 & 1 \\ 1 & -1 \end{bmatrix} \begin{bmatrix} e^4 & 0 \\ 0 & e^6 \end{bmatrix} \begin{bmatrix} 1 & 1 \\ 1 & -1 \end{bmatrix}^{-1}= \begin{bmatrix} \frac{e^4+e^6}{2} & \frac{e^4-e^6}{2} \\ \frac{e^4-e^6}{2} & \frac{e^4+e^6}{2} \\ \end{bmatrix} . $$ --- ##### Exercise 4(b) 已知 $$ \begin{bmatrix} 4 & 0 \\ 0 & -6 \end{bmatrix} = \begin{bmatrix} 1 & 1 \\ 1 & -1 \end{bmatrix}^{-1} \begin{bmatrix} -1 & 5 \\ 5 & -1 \end{bmatrix} \begin{bmatrix} 1 & 1 \\ 1 & -1 \end{bmatrix}. $$ 令 $$ A = \begin{bmatrix} -1 & 5 \\ 5 & -1 \end{bmatrix}. $$  It is known that $$ \begin{bmatrix} 4 & 0 \\ 0 & -6 \end{bmatrix} = \begin{bmatrix} 1 & 1 \\ 1 & -1 \end{bmatrix}^{-1} \begin{bmatrix} -1 & 5 \\ 5 & -1 \end{bmatrix} \begin{bmatrix} 1 & 1 \\ 1 & -1 \end{bmatrix}. $$ Let $$ A = \begin{bmatrix} -1 & 5 \\ 5 & -1 \end{bmatrix}. $$  ##### **Answer:** Known from the given diagonal matrix equation: $$ A= \begin{bmatrix} -1 & 5 \\ 5 & -1 \end{bmatrix}= \begin{bmatrix} 1 & 1 \\ 1 & -1 \end{bmatrix} \begin{bmatrix} 4 & 0 \\ 0 & -6 \end{bmatrix} \begin{bmatrix} 1 & 1 \\ 1 & -1 \end{bmatrix}^{-1}. $$ Let $Q= \begin{bmatrix} 1 & 1 \\ 1 & -1 \end{bmatrix}$ and we apply $e^A=Qe^DQ^{-1}$ $$ e^A= \begin{bmatrix} 1 & 1 \\ 1 & -1 \end{bmatrix} \begin{bmatrix} e^4 & 0 \\ 0 & e^{-6} \end{bmatrix} \begin{bmatrix} 1 & 1 \\ 1 & -1 \end{bmatrix}^{-1}= \begin{bmatrix} \frac{e^4+e^{-6}}{2} & \frac{e^4-e^{-6}}{2} \\ \frac{e^4-e^{-6}}{2} & \frac{e^4+e^{-6}}{2} \\ \end{bmatrix} . $$ --- ##### Exercise 4(c) 已知 $$ \begin{bmatrix} 2 & 0 \\ 0 & 0 \end{bmatrix} = \begin{bmatrix} 1 & 1 \\ 1 & -1 \end{bmatrix}^{-1} \begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix} \begin{bmatrix} 1 & 1 \\ 1 & -1 \end{bmatrix}. $$ 令 $$ A = \begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix}. $$  It is known that $$ \begin{bmatrix} 2 & 0 \\ 0 & 0 \end{bmatrix} = \begin{bmatrix} 1 & 1 \\ 1 & -1 \end{bmatrix}^{-1} \begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix} \begin{bmatrix} 1 & 1 \\ 1 & -1 \end{bmatrix}. $$ Let $$ A = \begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix}. $$  ##### **Answer:** Known from the given diagonal matrix equation: $$ A= \begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix}= \begin{bmatrix} 1 & 1 \\ 1 & -1 \end{bmatrix} \begin{bmatrix} 2 & 0 \\ 0 & 0 \end{bmatrix} \begin{bmatrix} 1 & 1 \\ 1 & -1 \end{bmatrix}^{-1}. $$ Let $Q= \begin{bmatrix} 1 & 1 \\ 1 & -1 \end{bmatrix}$ and we apply $e^A=Qe^DQ^{-1}$ $$ e^A= \begin{bmatrix} 1 & 1 \\ 1 & -1 \end{bmatrix} \begin{bmatrix} e^2 & 0 \\ 0 & e^0 \end{bmatrix} \begin{bmatrix} 1 & 1 \\ 1 & -1 \end{bmatrix}^{-1}= \begin{bmatrix} \frac{e^2+1}{2} & \frac{e^2-1}{2} \\ \frac{e^2-1}{2} & \frac{e^2+1}{2} \\ \end{bmatrix} . $$ --- ##### Exercise 5 對以下矩陣 $A$ 求出 $e^A$。  