3.3 方阵函数及其计算

高等工程数学 讲义 2025AU

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3.3.1 方阵函数的概念

设

$$f(z)=\sum_{k=0}^{\infty} c_{k} z^{k}, \quad (|z|<R).$$

若方阵 $\boldsymbol{A}$ 满足： $\rho(\boldsymbol{A})<R$ ，则方阵幂级数 $\sum_{k=0}^{\infty} c_{k} \boldsymbol{A}^{k}$ 收敛.

• 记 $f(\boldsymbol{A})=\sum_{k=0}^{\infty} c_{k} \boldsymbol{A}^{k}$ ， $f(\boldsymbol{A})$ 称为 方阵 $\boldsymbol{A}$ 的函数.
• 注：幂级数的和函数一般也称为 解析函数(Analytic Function)

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例 $f(z)=\mathrm{e}^{z}$ 的Taylor展开为

$$f(z)=\mathrm{e}^{z}=\sum_{k=0}^{\infty} \frac{1}{k !} z^{k}, \quad|z|<\infty$$

• 对任意方阵 $\boldsymbol{A}$ ，定义 $f(\boldsymbol{A})=\mathrm{e}^{\boldsymbol{A}} \triangleq \sum_{k=0}^{\infty} \frac{1}{k !} \boldsymbol{A}^{k}$
• 若 $\boldsymbol{A}=\left[\begin{array}{ll}{0} & {0} \\ {0} & {1}\end{array}\right]$
  
  • $f(\boldsymbol{A})=\mathrm{e}^{\boldsymbol{A}}=\sum_{k=0}^{\infty} \frac{1}{k !} \boldsymbol{A}^{k} =\sum_{k=0}^{\infty} \frac{1}{k !}\left[\begin{array}{cc}{0} & {0} \\ {0} & {1}\end{array}\right]^{k}$
    
    • $=I+\sum_{k=1}^{\infty} \frac{1}{k !}\left[\begin{array}{ll}{0} & {0} \\ {0} & {1}\end{array}\right] =\left[\begin{array}{cc}{1} & {0} \\ {0} & {\sum_{k=0}^{\infty} \frac{1}{k !}}\end{array}\right]=\left[\begin{array}{ll}{1} & {0} \\ {0} & \mathrm{e}\end{array}\right]$

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方阵函数的性质

定理 设方阵 $\boldsymbol{A}$ 与 $\boldsymbol{B}$ 相似，即存在可逆矩阵 $\boldsymbol{P}$ ，使得

$$\boldsymbol{A}=\boldsymbol{P B P}^{-1},$$

$f(\cdot)$ 是某个解析函数，则

$$f(\boldsymbol{A})={\boldsymbol{P}f}(\boldsymbol{B}) \boldsymbol{P}^{-1}.$$

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例 设 $\boldsymbol{A}=\left[\begin{array}{ll}{1} & {2} \\ {0} & {2}\end{array}\right]$ ，求 $\sin (\boldsymbol{A})$ .

• 提示： $\sin z=\sum_{k=0}^{\infty} \frac{(-1)^{k}}{(2 k+1) !} z^{2 k+1}, \quad(|z|<\infty)$
• $\sin (\boldsymbol{A})=\sum_{k=0}^{\infty} \frac{(-1)^{k}}{(2 k+1) !} \boldsymbol{A}^{2 k+1}$
• $\boldsymbol{A}=\left[\begin{array}{ll}{1} & {2} \\ {0} & {2}\end{array}\right]=\boldsymbol{P}\left[\begin{array}{ll}{1} & {0} \\ {0} & {2}\end{array}\right] \boldsymbol{P}^{-1}$
  
  • $\boldsymbol{P}=\left[\begin{array}{ll}{1} & {2} \\ {0} & {1}\end{array}\right], \quad \boldsymbol{P}^{-1}=\left[\begin{array}{cc}{1} & {-2} \\ {0} & {1}\end{array}\right]$
• $\displaystyle\sin \boldsymbol{A}=\left[\begin{array}{ll}{1} & {2} \\ {0} & {1}\end{array}\right] \sin \left(\left[\begin{array}{ll}{1} & {0} \\ {0} & {2}\end{array}\right]\right)\left[\begin{array}{ll}{1} & {-2} \\ {0} & {1}\end{array}\right]$
  
