利用 Cayley-Hamilton 定理計算矩陣函數

$A=\begin{bmatrix} 1&1\\ 0&1 \end{bmatrix}\begin{bmatrix} 1&0\\ 0&2 \end{bmatrix}\begin{bmatrix} 1&1\\ 0&1 \end{bmatrix}^{-1}=S\Lambda S^{-1}$

$A^{100}=(S\Lambda S^{-1})^{100}=S\Lambda^{100}S^{-1}=\begin{bmatrix} 1&1\\ 0&1 \end{bmatrix}\begin{bmatrix} 1&0\\ 0&2^{100} \end{bmatrix}\begin{bmatrix} 1&1\\ 0&1 \end{bmatrix}^{-1}=\begin{bmatrix} 1&2^{100}-1\\ 0&2^{100} \end{bmatrix}$

$\displaystyle A_{\epsilon}=\begin{bmatrix} 1&2\\ 0&1+\epsilon \end{bmatrix}=\begin{bmatrix} 1&2\\ 0&\epsilon \end{bmatrix}\begin{bmatrix} 1&0\\ 0&1+\epsilon \end{bmatrix}\frac{1}{\epsilon}\begin{bmatrix} \epsilon &-2\\ 0&1 \end{bmatrix}$

\displaystyle\begin{aligned} A_{\epsilon}^{100}&=\begin{bmatrix} 1&2\\ 0&\epsilon \end{bmatrix}\begin{bmatrix} 1&0\\ 0&(1+\epsilon)^{100} \end{bmatrix}\frac{1}{\epsilon}\begin{bmatrix} \epsilon &-2\\ 0&1 \end{bmatrix}\\ &=\frac{1}{\epsilon}\begin{bmatrix} 1&2\\ 0&\epsilon \end{bmatrix}\begin{bmatrix} 1&0\\ 0&1+100\epsilon+o(\epsilon^2) \end{bmatrix}\begin{bmatrix} \epsilon &-2\\ 0&1 \end{bmatrix}\\ &=\frac{1}{\epsilon}\begin{bmatrix} 1&2+200\epsilon+o(\epsilon^2)\\ 0&\epsilon+100\epsilon^2+o(\epsilon^3) \end{bmatrix}\begin{bmatrix} \epsilon &-2\\ 0&1 \end{bmatrix}\\ &=\frac{1}{\epsilon}\begin{bmatrix} \epsilon&200\epsilon+o(\epsilon^2)\\ 0&\epsilon+100\epsilon^2+o(\epsilon^3) \end{bmatrix}\\ &=\begin{bmatrix} 1&200+o(\epsilon)\\ 0&1+100\epsilon+o(\epsilon^2) \end{bmatrix},\end{aligned}

$p(t)=\begin{vmatrix} 1-t&2\\ 0&1-t \end{vmatrix}=(1-t)^2=t^2-2t+1$

\begin{aligned} A^3&=A(A^2)=A(2A-I)=2A^2-A=2(2A-I)-A=3A-2I,\\ A^4&=A(A^3)=A(3A-2I)=3A^2-2A=3(2A-I)-2A=4A-3I,\\ \vdots \end{aligned}

$A^{100}=q(A)(A-I)^2+c_1A+c_0I$

$t^{100}=q(t)(t-1)^2+c_1t+c_0$

$100t^{99}=q^{\prime}(t)(t-1)^2+2q(t)(t-1)+c_1=\left(q^{\prime}(t)(t-1)+2q(t)\right)(t-1)+c_1$

$t=1$ 代入上面兩式可得

$c_1+c_0=1,~~c_1=100$

$f(A)=q(A)p(A)+c_{n-1}A^{n-1}+\cdots+c_1A+c_0I=c_{n-1}A^{n-1}+\cdots+c_1A+c_0I$

$A$ 有相異特徵值 $\lambda_1,\ldots,\lambda_m$，且 $\beta_1,\ldots,\beta_m$ 為對應的相重數 (亦稱代數重數)，則 $A$ 的特徵多項式為

