1,矩阵序列
1.1,矩阵序列
设 中的矩阵序列 ,其中 。若:
![\small \underset{k\rightarrow +\infty }{lim}a_{ij}^{(k)}=a_{ij},i=1,2,...,m;j=1,2,...,n;](https://latex.csdn.net/eq?%5Cdpi%7B120%7D%20%5Csmall%20%5Cunderset%7Bk%5Crightarrow%20+%5Cinfty%20%7D%7Blim%7Da_%7Bij%7D%5E%7B%28k%29%7D%3Da_%7Bij%7D%2Ci%3D1%2C2%2C...%2Cm%3Bj%3D1%2C2%2C...%2Cn%3B)
则称矩阵序列 收敛于 ,或称 为矩阵序列 的极限,记为:
或 ![\small A^{(k)}\rightarrow A(k\rightarrow +\infty )](https://latex.csdn.net/eq?%5Cdpi%7B120%7D%20%5Csmall%20A%5E%7B%28k%29%7D%5Crightarrow%20A%28k%5Crightarrow%20+%5Cinfty%20%29)
(1)不收敛的矩阵序列称为发散。
(2)矩阵序列收敛的本质是矩阵的所有元素都收敛。
(3) 上一个矩阵序列的收敛相当于 个数列同时收敛。
(4)用初等分析法太繁琐。
(5)与向量序列一样,可以利用矩阵范数来研究矩阵序列的收敛问题。
设 ,则:
![\small \underset{k\rightarrow \infty }{lim}A^{(k)}=A\Leftrightarrow \underset{k\rightarrow \infty }{lim}||A^{(k)}-A||=0](https://latex.csdn.net/eq?%5Cdpi%7B120%7D%20%5Csmall%20%5Cunderset%7Bk%5Crightarrow%20%5Cinfty%20%7D%7Blim%7DA%5E%7B%28k%29%7D%3DA%5CLeftrightarrow%20%5Cunderset%7Bk%5Crightarrow%20%5Cinfty%20%7D%7Blim%7D%7C%7CA%5E%7B%28k%29%7D-A%7C%7C%3D0)
其中 是 上的任一矩阵范数。
证明:
![\small \underset{k\rightarrow +\infty }{lim}A^{(k)}=A\Leftrightarrow \underset{k\rightarrow +\infty }{lim}||A^{(k)}-A||_G=0\Leftrightarrow \underset{k\rightarrow +\infty }{lim}||A^{(k)}-A||=0](https://latex.csdn.net/eq?%5Cdpi%7B120%7D%20%5Csmall%20%5Cunderset%7Bk%5Crightarrow%20+%5Cinfty%20%7D%7Blim%7DA%5E%7B%28k%29%7D%3DA%5CLeftrightarrow%20%5Cunderset%7Bk%5Crightarrow%20+%5Cinfty%20%7D%7Blim%7D%7C%7CA%5E%7B%28k%29%7D-A%7C%7C_G%3D0%5CLeftrightarrow%20%5Cunderset%7Bk%5Crightarrow%20+%5Cinfty%20%7D%7Blim%7D%7C%7CA%5E%7B%28k%29%7D-A%7C%7C%3D0)
当矩阵序列收敛时,矩阵序列的任意范数也收敛,反之不成立。
设 ,则:
![\small \underset{k\rightarrow +\infty }{lim}||A^{(k)}||=||A||](https://latex.csdn.net/eq?%5Cdpi%7B120%7D%20%5Csmall%20%5Cunderset%7Bk%5Crightarrow%20+%5Cinfty%20%7D%7Blim%7D%7C%7CA%5E%7B%28k%29%7D%7C%7C%3D%7C%7CA%7C%7C)
其中 是 上的任一矩阵范数。
收敛的矩阵序列的性质:
设 为适当阶的矩阵, ,则:
(1)![\small \underset{k\rightarrow +\infty }{lim}(sA^{(k)}+tB^{(k)})=sA+tB](https://latex.csdn.net/eq?%5Cdpi%7B120%7D%20%5Csmall%20%5Cunderset%7Bk%5Crightarrow%20+%5Cinfty%20%7D%7Blim%7D%28sA%5E%7B%28k%29%7D+tB%5E%7B%28k%29%7D%29%3DsA+tB)
(2)![\small \underset{k\rightarrow +\infty }{lim}(A^{(k)}B^{(k)})=AB](https://latex.csdn.net/eq?%5Cdpi%7B120%7D%20%5Csmall%20%5Cunderset%7Bk%5Crightarrow%20+%5Cinfty%20%7D%7Blim%7D%28A%5E%7B%28k%29%7DB%5E%7B%28k%29%7D%29%3DAB)
(3)当 和 均可逆时,![\small \underset{k\rightarrow +\infty }{lim}(A^{(k)})^{-1}=A^{-1}](https://latex.csdn.net/eq?%5Cdpi%7B120%7D%20%5Csmall%20%5Cunderset%7Bk%5Crightarrow%20+%5Cinfty%20%7D%7Blim%7D%28A%5E%7B%28k%29%7D%29%5E%7B-1%7D%3DA%5E%7B-1%7D)
证明: 可逆不能保证 可逆
![\small A^{(k)}=\begin{bmatrix} 1+\frac{1}{k} & 1\\ 1 & 1 \end{bmatrix}\rightarrow A=\begin{bmatrix} k &-k \\ -k &k+1 \end{bmatrix}](https://latex.csdn.net/eq?%5Cdpi%7B120%7D%20%5Csmall%20A%5E%7B%28k%29%7D%3D%5Cbegin%7Bbmatrix%7D%201+%5Cfrac%7B1%7D%7Bk%7D%20%26%201%5C%5C%201%20%26%201%20%5Cend%7Bbmatrix%7D%5Crightarrow%20A%3D%5Cbegin%7Bbmatrix%7D%20k%20%26-k%20%5C%5C%20-k%20%26k+1%20%5Cend%7Bbmatrix%7D)
对每一个 都有可逆矩阵 ,但 不可逆。
1.2,幂收敛矩阵
设 ,若 ,则称 为收敛矩阵。
设 ,则 是收敛矩阵的充要条件是 。
设 ,若对 上的某个矩阵范数 有 ,则 为收敛矩阵。
借助矩阵的谱半径或矩阵的范数来判断矩阵是否收敛。
【例1】判断下列矩阵是否为收敛矩阵
![\small A=\frac{1}{6}\begin{bmatrix} 1 &-8 \\ -2 &1 \end{bmatrix}](https://latex.csdn.net/eq?%5Cdpi%7B120%7D%20%5Csmall%20A%3D%5Cfrac%7B1%7D%7B6%7D%5Cbegin%7Bbmatrix%7D%201%20%26-8%20%5C%5C%20-2%20%261%20%5Cend%7Bbmatrix%7D)
可求得 ,故 是收敛矩阵。
![\small A=\begin{bmatrix} 0.2 & 0.1& 0.2\\ 0.5 &0.5 &0.4 \\ 0.1& 0.3 & 0.2 \end{bmatrix}](https://latex.csdn.net/eq?%5Cdpi%7B120%7D%20%5Csmall%20A%3D%5Cbegin%7Bbmatrix%7D%200.2%20%26%200.1%26%200.2%5C%5C%200.5%20%260.5%20%260.4%20%5C%5C%200.1%26%200.3%20%26%200.2%20%5Cend%7Bbmatrix%7D)
因为 ,故 是收敛矩阵。
2,矩阵级数
2.1,矩阵级数
由 中的矩阵序列 构成的无穷和:
![\small A^{(0)}+A^{(1)}+...+A^{(k)}+...](https://latex.csdn.net/eq?%5Cdpi%7B120%7D%20%5Csmall%20A%5E%7B%280%29%7D+A%5E%7B%281%29%7D+...+A%5E%7B%28k%29%7D+...)