  • $=\left[\begin{array}{ll}{1} & {2} \\ {0} & {1}\end{array}\right]\left[\begin{array}{cc}{\sin 1} & {0} \\ {0} & {\sin 2}\end{array}\right]\left[\begin{array}{cc}{1} & {-2} \\ {0} & {1}\end{array}\right] =\left[\begin{array}{cc}{\sin 1} & {2 \sin 2-2 \sin 1} \\ {0} & {\sin 2}\end{array}\right]$

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例 已知

$$\sin (\boldsymbol{A}t)= \begin{bmatrix} \frac{3}{4} \sin (4 t)+\frac{1}{4} \sin (20 t) & \frac{1}{2} \sin (20 t)-\frac{1}{2} \sin (4 t) & \frac{1}{4} \sin (20 t)-\frac{1}{4} \sin (4 t) \\ \frac{1}{4} \sin (20 t)-\frac{1}{4} \sin (4 t) & \frac{1}{2} \sin (4 t)+\frac{1}{2} \sin (20 t) & \frac{1}{4} \sin (20 t)-\frac{1}{4} \sin (4 t) \\ \frac{1}{4} \sin (20 t)-\frac{1}{4} \sin (4 t) & \frac{1}{2} \sin (20 t)-\frac{1}{2} \sin (4 t) & \frac{3}{4} \sin (4 t)+\frac{1}{4} \sin (20 t) \\ \end{bmatrix},$$
求 $\boldsymbol{A}$.

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提示：

• $\dfrac{3}{4} \sin (4 t)+\dfrac{1}{4} \sin (20 t)=\dfrac{3}{4}4t+\dfrac{1}{4}20t+\circ(t)=8t+\circ(t)$.
• $\sin (\boldsymbol{A}t)= \begin{bmatrix}8 & 8 & 4 \\ 4 & 12 & 4 \\ 4 & 8 & 8 \end{bmatrix}t+\circ(t)=\boldsymbol{A}t-\dfrac{\boldsymbol{A}^3}{3!}t^3+\circ(t^3),$
• 故 $\boldsymbol{A}=\begin{bmatrix}8 & 8 & 4 \\ 4 & 12 & 4 \\ 4 & 8 & 8 \end{bmatrix}$.

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分块对角阵的函数

定理 若方阵 $\boldsymbol{A}$ 为对角块矩阵

$$\boldsymbol{A}=\mathrm{diag}\{\boldsymbol{A}_1,\boldsymbol{A}_2,...,\boldsymbol{A}_s\},$$

则

$$f(\boldsymbol{A})=\mathrm{diag}\{f(\boldsymbol{A}_1),f(\boldsymbol{A}_2),...,f(\boldsymbol{A}_s)\}.$$

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Jordan 块的函数

定理 若 $\boldsymbol{J}$ 是对角元为 $\lambda_0$ 的 $r$ 阶 Jordan 块，则对解析函数 $f(\lambda)$ 有

$$f(\boldsymbol{J})=\left[\begin{array}{cccc}{f\left(\lambda_{0}\right)} & {f^{\prime}\left(\lambda_{0}\right)} & {\frac{f^{\prime \prime}\left(\lambda_{0}\right)}{2 !}} & {\cdots} & {\frac{f^{(r-1)}\left(\lambda_{0}\right)}{(r-1) !}} \\ {} & {f\left(\lambda_{0}\right)} & {f^{\prime}\left(\lambda_{0}\right)} & {\ddots} & {\vdots} \\ {} & {} & {f\left(\lambda_{0}\right)} & {\ddots} & {\frac{f^{\prime}\left(\lambda_{0}\right)}{2 !}} \\ &{} & {} & {\ddots} & {f^{\prime}\left(\lambda_{0}\right)} \\ & {} & {} & {} & {f\left(\lambda_{0}\right)}\end{array}\right]_{r \times r}$$