For the following matrices $A$, find $e^A$.  ##### Exercise 5(a) 已知 $$ \begin{bmatrix} 3 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & 6 \end{bmatrix} = \begin{bmatrix} 1 & 1 & 1 \\ 1 & -1 & 1 \\ 1 & 0 & -2 \end{bmatrix}^{-1} \begin{bmatrix} 4 & 0 & -1 \\ 0 & 4 & -1 \\ -1 & -1 & 5 \end{bmatrix} \begin{bmatrix} 1 & 1 & 1 \\ 1 & -1 & 1 \\ 1 & 0 & -2 \end{bmatrix}. $$ 令 $$ A = \begin{bmatrix} 4 & 0 & -1 \\ 0 & 4 & -1 \\ -1 & -1 & 5 \end{bmatrix}. $$  It is known that $$ \begin{bmatrix} 3 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & 6 \end{bmatrix} = \begin{bmatrix} 1 & 1 & 1 \\ 1 & -1 & 1 \\ 1 & 0 & -2 \end{bmatrix}^{-1} \begin{bmatrix} 4 & 0 & -1 \\ 0 & 4 & -1 \\ -1 & -1 & 5 \end{bmatrix} \begin{bmatrix} 1 & 1 & 1 \\ 1 & -1 & 1 \\ 1 & 0 & -2 \end{bmatrix}. $$ Let $$ A = \begin{bmatrix} 4 & 0 & -1 \\ 0 & 4 & -1 \\ -1 & -1 & 5 \end{bmatrix}. $$  ##### **Answer:** Known from the given diagonal matrix equation: $$ A= \begin{bmatrix} 4 & 0 & -1 \\ 0 & 4 & -1 \\ -1 & -1 & 5 \end{bmatrix}= \begin{bmatrix} 1 & 1 & 1 \\ 1 & -1 & 1 \\ 1 & 0 & -2 \end{bmatrix} \begin{bmatrix} 3 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & 6 \end{bmatrix} \begin{bmatrix} 1 & 1 & 1 \\ 1 & -1 & 1 \\ 1 & 0 & -2 \end{bmatrix}^{-1}. $$ Let $Q= \begin{bmatrix} 1 & 1 & 1 \\ 1 & -1 & 1 \\ 1 & 0 & -2 \end{bmatrix}$ and we apply $e^A=Qe^DQ^{-1}$ $$ e^A= \begin{bmatrix} 1 & 1 & 1 \\ 1 & -1 & 1 \\ 1 & 0 & -2 \end{bmatrix} \begin{bmatrix} e^{3} & 0 & 0 \\ 0 & e^{4} & 0 \\ 0 & 0 & e^{6} \end{bmatrix} \begin{bmatrix} 1 & 1 & 1 \\ 1 & -1 & 1 \\ 1 & 0 & -2 \end{bmatrix}^{-1}= \begin{bmatrix} \frac{2e^3+3e^4+e^6}{6} & \frac{2e^3-3e^4+e^6}{6} & \frac{e^3-e^6}{3} \\ \frac{2e^3-3e^4+e^6}{6} & \frac{2e^3+3e^4+e^6}{6} & \frac{e^3-e^6}{3} \\ \frac{e^3-e^6}{3} & \frac{e^3-e^6}{3} & \frac{e^3+2e^6}{3} \\ \end{bmatrix} . $$ --- ##### Exercise 5(b) 已知 $$ \begin{bmatrix} 3 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & -6 \end{bmatrix} = \begin{bmatrix} 1 & 1 & 1 \\ 1 & -1 & 1 \\ 1 & 0 & -2 \end{bmatrix}^{-1} \begin{bmatrix} 2 & -2 & 3 \\ -2 & 2 & 3 \\ 3 & 3 & -3 \end{bmatrix} \begin{bmatrix} 1 & 1 & 1 \\ 1 & -1 & 1 \\ 1 & 0 & -2 \end{bmatrix}. $$ 令 $$ A = \begin{bmatrix} 2 & -2 & 3 \\ -2 & 2 & 3 \\ 3 & 3 & -3 \end{bmatrix}. $$  It is known that $$ \begin{bmatrix} 3 