方法一
根据矩阵指数函数的定义直接计算：
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e^{At}\overset{def}{=}I+At+\frac{1}{2!}A^{2}t^{2}+...=\sum_{k=0}^{\infty}\frac{1}{k!}A^{k}t{^k}
eAt=defI+At+2!1A2t2+...=k=0∑∞k!1Aktk
方法二
将 A A A阵化为对角标准型或约当标准型求解
1.
A
A
A的特征值不存在重根
若
A
A
A的特征值
λ
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λ
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,
λ
n
\lambda_{1},\lambda_{2},...,\lambda_{n}
λ1,λ2,...,λn不存在重根，则在求出使
A
A
A阵实现对角化
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A
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T^{-1}AT= \begin{bmatrix} \lambda_{1} & & & \\ & \lambda_{2} & & \\ & & \ddots & \\ & & & \lambda_{n} \end{bmatrix}
T−1AT=⎣⎢⎢⎡λ1λ2⋱λn⎦⎥⎥⎤
的变换阵
T
−
1
T^{-1}
T−1，
T
T
T后，即有指数函数矩阵：
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e^{At}= T\begin{bmatrix} e^{\lambda_{1}t} & & & \\ & e^{\lambda_{2}t} & & \\ & & \ddots & \\ & & & e^{\lambda_{n}t} \end{bmatrix}T^{-1}
eAt=T⎣⎢⎢⎡eλ1teλ2t⋱eλnt⎦⎥⎥⎤T−1
证明：
由
:
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⋱
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]
可
得
:
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T
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由: T^{-1}AT= \begin{bmatrix} \lambda_{1} & & & \\ & \lambda_{2} & & \\ & & \ddots & \\ & & & \lambda_{n} \end{bmatrix} \qquad可得: A= T\begin{bmatrix} \lambda_{1} & & & \\ & \lambda_{2} & & \\ & & \ddots & \\ & & & \lambda_{n} \end{bmatrix} T^{-1}
由:T−1AT=⎣⎢⎢⎡λ1λ2⋱λn⎦⎥⎥⎤可得:A=T⎣⎢⎢⎡λ1λ2⋱λn⎦⎥⎥⎤T−1
而:
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\begin{aligned} e^{At}&=\sum_{k=0}^{\infty}\frac{1}{k!}A^{k}t{^k}\\ &=\sum_{k=0}^{\infty}\frac{1}{k!} \left ( T \begin{bmatrix} \lambda_{1} & & & \\ & \lambda_{2} & & \\ & & \ddots & \\ & & & \lambda_{n} \end{bmatrix} T^{-1} \right )^{k} t^{k} \end{aligned}
eAt=k=0∑∞k!1Aktk=k=0∑∞k!1⎝⎜⎜⎛T⎣⎢⎢⎡λ1λ2⋱λn⎦⎥⎥⎤T−1⎠⎟⎟⎞ktk
接下来，先考虑
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k=2
k=2的情况：
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\begin{aligned} \left ( T \begin{bmatrix} \lambda_{1} & & & \\ & \lambda_{2} & & \\ & & \ddots & \\ & & & \lambda_{n} \end{bmatrix} T^{-1} \right )^{2} &= T\begin{bmatrix} \lambda_{1} & & & \\ & \lambda_{2} & & \\ & & \ddots & \\ & & & \lambda_{n} \end{bmatrix} T^{-1} T\begin{bmatrix} \lambda_{1} & & & \\ & \lambda_{2} & & \\ & & \ddots & \\ & & & \lambda_{n} \end{bmatrix} T^{-1}\\ &= T\begin{bmatrix} \lambda_{1} & & & \\ & \lambda_{2} & & \\ & & \ddots & \\ & & & \lambda_{n} \end{bmatrix}^{2} T^{-1} \end{aligned}
⎝⎜⎜⎛T⎣⎢⎢⎡λ1λ2⋱λn⎦⎥⎥⎤T−1⎠⎟⎟⎞2=T⎣⎢⎢⎡λ1λ2⋱λn⎦⎥⎥⎤T−1T⎣⎢⎢⎡λ1λ2⋱λn⎦⎥⎥⎤T−1=T⎣⎢⎢⎡λ1λ2⋱λn⎦⎥⎥⎤2T−1
将结果带入上式中有