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证明思路

• $\boldsymbol{J}=\left[\begin{array}{cccc}{\lambda_{0}} & {1} & {} & {} \\ {} & {\lambda_{0}} & {\ddots} & {} \\ {} & {} & {\ddots} & {1} \\ {} & {} & {} & {\lambda_{0}}\end{array}\right]_{r \times r}$
• $\boldsymbol{J}^{k}=\left[\begin{array}{cccc}{\lambda_{0}^{k}} & {\left(\lambda_{0}^{k}\right)^{\prime}} & {\frac{1}{2 !}\left(\lambda_{0}^{k}\right)^{\prime \prime}} & {\cdots} & {\frac{1}{(r-1) !}\left(\lambda_{0}^{k}\right)^{(r-1)}} \\ {} & {\lambda_{0}^{k}} & {\left(\lambda_{0}^{k}\right)^{\prime}} & {\ddots} & {\vdots} \\ {} & {} & {\lambda_{0}^{k}} & {\ddots} & {\frac{1}{2 !}\left(\lambda_{0}^{k}\right)^{\prime \prime}} \\ & {} & {} & {\ddots} & {\left(\lambda_{0}^{k}\right)^{ \prime}} \\ &{} & {} & {} & {\lambda_{0}^{k}}\end{array}\right]_{r\times r}$

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• $f(\boldsymbol{J})=\sum_{k=0}^{\infty} c_{k} \boldsymbol{J}^{k} =\sum_{k=0}^{\infty} c_{k}\left[\begin{array}{cccc}{\lambda_{0}^{k}} & {\left(\lambda_{0}^{k}\right)^{\prime}} & {\frac{1}{2 !}\left(\lambda_{0}^{k}\right)^{\prime \prime}} & {\cdots} & {\frac{1}{(r-1) !}\left(\lambda_{0}^{k}\right)^{(r-1)}} \\ {} & {\lambda_{0}^{k}} & {\left(\lambda_{0}^{k}\right)^{\prime}} & {\ddots} & {\vdots} \\ &{} & {\lambda_{0}^{k}} & {\ddots} & {\frac{1}{2 !}\left(\lambda_{0}^{k}\right)^{\prime \prime}} \\ &{} & {} & {\ddots} & {\left(\lambda_{0}^{k}\right)^{\prime}} \\ &{} & {} & {} & {\lambda_{0}^{k}}\end{array}\right]_{r \times r}$
  
  • $\displaystyle=\left[\begin{array}{ccccc}{\sum\limits_{k=0}^{\infty} c_{k} \lambda_{0}^{k}} & {\sum\limits_{k=0}^{\infty} c_{k}\left(\lambda_{0}^{k}\right)^{\prime}} & {\sum\limits_{k=0}^{\infty} c_{k} \frac{1}{2 !}\left(\lambda_{0}^{k}\right)^{\prime \prime}} & {\cdots} & {\sum\limits_{k=0}^{\infty} c_{k} \frac{1}{(r-1) !}\left(\lambda_{0}^{k}\right)^{(r-1)}} \\ {} & {\sum\limits_{k=0}^{\infty} c_{k} \lambda_{0}^{k}} & {\sum\limits_{k=0}^{\infty} c_{k}\left(\lambda_{0}^{k}\right)^{\prime}} & {\ddots} & {\vdots} \\ {} & {} & {\sum\limits_{k=0}^{\infty} c_{k} \lambda_{0}^{k}} & {\ddots} & {\sum\limits_{k=0}^{\infty} c_{k} \frac{1}{2 !}\left(\lambda_{0}^{k}\right)^{\prime \prime}} \\ &{} & {} & {\ddots} & {\sum\limits_{k=0}^{\infty} c_{k}\left(\lambda_{0}^{k}\right)^{\prime}} \\& {} & {} & {} & {\sum\limits_{k=0}^{\infty} c_{k} \lambda_{0}^{k}} \end{array}\right]_{r\times r}$