称为矩阵级数,记为 ,对任一正整数 ,称为矩阵级数,记为 。
对任一正整数 ,称 为矩阵级数的部分和,如果由部分和构成的矩阵序列 收敛,且有极限 ,即 ,则称矩阵级数 收敛,
且有和 ,记为 不收敛的矩阵级数称为发散的。
如果记由 ,显然 相当于
,即 个数项级数都收敛。
【例1】已知 ,研究矩阵级数 的敛散性。
因为 ![\small S^{(N)}=\sum_{k=0}^{N}A^{(k)}=\begin{bmatrix} \sum_{k=0}^{N}\frac{1}{2^k} & \sum_{k=0}^{N}\frac{\pi}{4^k} \\ 0 &\sum_{k=0}^{N}\frac{1}{(k+1)(k+2)} \end{bmatrix}=\begin{bmatrix} 2-\frac{1}{2^N} &\frac{\pi}{3}(4-\frac{1}{4^N}) \\ 0 & 1-\frac{1}{N+2} \end{bmatrix}](https://latex.csdn.net/eq?%5Cdpi%7B120%7D%20%5Csmall%20S%5E%7B%28N%29%7D%3D%5Csum_%7Bk%3D0%7D%5E%7BN%7DA%5E%7B%28k%29%7D%3D%5Cbegin%7Bbmatrix%7D%20%5Csum_%7Bk%3D0%7D%5E%7BN%7D%5Cfrac%7B1%7D%7B2%5Ek%7D%20%26%20%5Csum_%7Bk%3D0%7D%5E%7BN%7D%5Cfrac%7B%5Cpi%7D%7B4%5Ek%7D%20%5C%5C%200%20%26%5Csum_%7Bk%3D0%7D%5E%7BN%7D%5Cfrac%7B1%7D%7B%28k+1%29%28k+2%29%7D%20%5Cend%7Bbmatrix%7D%3D%5Cbegin%7Bbmatrix%7D%202-%5Cfrac%7B1%7D%7B2%5EN%7D%20%26%5Cfrac%7B%5Cpi%7D%7B3%7D%284-%5Cfrac%7B1%7D%7B4%5EN%7D%29%20%5C%5C%200%20%26%201-%5Cfrac%7B1%7D%7BN+2%7D%20%5Cend%7Bbmatrix%7D)
所以 ![\small S=\underset{N\rightarrow +\infty }{lim}S^{(N)}=\begin{bmatrix} 2 & \frac{4\pi}{3}\\ 0&1 \end{bmatrix}](https://latex.csdn.net/eq?%5Cdpi%7B120%7D%20%5Csmall%20S%3D%5Cunderset%7BN%5Crightarrow%20+%5Cinfty%20%7D%7Blim%7DS%5E%7B%28N%29%7D%3D%5Cbegin%7Bbmatrix%7D%202%20%26%20%5Cfrac%7B4%5Cpi%7D%7B3%7D%5C%5C%200%261%20%5Cend%7Bbmatrix%7D)
故所给矩阵级数收敛,且和为 。
设 ,如果 个数项级数 都绝对收敛,即 都收敛,则称矩阵级数 绝对收敛。
利用矩阵范数,可以将判定矩阵级数的绝对收敛问题转化为判定正项级数是否收敛的问题。
设 ,则矩阵级数 绝对收敛的充分必要条件是正项级数 绝对收敛的充分必要条件是正项级数 收敛,其中 是 上任一矩阵范数。
矩阵级数性质:设 ,其中 是适当阶的矩阵,则:
(1)![\small \sum_{k=0}^{+\infty }(cA^{(k)}+dB^{(k)})=cA+dB,\forall c,d\in \mathbb{C}](https://latex.csdn.net/eq?%5Cdpi%7B120%7D%20%5Csmall%20%5Csum_%7Bk%3D0%7D%5E%7B+%5Cinfty%20%7D%28cA%5E%7B%28k%29%7D+dB%5E%7B%28k%29%7D%29%3DcA+dB%2C%5Cforall%20c%2Cd%5Cin%20%5Cmathbb%7BC%7D)
(2)绝对收敛的矩阵级数必收敛,并且任一调换其项的顺序所得的矩阵级数仍收敛,且其和不变。
(3)若矩阵级数 收敛(或绝对收敛),则矩阵级数 也收敛(或绝对收敛),并且有: 。
(4)若 和 均绝对收敛,则它们的项数相乘所得的矩阵级数:
![\small A^{(0)}B^{(0)}+(A^{(0)}B^{(1)}+A^{(1)}B^{(0)})+...+(A^{(0)}B^{(k)}+A^{(1)}B^{(k-1)}+...+A^{(k)}B^{(0)})+...](https://latex.csdn.net/eq?%5Cdpi%7B120%7D%20%5Csmall%20A%5E%7B%280%29%7DB%5E%7B%280%29%7D+%28A%5E%7B%280%29%7DB%5E%7B%281%29%7D+A%5E%7B%281%29%7DB%5E%7B%280%29%7D%29+...+%28A%5E%7B%280%29%7DB%5E%7B%28k%29%7D+A%5E%7B%281%29%7DB%5E%7B%28k-1%29%7D+...+A%5E%7B%28k%29%7DB%5E%7B%280%29%7D%29+...)
