※注意：請於試卷上「非選擇題作答區」標明題號並依序作答。
- (25 points) Let $A=\begin{pmatrix}-2&-1&2&-1\\11&7&-1&5\\-8&-2&6&-2\\4&1&-2&3\end{pmatrix}$. Find the Jordan canonical form of $A$. Compute $\exp\left(tA\right)$ and derive the general solution to $x'\left(t\right)=Ax\left(t\right)$, where $x\left(t\right)$ is a $4$-dimensional column vector.
- 若 $x=2$，則 $A-2I=\begin{bmatrix}-4&-1&2&-1\\11&5&-1&5\\-8&-2&4&-2\\4&1&-2&1\end{bmatrix}$，如此可取 $v_1=\begin{bmatrix}1\\-2\\1\\0\end{bmatrix}$ 與 $v_2=\begin{bmatrix}0\\-1\\0\\1\end{bmatrix}$ 滿足 $Av_1=2v_1$ 且 $Av_2=2v_2$。
- 若 $x=5$，則 $A-5I=\begin{bmatrix}-7&-1&2&-1\\11&2&-1&5\\-8&-2&1&-2\\4&1&-2&-2\end{bmatrix}$，如此可取 $v_3=\begin{bmatrix}-1\\2\\-2\\1\end{bmatrix}$ 滿足 $Av_3=5v_3$。進一步地，我們可取 $v_4=\begin{bmatrix}0\\1\\0\\0\end{bmatrix}$ 滿足 $\left(A-5I\right)v_4=v_3$。
- (25 points) Let $A$ and $B$ be any $n\times n$ complex matrices. Show that $\exp\left(A\right)\exp\left(B\right)=\exp\left(A+B\right)$ if $A$ and $B$ commute.
Hint: You may consider the norm $\left\|C\right\|=\max\left\{\left|C_{ij}\right|\right\}$ for $C=\left(C_{ij}\right)\in M_n\left(\mathbb{C}\right)$ and the remainder term$\displaystyle R_p=\sum_{i=0}^{2p}\frac{\left(A+B\right)^i}{i!}-\sum_{j=0}^p\frac{A^j}{j!}\sum_{k=0}^p\frac{B^k}{k!}$.
- (25 points) Show that
$\displaystyle\det\begin{pmatrix}X_0&X_1&X_2&\cdots&X_{n-1}\\X_{n-1}&X_0&X_1&\cdots&X_{n-2}\\\vdots&\vdots&\vdots&\ddots&\vdots\\X_1&X_2&X_3&\cdots&X_0\end{pmatrix}=\prod_{j=0}^{n-1}\left(\sum_{k=0}^{n-1}\zeta^{jk}X_k\right)$
where $\zeta$ is a primitive $n$-th root of unity.
Hint: You may first compute, for example, $\begin{pmatrix}X_0&X_1&X_2&X_3\\X_3&X_0&X_1&X_2\\X_2&X_3&X_0&X_1\\X_1&X_2&X_3&X_0\end{pmatrix}\begin{pmatrix}1&1&1&1\\1&\zeta&\zeta^2&\zeta^3\\1&\zeta^2&\zeta^4&\zeta^6\\1&\zeta^3&\zeta^6&\zeta^9\end{pmatrix}$. - (25 points) Let $V$ and $W$ be vector spaces over the same field $F$, and let $T:V\to W$ be a linear transformation. Let $\mathcal{B}\left(V\right)$ and $\mathcal{B}\left(W\right)$ denote the spaces of bilinear forms on $V$ and $W$ respectively. For any $H\in\mathcal{B}\left(W\right)$, define $\hat{T}\left(H\right):V\times V\to F$ by $\hat{T}\left(H\right)\left(x,y\right)=H\left(T\left(x\right),T\left(y\right)\right)$ for any $x,y\in V$.
- Show that $\hat{T}\left(H\right)$ is a bilinear form on $V$.
- Show that if $T$ is an isomorphism, then so is $\hat{T}:\mathcal{B}\left(W\right)\to\mathcal{B}\left(V\right)$.