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$$f(\boldsymbol{J})=\left[\begin{array}{cccc}{f\left(\lambda_{0}\right)} & {f^{\prime}\left(\lambda_{0}\right)} & {\frac{f^{\prime \prime}\left(\lambda_{0}\right)}{2 !}} & {\cdots} & {\frac{f^{(r-1)}\left(\lambda_{0}\right)}{(r-1) !}} \\ {} & {f\left(\lambda_{0}\right)} & {f^{\prime}\left(\lambda_{0}\right)} & {\ddots} & {\vdots} \\ {} & {} & {f\left(\lambda_{0}\right)} & {\ddots} & {\frac{f^{\prime \prime}\left(\lambda_{0}\right)}{2 !}} \\ &{} & {} & {\ddots} & {f^{\prime}\left(\lambda_{0}\right)} \\ & {} & {} & {} & {f\left(\lambda_{0}\right)}\end{array}\right]_{r \times r}$$

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例 设 $\boldsymbol{A}=\left[\begin{array}{ll}{2} & {1} \\ {0} & {2}\end{array}\right]$ ，求 $\cos(\boldsymbol{A}).$

• $\cos (\boldsymbol{A})=\left[\begin{array}{cc}{\cos 2} & {\cos ^{\prime} 2} \\ {0} & {\cos 2}\end{array}\right]=\left[\begin{array}{cc}{\cos 2} & {-\sin 2} \\ {0} & {\cos 2}\end{array}\right]$
• 思考：若 $\boldsymbol{A}=\left[\begin{array}{cc}{2} & {1} \\ {2} & {2}\end{array}\right]$ ，如何求 $\cos(\boldsymbol{A})$ ？

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3.3.2 方阵函数的计算

• 方法一：Jordan标准形法
• 方法二：最小多项式法

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方法一：Jordan标准形法

• 求出方阵 $\boldsymbol{A}$ 的 Jordan 标准形 $\boldsymbol{J}$ 及其相似变换矩阵 $\boldsymbol{P}$
  
  • $\boldsymbol{A}=\boldsymbol{P}\mathrm{diag}\{\boldsymbol{J}_1,\boldsymbol{J}_2,\ldots,\boldsymbol{J}_s\}\boldsymbol{P}^{-1}$
  • 其中 $\boldsymbol{J}_i$ 是对角元为 $\lambda_i$ 的 $r_i$ 阶 Jordan 块 $(i=1,2,...,s)$
• 计算 Jordan 块 $\boldsymbol{J}_i$ 的方阵函数 $f(\boldsymbol{J}_i)$
• $f(\boldsymbol{A})=\boldsymbol{P}\mathrm{diag}\{f(\boldsymbol{J}_{1}),f(\boldsymbol{J}_2),...,f(\boldsymbol{J}_n)\}\boldsymbol{P}^{-1}$

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例 设

$$\boldsymbol{A}=\left[\begin{array}{ccc}{0} & {1} & {0} \\ {0} & {0} & {1} \\ {2} & {-5} & {4}\end{array}\right],$$

求 $\sin(\boldsymbol{A})$.

• $|\lambda \boldsymbol{I}-\boldsymbol{A}|=\left|\begin{array}{rrr}{\lambda} & {-1} & {0} \\ {0} & {\lambda} & {-1} \\ {-2} & {5} & {\lambda-4}\end{array}\right| =(\lambda-1)^{2}(\lambda-2)$
• $\boldsymbol{A}$ 的特征值 $\lambda_{1}=2, \lambda_{2}=\lambda_{3}=1$

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• $\boldsymbol{A}-2 \boldsymbol{I}=\left[\begin{array}{rrr}{-2} & {1} & {0} \\ {0} & {-2} & {1} \\ {2} & {-5} & {2}\end{array}\right] \rightarrow\left[\begin{array}{rrr}{-2} & {1} & {0} \\ {0} & {-2} & {1} \\ {0} & {0} & {0}\end{array}\right]$
  
  • $\lambda_1=2$ 对应的特征向量 $\boldsymbol{X}_{1}=[1\;\;2\;\;4]^{\mathrm{T}}$
• $A-I=\left[\begin{array}{rrr}{-1} & {1} & {0} \\ {0} & {-1} & {1} \\ {2} & {-5} & {3}\end{array}\right] \rightarrow\left[\begin{array}{rrr}{-1} & {1} & {0} \\ {0} & {-1} & {1} \\ {0} & {0} & {0}\end{array}\right]$
  