也绝对首先,且其和为 。
2.2,矩阵幂级数
设 , ,称矩阵级数 为矩阵 的幂级数。
(1)利用定义判断矩阵幂级数的收敛性很不方便。
(2)矩阵幂级数是复变量 的幂级数 的推广。如果幂级数 的收敛半径为 ,则对收敛圆 内的所有 , 都是绝对收敛的。因此,讨论 的敛散性问题可以自然的联系到 的收敛半径。
设幂级数 的收敛半径为 , ,则:
(1)当 时,矩阵幂级数 绝对收敛。
(2)当 时,矩阵幂级数 发散。
推论:设幂级数 的收敛半径是 , ,若存在 上的某个矩阵范数 使得 ,则矩阵幂级数 绝对收敛。
【例2】判断矩阵幂级数 的敛散性。
令 ,可求得 ,由于幂级数收敛半径 ,所以 收敛。
收敛半径:![\small R=\underset{k\rightarrow +\infty }{lim}\frac{|a_{k+1}|}{|a_k|}=\underset{k\rightarrow +\infty }{lim}|a_k|^{\frac{1}{k}}=\frac{1}{r}](https://latex.csdn.net/eq?%5Cdpi%7B120%7D%20%5Csmall%20R%3D%5Cunderset%7Bk%5Crightarrow%20+%5Cinfty%20%7D%7Blim%7D%5Cfrac%7B%7Ca_%7Bk+1%7D%7C%7D%7B%7Ca_k%7C%7D%3D%5Cunderset%7Bk%5Crightarrow%20+%5Cinfty%20%7D%7Blim%7D%7Ca_k%7C%5E%7B%5Cfrac%7B1%7D%7Bk%7D%7D%3D%5Cfrac%7B1%7D%7Br%7D)
2.3,Neumann级数
设 ,矩阵幂级数 (称为Neumann级数)收敛 ,并且在收敛时其和为 。
【例3】已知 ,判定矩阵幂级数 的收敛性,试求其和。
因为 ,所以 收敛,且
![\small \sum_{k=0}^{+\infty }A^k=(I-A)^{-1}=\frac{1}{14}\begin{bmatrix} 28 &14 &14 \\ 44& 62 &62 \\ 20& 25&35 \end{bmatrix}](https://latex.csdn.net/eq?%5Cdpi%7B120%7D%20%5Csmall%20%5Csum_%7Bk%3D0%7D%5E%7B+%5Cinfty%20%7DA%5Ek%3D%28I-A%29%5E%7B-1%7D%3D%5Cfrac%7B1%7D%7B14%7D%5Cbegin%7Bbmatrix%7D%2028%20%2614%20%2614%20%5C%5C%2044%26%2062%20%2662%20%5C%5C%2020%26%2025%2635%20%5Cend%7Bbmatrix%7D)
3,矩阵函数
3.1,矩阵函数的概念
设幂级数 的收敛半径为 ,且当 时,幂级数收敛于函数 ,即:
![\small f(z)=\sum_{k=0}^{+\infty }a_kz^k(|z|r)](https://latex.csdn.net/eq?%5Cdpi%7B120%7D%20%5Csmall%20f%28z%29%3D%5Csum_%7Bk%3D0%7D%5E%7B+%5Cinfty%20%7Da_kz%5Ek%28%7Cz%7C%3Cr%29)
如果 满足 ,则称收敛的矩阵幂级数 的和为矩阵函数,记为: ,即
![\small f(A)=\sum_{k=0}^{+\infty }a_kA^k](https://latex.csdn.net/eq?%5Cdpi%7B120%7D%20%5Csmall%20f%28A%29%3D%5Csum_%7Bk%3D0%7D%5E%7B+%5Cinfty%20%7Da_kA%5Ek)
常见矩阵函数:
![\small sinA=\sum_{k=0}^{+\infty }\frac{(-1)^k}{(2k+1)!}A^{2k+1}(\forall A\in \mathbb{C}^{n\times n})](https://latex.csdn.net/eq?%5Cdpi%7B120%7D%20%5Csmall%20sinA%3D%5Csum_%7Bk%3D0%7D%5E%7B+%5Cinfty%20%7D%5Cfrac%7B%28-1%29%5Ek%7D%7B%282k+1%29%21%7DA%5E%7B2k+1%7D%28%5Cforall%20A%5Cin%20%5Cmathbb%7BC%7D%5E%7Bn%5Ctimes%20n%7D%29) ![\small cosA=\sum_{k=0}^{+\infty }\frac{(-1)^k}{(2k)!}A^{2k}(\forall A\in \mathbb{C}^{n\times n})](https://latex.csdn.net/eq?%5Cdpi%7B120%7D%20%5Csmall%20cosA%3D%5Csum_%7Bk%3D0%7D%5E%7B+%5Cinfty%20%7D%5Cfrac%7B%28-1%29%5Ek%7D%7B%282k%29%21%7DA%5E%7B2k%7D%28%5Cforall%20A%5Cin%20%5Cmathbb%7BC%7D%5E%7Bn%5Ctimes%20n%7D%29) ![\small (I-A)^{-1}=\sum_{k=0}^{+\infty }A^k(\rho(A)1)](https://latex.csdn.net/eq?%5Cdpi%7B120%7D%20%5Csmall%20%28I-A%29%5E%7B-1%7D%3D%5Csum_%7Bk%3D0%7D%5E%7B+%5Cinfty%20%7DA%5Ek%28%5Crho%28A%29%3C1%29)
如果把矩阵函数 的变元 换成 其中 为参数,则相应得到:
![\small f(At)=\sum_{k=0}^{+\infty }a_k(At)^k](https://latex.csdn.net/eq?%5Cdpi%7B120%7D%20%5Csmall%20f%28At%29%3D%5Csum_%7Bk%3D0%7D%5E%7B+%5Cinfty%20%7Da_k%28At%29%5Ek)
3.2,矩阵函数的计算
方法一:利用Hamilton-Cayley定理:利用Hamilton-Cayley定理找出矩阵方幂之间的关系,然后化简矩阵幂级数,求出矩阵函数的值。
【例4】已知 ,求 。
,由Hamilton-Cayley定理知 ![\small A^2=-I](https://latex.csdn.net/eq?%5Cdpi%7B120%7D%20%5Csmall%20A%5E2%3D-I)
![\small A^{2k}=(-1)^kI,A^{2k+1}=(-1)^kA,(k=1,2,...)](https://latex.csdn.net/eq?%5Cdpi%7B120%7D%20%5Csmall%20A%5E%7B2k%7D%3D%28-1%29%5EkI%2CA%5E%7B2k+1%7D%3D%28-1%29%5EkA%2C%28k%3D1%2C2%2C...%29)
![\small e^{At}=\sum_{k=0}^{+\infty }\frac{1}{k!}(At)^k=(1-\frac{t^2}{2!}+\frac{t^4}{4!}+...)I+(t-\frac{t^3}{3!}+\frac{t^5}{5!}+...)A](https://latex.csdn.net/eq?%5Cdpi%7B120%7D%20%5Csmall%20e%5E%7BAt%7D%3D%5Csum_%7Bk%3D0%7D%5E%7B+%5Cinfty%20%7D%5Cfrac%7B1%7D%7Bk%21%7D%28At%29%5Ek%3D%281-%5Cfrac%7Bt%5E2%7D%7B2%21%7D+%5Cfrac%7Bt%5E4%7D%7B4%21%7D+...%29I+%28t-%5Cfrac%7Bt%5E3%7D%7B3%21%7D+%5Cfrac%7Bt%5E5%7D%7B5%21%7D+...%29A)
![\small =(cost)I+(sint)A=\begin{bmatrix} cost & sint\\ -sint & cost \end{bmatrix}](https://latex.csdn.net/eq?%5Cdpi%7B120%7D%20%5Csmall%20%3D%28cost%29I+%28sint%29A%3D%5Cbegin%7Bbmatrix%7D%20cost%20%26%20sint%5C%5C%20-sint%20%26%20cost%20%5Cend%7Bbmatrix%7D)