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\begin{aligned} e^{At}&=\sum_{k=0}^{\infty}\frac{1}{k!}A^{k}t{^k}\\ &=\sum_{k=0}^{\infty}\frac{1}{k!} \left ( \begin{bmatrix} \lambda_{1} & & & \\ & \lambda_{2} & & \\ & & \ddots & \\ & & & \lambda_{n} \end{bmatrix} T^{-1} \right )^{k} t^{k}\\ &=\sum_{k=0}^{\infty}\frac{1}{k!} T \begin{bmatrix} \lambda_{1} & & & \\ & \lambda_{2} & & \\ & & \ddots & \\ & & & \lambda_{n} \end{bmatrix} ^{k} T^{-1} t^{k} \end{aligned}
eAt=k=0∑∞k!1Aktk=k=0∑∞k!1⎝⎜⎜⎛⎣⎢⎢⎡λ1λ2⋱λn⎦⎥⎥⎤T−1⎠⎟⎟⎞ktk=k=0∑∞k!1T⎣⎢⎢⎡λ1λ2⋱λn⎦⎥⎥⎤kT−1tk
而
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\frac{1}{k!}
k!1和
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t^{k}
tk都是常数，将其乘到对角阵中去
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\begin{aligned} &=T \begin{bmatrix} \sum_{k=0}^{\infty}\frac{1}{k!}\lambda_{1}^{k}t^{k}& & & \\ & \sum_{k=0}^{\infty}\frac{1}{k!}\lambda_{2}^{k}t^{k}& & \\ & & \ddots & \\ & & & \sum_{k=0}^{\infty}\frac{1}{k!}\lambda_{n}^{k}t^{k} \end{bmatrix} T^{-1}\\ &=T \begin{bmatrix} e^{\lambda_{1}t}& & & \\ &e^{\lambda_{2}t} & & \\ & & \ddots & \\ & & &e^{\lambda_{k}t} \end{bmatrix} T^{-1}\\ \end{aligned}
=T⎣⎢⎢⎡∑k=0∞k!1λ1ktk∑k=0∞k!1λ2ktk⋱∑k=0∞k!1λnktk⎦⎥⎥⎤T−1=T⎣⎢⎢⎡eλ1teλ2t⋱eλkt⎦⎥⎥⎤T−1
证毕
2.
A
A
A的特征值存在重根时
暂时略
方法三
拉氏变换法:
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e^{At}=L^{-1}[(sI-A)^{-1}]
eAt=L−1[(sI−A)−1]
证明:
由矩阵指数函数的定义：
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+
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+
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e^{At}=I+At+\frac{1}{2!}A^{2}t^{2}+...=\sum_{k=0}^{\infty}\frac{1}{k!}A^{k}t{^k}
eAt=I+At+2!1A2t2+...=k=0∑∞k!1Aktk
两边取拉氏变换:
根据:
L
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L(\frac{t^{k}}{k!})=\frac{1}{s^{k+1}}
L(k!tk)=sk+11
L ( e A t ) = 1 s I + 1 s 2 A + 1 s 3 A 2 + . . . = ∑ k = 0 ∞ 1 s k + 1 A k = s − 1 ∑ k = 0 + ∞ ( s − 1 A ) k \begin{aligned} L(e^{At})&=\frac{1}{s}I+\frac{1}{s^2}A+\frac{1}{s^3}A^2+...\\ &=\sum_{k=0}^{\infty}\frac{1}{s^{k+1}}A^{k}\\ &=s^{-1}\sum_{k=0}^{+\infty}(s^{-1}A)^k \end{aligned} L(eAt)=s1I+s21A+s31A2+...=k=0∑∞sk+11Ak=s−1k=0∑+∞(s−1A)k
根据：
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\begin{aligned} 1+x+x^2+...+x^k+...&=\sum_{k=0}^{+\infty}x^k\\ &=\frac{1}{1-x}=(1-x)^{-1} \end{aligned}
1+x+x2+...+xk+...=k=0∑+∞xk=1−x1=(1−x)−1
有：
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\begin{aligned} L(e^{At})&=s^{-1}\sum_{k=0}^{+\infty}(s^{-1}A)^k\\ &=s^{-1}(I-s^{-1}A)^{-1}\\ &=(sI-A)^{-1} \end{aligned}
L(eAt)=s−1k=0∑+∞(s−1A)k=s−1(I−s−1A)−1=(sI−A)−1
两边取拉式反变换：
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e^{At}=L^{-1}[(sI-A)^{-1}]
eAt=L−1[(sI−A)−1]
证毕
凯莱-哈米尔顿定理：
设
A
∈
R
x
×
x
A\in R^{x \times x}
A∈Rx×x，其特征多项式为:
D
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\begin{aligned} D(\lambda)&=|\lambda I-A|\\ &=\lambda^n+a_{n-1}\lambda^{n-1}+...+a_{1}\lambda+a_{0}\\ &=0\\ \end{aligned}
D(λ)=∣λI−A∣=λn+an−1λn−1+...+a1λ+a0=0
则矩阵A必须满足其特征多项式，即:
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A^n+a_{n-1}A^{n-1}+...+a_1A+a_0I=0
An+an−1An−1+...+a1A+a0I=0