  • $\lambda_{2}=\lambda_{3}=1$ 对应的特征向量 $\boldsymbol{X}_{2}=[1\;\;1\;\;1]^{\mathrm{T}}$
• 因为特征值 $\lambda_{2}=\lambda_{3}=1$ 的几何重数为 $1$ ，小于其代数重数，故 $\boldsymbol{A}$ 不能相似对角化

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• 考虑 $(\boldsymbol{A}-\boldsymbol{I}) \boldsymbol{X}=\boldsymbol{X}_{2}$
  
  • $\left[\boldsymbol{A}-\boldsymbol{I}\; \boldsymbol{X}_{2}\right]=\left[\begin{array}{rrrr}{-1} & {1} & {0} & {1} \\ {0} & {-1} & {1} & {1} \\ {2} & {-5} & {3} & {1}\end{array}\right] \rightarrow\left[\begin{array}{cccc}{-1} & {1} & {0} & {1} \\ {0} & {-1} & {1} & {1} \\ {0} & {0} & {0} & {0}\end{array}\right]$
  • 解得 $\boldsymbol{X}_{3}=[0\;\;1\;\;2]^{\mathrm{T}}$
• $\boldsymbol{A}$ 的 Jordan标准形 $\boldsymbol{J}=\left[\begin{array}{lll}{2} & {0} & {0} \\ {0} & {1} & {1} \\ {0} & {0} & {1}\end{array}\right]$
  
  • 相似变换矩阵 $\boldsymbol{P}=\left[\begin{array}{lll}{\boldsymbol{X}_{1}} & {\boldsymbol{X}_{2}} & {\boldsymbol{X}_{3}}\end{array}\right]=\left[\begin{array}{ccc}{1} & {1} & {0} \\ {2} & {1} & {1} \\ {4} & {1} & {2}\end{array}\right]$

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• $\sin (\boldsymbol{J})=\left[\begin{array}{ccc}{\sin 2} & {0} & {0} \\ {0} & {\sin 1} & {\cos 1} \\ {0} & {0} & {\sin 1}\end{array}\right]$
• $\sin (\boldsymbol{A})=\boldsymbol{P} \sin (\boldsymbol{J}) \boldsymbol{P}^{-1}$
• $=\left[\begin{array}{lll}{1} & {1} & {0} \\ {2} & {1} & {1} \\ {4} & {1} & {2}\end{array}\right]\left[\begin{array}{ccc}{\sin 2} & {0} & {0} \\ {0} & {\sin 1} & {\cos 1} \\ {0} & {0} & {\sin 1}\end{array}\right]\left[\begin{array}{ccc}{1} & {-2} & {1} \\ {0} & {2} & {-1} \\ {-2} & {3} & {-1}\end{array}\right]$
• $=\left[\begin{array}{ccc}{\sin 2-2 \cos 1} & {-2 \sin 2+2 \sin 1+3 \cos 1} & {\sin 2-\sin 1-\cos 1} \\ {2 \sin 2-2 \sin 1-2 \cos 1} & {-4 \sin 2+5 \sin 1+3 \cos 1} & {2 \sin 2-2 \sin 1-\cos 1} \\ {4 \sin 2-4 \sin 1-2 \cos 1} & {-8 \sin 2+8 \sin 1+3 \cos 1} & {4 \sin 2-3 \sin 1-\cos 1}\end{array}\right]$

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注： 用 Jordan 标准形法求方阵函数的困难之处：

• 相似变换矩阵 $\boldsymbol{P}$ 及 $\boldsymbol{P}^{-1}$ 难以计算
• 矩阵乘法 $\boldsymbol{P} f(\boldsymbol{J}) \boldsymbol{P}^{-1}$ 计算繁琐

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方法二：最小多项式法

设方阵 $\boldsymbol{A}$ 的最小多项式为 $m_A(\lambda)=\prod_{i=1}^{s}(\lambda-\lambda_i)^{m_i}$，若函数 $f(\lambda)$ 和 $g(\lambda)$ 满足：对任意 $i=1,2,\ldots,s$

$$\left\{\begin{array}{c}{f\left(\lambda_{i}\right)=g\left(\lambda_{i}\right)} \\ {f^{\prime}\left(\lambda_{i}\right)=g^{\prime}\left(\lambda_{i}\right)} \\ {\vdots} \\ {f^{\left(m_{i}-1\right)}\left(\lambda_{i}\right)=g^{\left(m_{i}-1\right)}\left(\lambda_{i}\right)}\end{array},\right.$$

则称二者 在方阵 $\boldsymbol{A}$ 的谱上是一致的.