【例5】已知 阶矩阵 的特征值为 ,求 ![\small sinA,cosA](https://latex.csdn.net/eq?%5Cdpi%7B120%7D%20%5Csmall%20sinA%2CcosA)
由于 ,因此 ![\small A^4=\pi^2A^2](https://latex.csdn.net/eq?%5Cdpi%7B120%7D%20%5Csmall%20A%5E4%3D%5Cpi%5E2A%5E2)
进而 ![\small A^{2k}=\pi^{2k-2}A^2,A^{2k+1}=\pi^{2k-2}A^3,(k=2,3,...)](https://latex.csdn.net/eq?%5Cdpi%7B120%7D%20%5Csmall%20A%5E%7B2k%7D%3D%5Cpi%5E%7B2k-2%7DA%5E2%2CA%5E%7B2k+1%7D%3D%5Cpi%5E%7B2k-2%7DA%5E3%2C%28k%3D2%2C3%2C...%29)
![\small sinA=\sum_{k=0}^{+\infty }\frac{(-1)^k}{(2k+1)!}A^{2k+1}=A-\frac{A^3}{3!}+\sum_{k=2}^{+\infty }\frac{(-1)^k}{(2k+1)!}\pi^{2k-2}A^3](https://latex.csdn.net/eq?%5Cdpi%7B120%7D%20%5Csmall%20sinA%3D%5Csum_%7Bk%3D0%7D%5E%7B+%5Cinfty%20%7D%5Cfrac%7B%28-1%29%5Ek%7D%7B%282k+1%29%21%7DA%5E%7B2k+1%7D%3DA-%5Cfrac%7BA%5E3%7D%7B3%21%7D+%5Csum_%7Bk%3D2%7D%5E%7B+%5Cinfty%20%7D%5Cfrac%7B%28-1%29%5Ek%7D%7B%282k+1%29%21%7D%5Cpi%5E%7B2k-2%7DA%5E3)
![\small =A-\frac{A^3}{3!}+\frac{A^3}{\pi^3}\sum_{k=2}^{+ \infty }\frac{(-1)^k}{(2k+1)!}\pi^{2k+1}](https://latex.csdn.net/eq?%5Cdpi%7B120%7D%20%5Csmall%20%3DA-%5Cfrac%7BA%5E3%7D%7B3%21%7D+%5Cfrac%7BA%5E3%7D%7B%5Cpi%5E3%7D%5Csum_%7Bk%3D2%7D%5E%7B+%20%5Cinfty%20%7D%5Cfrac%7B%28-1%29%5Ek%7D%7B%282k+1%29%21%7D%5Cpi%5E%7B2k+1%7D)
![\small =A-\frac{A^3}{3!}+\frac{A^3}{\pi^3}(sin\pi-\pi+\frac{\pi^3}{3!})=A-\frac{A^3}{\pi^2}](https://latex.csdn.net/eq?%5Cdpi%7B120%7D%20%5Csmall%20%3DA-%5Cfrac%7BA%5E3%7D%7B3%21%7D+%5Cfrac%7BA%5E3%7D%7B%5Cpi%5E3%7D%28sin%5Cpi-%5Cpi+%5Cfrac%7B%5Cpi%5E3%7D%7B3%21%7D%29%3DA-%5Cfrac%7BA%5E3%7D%7B%5Cpi%5E2%7D)
![\small cosA=\sum_{k=0}^{+\infty }\frac{(-1)^k}{(2k)!}A^{2k}=A-\frac{A^2}{2!}+\sum_{k=2}^{+\infty }\frac{(-1)^k}{(2k)!}\pi^{2k-2}A^2](https://latex.csdn.net/eq?%5Cdpi%7B120%7D%20%5Csmall%20cosA%3D%5Csum_%7Bk%3D0%7D%5E%7B+%5Cinfty%20%7D%5Cfrac%7B%28-1%29%5Ek%7D%7B%282k%29%21%7DA%5E%7B2k%7D%3DA-%5Cfrac%7BA%5E2%7D%7B2%21%7D+%5Csum_%7Bk%3D2%7D%5E%7B+%5Cinfty%20%7D%5Cfrac%7B%28-1%29%5Ek%7D%7B%282k%29%21%7D%5Cpi%5E%7B2k-2%7DA%5E2)