- 按照定義檢驗 $\hat{T}\left(H\right)$ 是雙線性形式：對於任何 $x,y,z\in V$ 與 $c\in F$，我們有
$\begin{aligned}\hat{T}\left(H\right)\left(cx+y,z\right)&=H\left(T\left(cx+y\right),T\left(z\right)\right)\\&=cH\left(T\left(x\right),T\left(z\right)\right)+H\left(T\left(y\right),T\left(z\right)\right)=c\hat{T}\left(H\right)\left(x,z\right)+\hat{T}\left(H\right)\left(y,z\right),\end{aligned}$
$\begin{aligned}\hat{T}\left(H\right)\left(x,cy+z\right)&=H\left(T\left(x\right),T\left(cy+z\right)\right)\\&=cH\left(T\left(x\right),T\left(y\right)\right)+H\left(T\left(x\right),T\left(z\right)\right)=c\hat{T}\left(H\right)\left(x,y\right)+\hat{T}\left(H\right)\left(x,z\right).\end{aligned}$
因此 $\hat{T}\left(H\right)$ 為 $V$ 上的雙線性形式。 - 如果 $T$ 為同構函數，亦即 $T$ 單射且滿射，那麼我們可以證明 $\hat{T}$ 為單射與雙射如下：
- 若 $\hat{T}\left(H_1\right)=\hat{T}\left(H_2\right)$，那麼對任何 $x',y'\in W$，由 $T$ 滿射可取 $x,y\in V$ 有 $T\left(x\right)=x'$ 與 $T\left(y\right)=y'$，從而對這樣的 $x,y$ 有
$H_1\left(T\left(x\right),T\left(y\right)\right)=\hat{T}\left(H_1\right)\left(x,y\right)=\hat{T}\left(H_2\right)\left(x,y\right)=H_2\left(T\left(x\right),T\left(y\right)\right)$
此即$H_1\left(x',y'\right)=H_2\left(x',y'\right)$
故由 $x',y'\in W$ 的任意性可知 $H_1=H_2$，故 $\hat{T}$ 單射。 - 對任何 $K\in\mathcal{B}\left(V\right)$，我們可取 $H_s\in\mathcal{B}\left(W\right)$ 如下
$H_s\left(x',y'\right)=K\left(T^{-1}\left(x'\right),T^{-1}\left(y'\right)\right)~~~\mbox{for}~x',y'\in W$.
那麼容易檢驗發現$\hat{T}\left(H_s\right)\left(x,y\right)=H_s\left(T\left(x\right),T\left(y\right)\right)=K\left(x,y\right)$
至此有 $\hat{T}\left(H_s\right)=K$，即 $\hat{T}$ 滿射。
- 若 $\hat{T}\left(H_1\right)=\hat{T}\left(H_2\right)$，那麼對任何 $x',y'\in W$，由 $T$ 滿射可取 $x,y\in V$ 有 $T\left(x\right)=x'$ 與 $T\left(y\right)=y'$，從而對這樣的 $x,y$ 有
訣竅
按照 Jordan canonical form 的求法計算之，其中注意到每行加總的值相同，故可據此簡化特徵多項式的計算。解法
首先計算 $A$ 的特徵多項式如下$\begin{aligned}\mbox{ch}_A\left(x\right)&=\det\left(A-xI\right)=\begin{vmatrix}-2-x&-1&2&-1\\11&7-x&-1&5\\-8&-2&6-x&-2\\4&1&-2&3-x\end{vmatrix}=\begin{vmatrix}-2-x&-1&2&-1\\11&7-x&-1&5\\-8&-2&6-x&-2\\5-x&5-x&5-x&5-x\end{vmatrix}\\&=\left(5-x\right)\begin{vmatrix}-2-x&-1&2&-1\\11&7-x&-1&5\\-8&-2&6-x&-2\\1&1&1&1\end{vmatrix}=\left(5-x\right)\begin{vmatrix}-1-x&0&3&0\\6&2-x&-6&0\\-6&0&8-x&0\\1&1&1&1\end{vmatrix}\\&=\left(5-x\right)\left(2-x\right)\begin{vmatrix}-1-x&3\\-6&8-x\end{vmatrix}=\left(x-2\right)\left(x-5\right)\left[\left(x-8\right)\left(x+1\right)+18\right]\\&=\left(x-2\right)^2\left(x-5\right)^2.\end{aligned}$