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例 设 $\boldsymbol{A}=\left[\begin{array}{ll}{1} & {0} \\ {1} & {1}\end{array}\right]$ ，验证 $f(\lambda)=\sin\lambda$ 与多项式函数

$$g(\lambda)=\lambda \cos 1+(\sin 1-\cos 1)$$

在 $\boldsymbol{A}$ 的谱上是一致的.

• $\boldsymbol{A}$ 的最小多项式 $m_{A}(\lambda)=(\lambda-1)^{2}$
• $g(\lambda)=\lambda \cos 1+(\sin 1-\cos 1)$
  
  • $g(1)=(\cos 1)+(\sin 1-\cos 1)=\sin 1=f(1)$
  • $g^{\prime}(1)=\cos 1=f^{\prime}(1)$

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定理 若 $f(\lambda)$ 和 $g(\lambda)$ 在方阵 $\boldsymbol{A}$ 的谱上是一致的，则对 $\boldsymbol{A}$ 的 Jordan 标准形中的每个 Jordan 块 $\boldsymbol{J}_i$，有

$$f\left(\boldsymbol{J}_{i}\right)=g\left(\boldsymbol{J}_{i}\right).$$

• 提示：$f(\boldsymbol{J}_i)$ 完全由 $f(\lambda_i),f'(\lambda_i),\ldots,$ $f^{(m_i-1)}(\lambda_i)$ 所决定.

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定理 若 $f(\lambda)$ 和 $g(\lambda)$ 在方阵 $\boldsymbol{A}$ 的谱上是一致的，则

$$f(\boldsymbol{A})=g(\boldsymbol{A}).$$

• 计算方阵函数的最小(零化)多项式法：
  
  • 构造一个多项式 $g(\lambda)$ ，使之与 $f(\lambda)$ 在方阵 $\boldsymbol{A}$ 的谱上一致
    
    • 则 $f(\boldsymbol{A})=g(\boldsymbol{A})$
  • 计算 $g(\boldsymbol{A})$ .

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证明思路：

• 设 $\boldsymbol{A}$ 的最小多项式
  
  • $m_{A}(\lambda)=\left(\lambda-\lambda_{1}\right)^{m_{1}}\left(\lambda-\lambda_{2}\right)^{m_{2}} \cdots\left(\lambda-\lambda_{s}\right)^{m_{s}}$,
  • $\sum_{i=1}^{s} m_{i}=m$
• 设 $\boldsymbol{A}$ 的Jordan标准形 $\boldsymbol{J}=\mathrm{diag}\{\boldsymbol{J}_1,...,\boldsymbol{J}_s\}$
  
  • $\boldsymbol{J}_{i}=\left[\begin{array}{cccc}{\lambda_{i}} & {1} & {} & {} \\ {} & {\lambda_{i}} & {\ddots} & {} \\ {} & {} & {\ddots} & {1} \\ {} & {} & {} & {\lambda_{i}}\end{array}\right]_{k \times k}$

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• $f(\lambda)$ 和 $g(\lambda)$ 在方阵 $A$ 的谱上是一致的
  
  • $f\left(J_{i_{i}}\right)=\left[\begin{array}{cccc}{f\left(\lambda_{i}\right)} & {f^{\prime}\left(\lambda_{i}\right)} & {\cdots} & {\frac{f^{(k-1)}\left(\lambda_{i}\right)}{(k-1) !}} \\ {} & {f\left(\lambda_{i}\right)} & {\ddots} & {\vdots} \\ {} & {} & {\ddots} & {f^{\prime}\left(\lambda_{i}\right)} \\ {} & {} & {} & {f\left(\lambda_{i}\right)}\end{array}\right]=g\left(J_{i_{j}}\right)$
• $f(A)=P f(J) P^{-1}=P g(J) P^{-1}=g(A)$