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & -6 \end{bmatrix} = \begin{bmatrix} 1 & 1 & 1 \\ 1 & -1 & 1 \\ 1 & 0 & -2 \end{bmatrix}^{-1} \begin{bmatrix} 2 & -2 & 3 \\ -2 & 2 & 3 \\ 3 & 3 & -3 \end{bmatrix} \begin{bmatrix} 1 & 1 & 1 \\ 1 & -1 & 1 \\ 1 & 0 & -2 \end{bmatrix}. $$ Let $$ A = \begin{bmatrix} 2 & -2 & 3 \\ -2 & 2 & 3 \\ 3 & 3 & -3 \end{bmatrix}. $$  ##### **Answer:** Known from the given diagonal matrix equation: $$ A= \begin{bmatrix} 2 & -2 & 3 \\ -2 & 2 & 3 \\ 3 & 3 & -3 \end{bmatrix}= \begin{bmatrix} 1 & 1 & 1 \\ 1 & -1 & 1 \\ 1 & 0 & -2 \end{bmatrix} \begin{bmatrix} 3 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & -6 \end{bmatrix} \begin{bmatrix} 1 & 1 & 1 \\ 1 & -1 & 1 \\ 1 & 0 & -2 \end{bmatrix}^{-1}. $$ Let $Q= \begin{bmatrix} 1 & 1 & 1 \\ 1 & -1 & 1 \\ 1 & 0 & -2 \end{bmatrix}$ and we apply $e^A=Qe^DQ^{-1}$ $$ e^A= \begin{bmatrix} 1 & 1 & 1 \\ 1 & -1 & 1 \\ 1 & 0 & -2 \end{bmatrix} \begin{bmatrix} e^{3} & 0 & 0 \\ 0 & e^{4} & 0 \\ 0 & 0 & e^{-6} \end{bmatrix} \begin{bmatrix} 1 & 1 & 1 \\ 1 & -1 & 1 \\ 1 & 0 & -2 \end{bmatrix}^{-1}= \begin{bmatrix} \frac{2e^3+3e^4+e^{-6}}{6} & \frac{2e^3-3e^4+e^{-6}}{6} & \frac{e^3-e^{-6}}{3} \\ \frac{2e^3-3e^4+e^{-6}}{6} & \frac{2e^3+3e^4+e^{-6}}{6} & \frac{e^3-e^{-6}}{3} \\ \frac{e^3-e^{-6}}{3} & \frac{e^3-e^{-6}}{3} & \frac{e^3+2e^{-6}}{3} \\ \end{bmatrix} . $$ --- ##### Exercise 6 對以下矩陣 $A$ 求出 $e^A$。  For the following matrices $A$, find $e^A$.  ##### Exercise 6(a) 已知 $$ \begin{bmatrix} 2 & 0 \\ 0 & 3 \end{bmatrix} = \begin{bmatrix} 1 & 1 \\ 3 & 2 \end{bmatrix}^{-1} \begin{bmatrix} 0 & 1 \\ -6 & 5 \end{bmatrix} \begin{bmatrix} 1 & 1 \\ 3 & 2 \end{bmatrix}. $$ 令 $$ A = \begin{bmatrix} 0 & 1 \\ -6 & 5 \end{bmatrix}. $$  It is known that $$ \begin{bmatrix} 2 & 0 \\ 0 & 3 \end{bmatrix} = \begin{bmatrix} 1 & 1 \\ 3 & 2 \end{bmatrix}^{-1} \begin{bmatrix} 0 & 1 \\ -6 & 5 \end{bmatrix} \begin{bmatrix} 1 & 1 \\ 3 & 2 \end{bmatrix}. $$ Let $$ A = \begin{bmatrix} 0 & 1 \\ -6 & 5 \end{bmatrix}. $$  ##### **Answer:** Known from the given diagonal matrix equation: $$ A= \begin{bmatrix} 0 & 1 \\ -6 & 5 \end{bmatrix}= \begin{bmatrix} 1 & 1 \\ 3 & 2 \end{bmatrix} \begin{bmatrix} 2 & 0 \\ 0 & 3 \end{bmatrix} \begin{bmatrix} 1 & 1 \\ 3 & 2 \end{bmatrix}^{-1}. $$ Let $Q= \begin{bmatrix} 1 & 1 \\ 3 & 2 \end{bmatrix}$ and we apply $e^A=Qe^DQ^{-1}$ $$ e^A= \begin{bmatrix} 1 & 1 \\ 3 & 2 \end{bmatrix} \begin{bmatrix} e^2 & 0 \\ 0 & e^3 \end{bmatrix} \begin{bmatrix} 1 & 1 \\ 3 & 2 \end{bmatrix}^{-1}= \begin{bmatrix} -2e^2+3e^3 & e^2-e^3 \\ -6e^2+6e^3 & 3e^2-2e^3 \\ \end{bmatrix} . $$ --- ##### Exercise 6(b) 已知 $$ \begin{bmatrix} 2 & 0 \\ 0 & -2 \end{bmatrix} = \begin{bmatrix} 1 & 1 \\ 2 & -2 \end{bmatrix}^{-1} \begin{bmatrix} 0 & 1 \\ -4 & 0 \end{bmatrix} \begin{bmatrix} 1 & 1 \\ 2 & -2 \end{bmatrix}. $$ 令 $$ A = \begin{bmatrix} 0 & 1 \\ -4 & 0 \end{bmatrix}. $$  It is known that $$ \begin{bmatrix} 2 & 0 \\ 0 & -2 \end{bmatrix} = \begin{bmatrix} 1 & 1 \\ 2 & -2 \end{bmatrix}^{-1} \begin{bmatrix} 0 & 1 \\ -4 & 0 \end{bmatrix} \begin{bmatrix} 1 & 1 \\ 2 & -2 \end{bmatrix}. $$ Let $$ A = \begin{bmatrix} 0 & 1 \\ -4 & 0 \end{bmatrix}. $$  ##### **Answer:** Known from the given diagonal matrix equation: $$ A= \begin{bmatrix} 0 & 1 \\ -4 & 0 \end{bmatrix}= \begin{bmatrix} 1 & 1 \\ 2 & -2 \end{bmatrix} \begin{bmatrix} 2 & 0 \\ 0 & -2 \end{bmatrix} \begin{bmatrix} 1 & 1 \\ 2 & -2 \end{bmatrix}^{-1}. $$ Let $Q= \begin{bmatrix} 1 & 1 \\ 2 & -2 \end{bmatrix}$ and we apply $e^A=Qe^DQ^{-1}$ $$ e^A= \begin{bmatrix} 1 & 1 \\ 2 & -2 \end{bmatrix} \begin{bmatrix} e^2 & 0 \\ 0 & e^{-2} \end{bmatrix} \begin{bmatrix} 1 & 1 \\ 2 & -2 \end{bmatrix}^{-1}= \begin{bmatrix} \frac{e^4+1}{2e^2} & \frac{e^4-1}{4e^2} \\ \frac{e^4-1}{e^2} & \frac{e^4+1}{2e^2} \\ \end{bmatrix} . $$ --- ##### Exercise 6(c) 已知 $$ \begin{bmatrix} 2 & 1 \\ 0 & 2 \end{bmatrix} = \begin{bmatrix} 2 & -1 \\ 4 & 0 \end{bmatrix}^{-1} \begin{bmatrix} 0 & 1 \\ -4 & 4 \end{bmatrix} \begin{bmatrix} 2 & -1 \\ 4 & 0 \end{bmatrix}. $$ 令 $$ A = \begin{bmatrix} 0 & 1 \\ -4 & 4 \end{bmatrix}. $$  It is known that $$ \begin{bmatrix} 2 & 1 \\ 0 & 2 \end{bmatrix} = \begin{bmatrix} 2 & -1 \\ 4 & 0 \end{bmatrix}^{-1} \begin{bmatrix} 0 & 1 \\ -4 & 4 \end{bmatrix} \begin{bmatrix} 2 & -1 \\ 4 & 0 \end{bmatrix}. $$ Let $$ A = \begin{bmatrix} 0 & 1 \\ -4 & 4 \end{bmatrix}. $$  ##### **Answer:** Notice that this is a **Jordan block**, and it has following chacteristic： If $$ B= P\begin{bmatrix} \lambda & 1 \\ 0 & \lambda \end{bmatrix}P^{-1}, $$ then $$ B^n= \begin{bmatrix} \lambda^n & n\lambda^{n-1} \\ 0 & \lambda^n \end{bmatrix}. $$ Therefore we know $$ e^B=P^{-1}(\sum_{n=0}^{\infty} \frac{1}{n!} \begin{bmatrix} \lambda^n & n\lambda^{n-1} \\ 0 & \lambda^n \end{bmatrix})P =P \begin{bmatrix} e^\lambda & e^\lambda \\ 0 & e^\lambda \end{bmatrix}P^{-1}. $$ Let $Q= \begin{bmatrix} 2 & -1 \\ 4 & 0 \end{bmatrix}$, and apply this formula to get $$ e^A= Qe^DQ^{-1}= \begin{bmatrix} 2 & -1 \\ 4 & 0 \end{bmatrix} \begin{bmatrix} e^2 & e^2 \\ 0 & e^2 \end{bmatrix} \begin{bmatrix} 