計算特徵向量與廣義特徵向量如下現在我們來計算 $e^{tA}$ 如下：
$\begin{aligned}e^{tA}&=Qe^{tJ}Q^{-1}=\begin{bmatrix}1&0&-1&0\\-2&-1&2&1\\1&0&-2&0\\0&1&1&0\end{bmatrix}\begin{bmatrix}e^{2t}&0&0&0\\0&e^{2t}&0&0\\0&0&e^{5t}&te^{5t}\\0&0&0&e^{5t}\end{bmatrix}\begin{bmatrix}2&0&-1&0\\1&0&-1&1\\1&0&-1&0\\1&1&1&1\end{bmatrix}\\&=\begin{bmatrix}2e^{2t}-e^{5t}-te^{5t}&-te^{5t}&-e^{2t}+e^{5t}-te^{5t}&-te^{5t}\\-5e^{2t}+2e^{5t}+2te^{5t}&e^{5t}+2te^{5t}&-3e^{2t}-e^{5t}+2te^{5t}&-e^{2t}+e^{5t}+2te^{5t}\\2e^{2t}-2e^{5t}-2te^{5t}&-2te^{5t}&-e^{2t}+2e^{5t}-2te^{5t}&-2te^{5t}\\e^{2t}+e^{5t}+te^{5t}&te^{5t}&-e^{2t}-e^{5t}+te^{5t}&e^{2t}+te^{5t}\end{bmatrix}\end{aligned}$
因此聯立微分方程組的通解為$x\left(t\right)=e^{tA}c=c_1\begin{bmatrix}2e^{2t}-e^{5t}-te^{5t}\\-5e^{2t}+2e^{5t}+2te^{5t}\\2e^{2t}-2e^{5t}-2te^{5t}\\e^{2t}+e^{5t}+te^{5t}\end{bmatrix}+c_2\begin{bmatrix}-te^{5t}\\e^{5t}+2te^{5t}\\-2te^{5t}\\te^{5t}\end{bmatrix}+c_3\begin{bmatrix}-e^{2t}+e^{5t}-te^{5t}\\-3e^{2t}-e^{5t}+2te^{5t}\\-e^{2t}+2e^{5t}-2te^{5t}\\-e^{2t}-e^{5t}+te^{5t}\end{bmatrix}+c_4\begin{bmatrix}-te^{5t}\\-e^{2t}+e^{5t}+2te^{5t}\\-2t^{5t}\\e^{2t}+te^{5t}\end{bmatrix}$.
訣竅
按照提示估算矩陣指數的誤差，從而確認兩端的誤差隨次數的增加而趨於零。解法
考慮等式左右兩端的差的前 $2p+1$ 項的值為$\displaystyle R_p=\sum_{i=0}^{2p}\frac{\left(A+B\right)^i}{i!}-\sum_{j=0}^p\frac{A^j}{j!}\sum_{k=0}^p\frac{B^k}{k!}=\sum_{i=0}^{2p}\sum_{s=0}^i\frac{A^sB^{i-s}}{s!\left(i-s\right)!}-\sum_{j=0}^p\frac{A^j}{j!}\sum_{k=0}^p\frac{B^k}{k!}$
此時注意第一項運用變數變換（令 $s=j$ 而 $i=j+k$），可拆為三項如下$\displaystyle\sum_{i=0}^{2p}\sum_{s=0}^i\frac{A^sB^{i-s}}{s!\left(i-s\right)!}=\sum_{j=0}^{2p}\sum_{k=0}^{2p-j}\frac{A^jB^k}{j!k!}=\sum_{j=0}^p\sum_{k=0}^p\frac{A^jB^k}{j!k!}+\sum_{j=p+1}^{2p}\sum_{k=0}^{2p-j}\frac{A^jB^k}{j!k!}+\sum_{j=0}^p\sum_{k=p+1}^{2p-j}\frac{A^jB^k}{j!k!}$
因此$\displaystyle R_p=\sum_{j=p+1}^{2p}\frac{A^j}{j!}\sum_{k=0}^{2p-j}\frac{B^k}{k!}+\sum_{j=0}^p\frac{A^j}{j!}\sum_{k=p+1}^{2p-j}\frac{B^k}{k!}$
至此，取 $C=\max\left\{\left\|A\right\|,\left\|B\right\|,1\right\}$，那麼由 $\left\|AB\right\|\leq n\left\|A\right\|\left\|B\right\|$，可以發現$\displaystyle\left\|R_p\right\|\leq\sum_{j=p+1}^{2p}\sum_{k=0}^{2p-j}\frac{n^{j+k}C^{j+k}}{j!k!}+\sum_{j=0}^p\sum_{k=p+1}^{2p-j}\frac{n^{j+k}C^{j+k}}{j!k!}\leq\frac{p\left(p+1\right)n^{2p}C^{2p}}{\left(p+1\right)!}\leq\frac{n^{2p}C^{2p}}{\left(p-1\right)!}\to0~~\mbox{as}~p\to\infty$.