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• 设 $A$ 的最小多项式为
  
  • $m_{A}(\lambda)=\left(\lambda-\lambda_{1}\right)^{m_{1}}\left(\lambda-\lambda_{2}\right)^{m_{2}} \cdots\left(\lambda-\lambda_{s}\right)^{m_{s}}, \sum_{i=1}^{s} m_{i}=m$
• 设 $g(\lambda)=a_{0} \lambda^{m-1}+a_{1} \lambda^{m-2}+\cdots a_{m-1}$
• 令 $\left\{\begin{array}{c}{f\left(\lambda_{i}\right)=g\left(\lambda_{i}\right)} \\ {f^{\prime}\left(\lambda_{i}\right)=g^{\prime}\left(\lambda_{i}\right)} \\ {\vdots} \\ {f^{\left(m_{i}-1\right)}\left(\lambda_{i}\right)=g^{\left(m_{i}-1\right)}\left(\lambda_{i}\right)}\end{array} \quad, \quad i=1,2, \cdots, s\right.$
  
  • 解出 $a_{0}, a_{1}, \cdots, a_{m-1}$

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例 $\boldsymbol{A}=\left[\begin{array}{ccc}{0} & {1} & {0} \\ {0} & {0} & {1} \\ {2} & {-5} & {4}\end{array}\right]$ ，求 $\sin(\boldsymbol{A})$ .

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提示

• $\boldsymbol{A}$ 的最小多项式 $m_{A}(\lambda)=(\lambda-2)(\lambda-1)^{2}$
• 设 $g(\lambda)=a_{0} \lambda^{2}+a_{1} \lambda+a_{2}$
• 令 $\left\{\begin{array}{l}{f(2)=g(2)} \\ {f(1)=g(1)} \\ {f^{\prime}(1)=g^{\prime}(1)}\end{array}\right.$
  
  • 即 $\left\{\begin{array}{l}{4 a_{0}+2 a_{1}+a_{2}=\sin 2} \\ {a_{0}+a_{1}+a_{2}=\sin 1} \\ {2 a_{0}+a_{1}=\cos 1}\end{array}\right.$
  • 解得 $\left\{\begin{array}{l}{a_{0}=\sin 2-\sin 1-\cos 1} \\ {a_{1}=-2 \sin 2+2 \sin 1+3 \cos 1} \\ {a_{2}=\sin 2-2 \cos 1}\end{array}\right.$

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$f(\boldsymbol{A})=\sin(\boldsymbol{A})$

• $=g(\boldsymbol{A})=a_{0} \boldsymbol{A}^{2}+a_{1} \boldsymbol{A}+a_{2} \boldsymbol{I}$
• $=a_{0}\left[\begin{array}{ccc}{0} & {0} & {1} \\ {2} & {-5} & {4} \\ {8} & {-18} & {11}\end{array}\right]+a_{1}\left[\begin{array}{ccc}{0} & {1} & {0} \\ {0} & {0} & {1} \\ {2} & {-5} & {4}\end{array}\right]+a_{2}\left[\begin{array}{ccc}{1} & {0} & {0} \\ {0} & {1} & {0} \\ {0} & {0} & {1}\end{array}\right]$
• $=\left[\begin{array}{lll}{a_{2}} & {a_{1}} & {a_{0}} \\ {2 a_{0}} & {-5 a_{0}+a_{2}} & {4 a_{0}+a_{1}} \\ {8 a_{0}+2 a_{1}} & {-18 a_{0}-5 a_{1}} & {11 a_{0}+4 a_{1}+a_{2}}\end{array}\right]$
• $=\left[\begin{array}{ccc}{\sin 2-2 \cos 1} & {-2 \sin 2+2 \sin 1+3 \cos 1} & {\sin 2-\sin 1-\cos 1} \\ {2 \sin 2-2 \sin 1-2 \cos 1} & {-4 \sin 2+5 \sin 1+3 \cos 1} & {2 \sin 2-2 \sin 1-\cos 1} \\ {4 \sin 2-4 \sin 1-2 \cos 1} & {-8 \sin 2+8 \sin 1+3 \cos 1} & {4 \sin 2-3 \sin 1-\cos 1}\end{array}\right]$

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关于方阵函数存在的条件

方阵函数 $f(\boldsymbol{A})$ 存在的前提

1. 函数 $f(z)$ 为解析函数
   
   • 即: $f(z)=\sum_{k=0}^{\infty} c_{k} z^{k}\;(|z|<R)$
2. $\rho(\boldsymbol{A})<R$

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例 考虑函数 $f(z)=(1-z)^{-1}=\sum_{k=0}^{\infty} z^{k} \quad(|z|<1)$ ，方阵 $\boldsymbol{A}=\left[\begin{array}{ll}{2} & {0} \\ {0} & {2}\end{array}\right]$，求 $f(\boldsymbol{A})$.