2 & -1 \\ 4 & 0 \end{bmatrix}^{-1}= \begin{bmatrix} -e^2 & e^2 \\ -4e^2 & 3e^2 \end{bmatrix}. $$ :::success Well done! :warning: There are typos, where $e^n$ should be $e^\lambda$. ::: --- ##### Exercise 6(d) 已知 $$ \begin{bmatrix} 3 & 1 \\ 0 & 3 \end{bmatrix} = \begin{bmatrix} 3 & -1 \\ 9 & 0 \end{bmatrix}^{-1} \begin{bmatrix} 0 & 1 \\ -9 & 6 \end{bmatrix} \begin{bmatrix} 3 & -1 \\ 9 & 0 \end{bmatrix}. $$ 令 $$ A = \begin{bmatrix} 0 & 1 \\ -9 & 6 \end{bmatrix}. $$  It is known that $$ \begin{bmatrix} 3 & 1 \\ 0 & 3 \end{bmatrix} = \begin{bmatrix} 3 & -1 \\ 9 & 0 \end{bmatrix}^{-1} \begin{bmatrix} 0 & 1 \\ -9 & 6 \end{bmatrix} \begin{bmatrix} 3 & -1 \\ 9 & 0 \end{bmatrix}. $$ Let $$ A = \begin{bmatrix} 0 & 1 \\ -9 & 6 \end{bmatrix}. $$  ##### **Answer:** Similar as last question. Let $Q= \begin{bmatrix} 3 & -1 \\ 9 & 0 \end{bmatrix}$, and apply this formula to get $$ e^A= Qe^DQ^{-1}= \begin{bmatrix} 3 & -1 \\ 9 & 0 \end{bmatrix} \begin{bmatrix} e^3 & e^3 \\ 0 & e^3 \end{bmatrix} \begin{bmatrix} 3 & -1 \\ 9 & 0 \end{bmatrix}^{-1}= \begin{bmatrix} -2e^3 & e^3 \\ -9e^3 & 4e^3 \end{bmatrix}. $$ --- ##### Exercise 7 經由以下步驟說明 $\dot{\bx} = A\bx$ 的解就是 $e^{tA}\bc$，其中 $\bc = (c_1,\ldots, c_n)$ 是一個控制常數的向量。  Use the given instructions to show that $e^{tA}\bc$ represents all the solutions of the equation $\dot{\bx} = A\bx$, where $\bc = (c_1,\ldots, c_n)$ is a vector of constant terms.  ##### Exercise 7(a) 若 $A$ 可被對角化為 $D = Q^{-1}AQ$。令 $Q\by = \bx$。首先說明 $\dot{\by} = D\by$ 的解就是 $e^{Dt}\bd$，其中 $\bd = (d_1,\ldots, d_n)$ 可以是任何常數。  Suppose $A$ is diagonalizable and has $D = Q^{-1}AQ$ for some diagonal matrix $D$. Let $Q\by = \bx$. First, show that $e^{Dt}\bd$ represents all the solutions of the equation $\dot{\by} = D\by$, where $\bd = (d_1,\ldots, d_n)$ is a vector of constant terms.  ##### **Answer:** --- ##### Exercise 7(b) 藉由 $\bx = Q^{-1}\by$ 來得到 $\bx = e^{tA}\bc$，其中 $\bc = Q\bd$ 是一組獨立的常數（因為 $Q$ 可逆）。  Let $\bx = Q^{-1}\by$. Explain why $\bx = e^{tA}\bc$ represents all the solutions to the equation, where $\bc = Q\bd$ is composed of independent constants (since $Q$ is invertible).  ##### **Answer:** --- ##### Remark 如果 $A$ 不可對角化則必須仰賴喬丹標準型，但 $e^{tA}\bc$ 這個公式還是正確的。  If $A$ is not diagonalizable, then we need to find the Jordan canonical form of $A$, but the formula $e^{tA}\bc$ is still the general solution.  :::info Excellent! collaboration: 2 3 problems: 3 - done: 2, 3a, 3b extra: 4.5 - done: 4abc, 5ab, 6abcd moderator: 1 qc: 1 :::

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