$p(t)=(t-\lambda_1)^{\beta_1}\cdots(t-\lambda_m)^{\beta_m}$

\begin{aligned} f(t)&=q(t)p(t)+c_{n-1}t^{n-1}+\cdots+c_1t+c_0\\ &=q(t)(t-\lambda_1)^{\beta_1}\cdots(t-\lambda_m)^{\beta_m}+c_{n-1}t^{n-1}+\cdots+c_1t+c_0, \end{aligned}

$\begin{bmatrix} 1&\lambda_1&\lambda_1^2&\lambda_1^3&\cdots&\lambda_1^{n-1}\\ \vdots&\vdots&\vdots&\vdots&&\vdots\\ 1&\lambda_m&\lambda_m^2&\lambda_m^3&\cdots&\lambda_m^{n-1}\\ -&-&-&-&-&-\\ \vdots&\vdots&\vdots&\vdots&&\vdots\\ 0&1&2\lambda_i&3\lambda_i^2&\cdots&(n-1)\lambda_i^{n-2}\\ \vdots&\vdots&\vdots&\vdots&&\vdots\\ -&-&-&-&-&-\\ \vdots&\vdots&\vdots&\vdots&&\vdots\\ 0&0&2&6\lambda_i&\cdots&(n-1)(n-2)\lambda_i^{n-3}\\ \vdots&\vdots&\vdots&\vdots&&\vdots\\ -&-&-&-&-&-\\ \vdots&\vdots&\vdots&\vdots&&\vdots \end{bmatrix}\begin{bmatrix} c_0\\ c_1\\ c_2\\ c_3\\ \vdots\\ \vdots\\ \vdots\\ c_{n-1} \end{bmatrix}=\begin{bmatrix} f(\lambda_1)\\ \vdots\\ f(\lambda_m)\\ -\\ \vdots\\ f^{\prime}(\lambda_i)\\ \vdots\\ -\\ \vdots\\ f^{\prime\prime}(\lambda_i)\\ \vdots\\ -\\ \vdots \end{bmatrix}$

(1) 求出特徵多項式 $p(t)=(t-1)^2(t-3)$，可知 $A$ 有特徵值 $1, 1, 3$

(2) 令 $e^A=c_2A^2+c_1A+c_0I$。寫出函數 $f(t)=e^t$ 的多項式除法表達式：

$f(t)=e^{t}=q(t)(t-1)^2(t-3)+c_2t^2+c_1t+c_0$

$f(t)$ 的導數，

$f^{\prime}(t)=e^t=g(t)(t-1)+2c_2t+c_1$

(3) 將 $t=1$$t=3$ 代入 $f(t)$，並將 $t=1$ 代入 $f^{\prime}(t)$，可得限制條件式：

\begin{aligned} e&=c_2+c_1+c_0\\ e^3&=9c_2+3c_1+c_0\\ e&=2c_2+c_1, \end{aligned}

$\begin{bmatrix} 1&1&1\\ 1&3&9\\ 0&1&2 \end{bmatrix}\begin{bmatrix} c_0\\ c_1\\ c_2 \end{bmatrix}=\begin{bmatrix} e\\ e^3\\ e \end{bmatrix}$

$\displaystyle c_0=\frac{1}{4}e^3-\frac{3}{4}e,~~c_1=-\frac{1}{2}e^3+\frac{5}{2}e,~~c_2=\frac{1}{4}e^3-\frac{3}{4}e$

\displaystyle\begin{aligned} e^A&=\left(\frac{1}{4}e^3-\frac{3}{4}e\right)A^2+\left(-\frac{1}{2}e^3+\frac{5}{2}e\right)A+\left(\frac{1}{4}e^3-\frac{3}{4}e\right)I\\ &=\left(\frac{1}{4}e^3-\frac{3}{4}e\right)\begin{bmatrix} 1&4&0\\ 0&1&0\\ 0&0&9 \end{bmatrix}+\left(-\frac{1}{2}e^3+\frac{5}{2}e\right)\begin{bmatrix} 1&2&0\\ 0&1&0\\ 0&0&3 \end{bmatrix}+\left(\frac{1}{4}e^3-\frac{3}{4}e\right)\begin{bmatrix} 1&0&0\\ 0&1&0\\ 0&0&1 \end{bmatrix}\\ &=\begin{bmatrix} e&2e&0\\ 0&e&0\\ 0&0&e^3 \end{bmatrix}.\end{aligned}

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