• 因为 $\rho(\boldsymbol{A})=2>R=1$ ，所以 $\sum_{k=0}^{\infty} \boldsymbol{A}^{k}$ 发散
• 进而可知， $f(\boldsymbol{A})=(I-\boldsymbol{A})^{-1}$ 没有定义
• 但实际上
  
  • $f(\boldsymbol{A})=(I-\boldsymbol{A})^{-1}=\left[\begin{array}{cc}{-1} & {0} \\ {0} & {-1}\end{array}\right]^{-1}=\left[\begin{array}{cc}{-1} & {0} \\ {0} & {-1}\end{array}\right]$
• 这说明前述方阵函数的定义过于严格！

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方阵函数定义的推广

设 $\boldsymbol{A}$ 是 $n$ 阶方阵， $f(\lambda)$ 是给定的函数，若存在多项式 $g(\lambda)$ ，使得 $f(\lambda)$ 与 $g(\lambda)$ 在 $\boldsymbol{A}$ 的谱上是一致的，则可定义方阵函数

$$f(\boldsymbol{A})=g(\boldsymbol{A}).$$

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例 $\boldsymbol{A}=\left[\begin{array}{ccc}{5} & {-4} & {4} \\ {1} & {0} & {5} \\ {1} & {-1} & {6}\end{array}\right]$ ，计算 $\ln \boldsymbol{A}$ .

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提示

• $\boldsymbol{A}$ 的特征多项式：$f_{\boldsymbol{A}}(\lambda)=\left|\begin{array}{ccc}{\lambda-5} & {4} & {-4} \\ {-1} & {\lambda} & {-5} \\ {-1} & {1} & {\lambda-6}\end{array}\right|=(\lambda-1)(\lambda-5)^{2}$
• $\boldsymbol{A}$ 的最小多项式：$m_{A}(\lambda)=(\lambda-1)(\lambda-5)^{2}$
• 设 $g(\lambda)=a \lambda^{2}+b \lambda+c$
  
  • 令 $\left\{\begin{array}{l}{g(1)=a+b+c=f(1)=\ln 1=0} \\ {g(5)=25 a+5 b+c=f(5)=\ln 5} \\ {g^{\prime}(5)=10 a+b=f^{\prime}(5)=1 / 5}\end{array}\right.$
  • 解得 $\left\{\begin{array}{l}a=\frac1{20}-\frac{\ln5}{10}\\ b=-\frac3{10}+\frac{5\ln5}{8}\\ c=\frac{1}{4}+\frac{21\ln5}{40}\end{array}\right.$

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• $f(\boldsymbol{A})=\ln \boldsymbol{A}$
  
  • $=g(\boldsymbol{A})=a \boldsymbol{A}^{2}+b \boldsymbol{A}+c \boldsymbol{I}$
  • $=\left[\begin{array}{ccc}{25 a+5 b+c} & {-24 a-4 b} & {24 a+4 b} \\ {10 a+b} & {-9 a+c} & {34 a+5 b} \\ {10 a+b} & {-10 a-b} & {35 a+6 b+c}\end{array}\right]$
  • $=\left[\begin{array}{ccc}{\ln 5} & {-\ln 5} & {\ln 5} \\ {1 / 5} & {-1 / 5} & {\ln 5+1 / 5} \\ {1 / 5} & {-1 / 5} & {1 / 5}\end{array}\right]$

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小结

• 方阵函数：解析函数的推广
• 方阵函数的计算
  
  • Jordan 标准形法：计算复杂
  • 最小多项式法
    
    • $f$ 和 $g$ 在 $\boldsymbol{A}$ 的谱上一致