訣竅
藉由提示與所欲證的等式考慮到特徵值的乘積。解法
設欲求行列式的矩陣為 $X$ 以及 $Z=\left[z_{ij}\right]$，其中 $z_{ij}=\zeta^{\left(i-1\right)\left(j-1\right)}$。此外方便起見約定 $X_k:=X_{n+k}$，其中 $-\left(n-1\right)\leq k\leq-1$。那麼觀察 $XZ$ 的第 $j$ 行可知$\begin{aligned}X\begin{bmatrix}\zeta^{\left(1-1\right)\left(j-1\right)}\\\zeta^{\left(2-1\right)\left(j-1\right)}\\\zeta^{\left(3-1\right)\left(j-1\right)}\\\vdots\\\zeta^{\left(n-1\right)\left(j-1\right)}\end{bmatrix}&=\left[\begin{array}{c}\sum\limits_{i=1}^nX_{i-1}\zeta^{\left(i-1\right)\left(j-1\right)}\\\sum\limits_{i=1}^nX_{i-2}\zeta^{\left(i-1\right)\left(j-1\right)}\\\sum\limits_{i=1}^nX_{i-3}\zeta^{\left(i-1\right)\left(j-1\right)}\\\vdots\\\sum\limits_{i=1}^nX_{i-n}\zeta^{\left(i-1\right)\left(j-1\right)}\end{array}\right]=\left[\begin{array}{c}\sum\limits_{i=1}^nX_{i-1}\zeta^{\left(i-1\right)\left(j-1\right)}\\\sum\limits_{i=0}^{n-1}X_{i-1}\zeta^{\left(i-1\right)\left(j-1\right)}\zeta^{j-1}\\\sum\limits_{i=-1}^{n-2}X_{i-1}\zeta^{\left(i-1\right)\left(j-1\right)}\zeta^{2\left(j-1\right)}\\\vdots\\\sum\limits_{i=-n+2}^1X_{i-1}\zeta^{\left(i-1\right)\left(j-1\right)}\zeta^{\left(n-1\right)\left(j-1\right)}\end{array}\right]\\&=\left(\sum_{i=1}^nX_{i-1}\zeta^{\left(i-1\right)\left(j-1\right)}\right)\begin{bmatrix}\zeta^{\left(1-1\right)\left(j-1\right)}\\\zeta^{\left(2-1\right)\left(j-1\right)}\\\zeta^{\left(3-1\right)\left(j-1\right)}\\\vdots\\\zeta^{\left(n-1\right)\left(j-1\right)}\end{bmatrix}=\left(\sum_{i=0}^{n-1}X_i\zeta^{i\left(j-1\right)}\right)\begin{bmatrix}\zeta^{\left(1-1\right)\left(j-1\right)}\\\zeta^{\left(2-1\right)\left(j-1\right)}\\\zeta^{\left(3-1\right)\left(j-1\right)}\\\vdots\\\zeta^{\left(n-1\right)\left(j-1\right)}\end{bmatrix}\end{aligned}$
因此對每個 $1\leq j\leq n$ 而言，$\sum\limits_{i=0}^{n-1}X_i\zeta^{i\left(j-1\right)}$ 都是特徵值。類似地，改變編號可知對於 $0\leq j\leq n-1$ 而言，$\sum\limits_{i=0}^{n-1}X_i\zeta^{ij}$ 都是特徵值。又行列式值為所有特徵值之乘積，故$\displaystyle\det\left(X\right)=\prod_{j=0}^{n-1}\left(\sum_{i=0}^{n-1}\zeta^{ij}X_i\right)$.
證明完畢。
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