![\small =A-\frac{A^2}{2!}+\frac{A^2}{\pi^2}\sum_{k=2}^{+ \infty }\frac{(-1)^k}{(2k)!}\pi^{2k}](https://latex.csdn.net/eq?%5Cdpi%7B120%7D%20%5Csmall%20%3DA-%5Cfrac%7BA%5E2%7D%7B2%21%7D+%5Cfrac%7BA%5E2%7D%7B%5Cpi%5E2%7D%5Csum_%7Bk%3D2%7D%5E%7B+%20%5Cinfty%20%7D%5Cfrac%7B%28-1%29%5Ek%7D%7B%282k%29%21%7D%5Cpi%5E%7B2k%7D)
![\small =A-\frac{A^2}{2!}+\frac{A^2}{\pi^2}(cos\pi-1+\frac{\pi^2}{2!})=I-\frac{A^2}{\pi^2}](https://latex.csdn.net/eq?%5Cdpi%7B120%7D%20%5Csmall%20%3DA-%5Cfrac%7BA%5E2%7D%7B2%21%7D+%5Cfrac%7BA%5E2%7D%7B%5Cpi%5E2%7D%28cos%5Cpi-1+%5Cfrac%7B%5Cpi%5E2%7D%7B2%21%7D%29%3DI-%5Cfrac%7BA%5E2%7D%7B%5Cpi%5E2%7D)
方法而:利用相似对角化:已知 是可对角化的,计算函数 ,由于 可对角化,即存在 ,使得
![\small P^{-1}AP=\Lambda =diag(\lambda_1,\lambda_2,...,\lambda_n)](https://latex.csdn.net/eq?%5Cdpi%7B120%7D%20%5Csmall%20P%5E%7B-1%7DAP%3D%5CLambda%20%3Ddiag%28%5Clambda_1%2C%5Clambda_2%2C...%2C%5Clambda_n%29)
则有:
![\small f(A)=\sum_{k=0}^{+\infty }a_kA^k=\sum_{k=0}^{+\infty }a_k(P\Lambda P^{-1})^k=P(\sum_{k=0}^{+\infty }a_k\Lambda^k)P^{-1}](https://latex.csdn.net/eq?%5Cdpi%7B120%7D%20%5Csmall%20f%28A%29%3D%5Csum_%7Bk%3D0%7D%5E%7B+%5Cinfty%20%7Da_kA%5Ek%3D%5Csum_%7Bk%3D0%7D%5E%7B+%5Cinfty%20%7Da_k%28P%5CLambda%20P%5E%7B-1%7D%29%5Ek%3DP%28%5Csum_%7Bk%3D0%7D%5E%7B+%5Cinfty%20%7Da_k%5CLambda%5Ek%29P%5E%7B-1%7D)
![\small =Pdiag(\sum_{k=0}^{+\infty }a_k\lambda_1^k,\sum_{k=0}^{+\infty }a_k\lambda_2^k,...,\sum_{k=0}^{+\infty }a_k\lambda_n^k)P^{-1}](https://latex.csdn.net/eq?%5Cdpi%7B120%7D%20%5Csmall%20%3DPdiag%28%5Csum_%7Bk%3D0%7D%5E%7B+%5Cinfty%20%7Da_k%5Clambda_1%5Ek%2C%5Csum_%7Bk%3D0%7D%5E%7B+%5Cinfty%20%7Da_k%5Clambda_2%5Ek%2C...%2C%5Csum_%7Bk%3D0%7D%5E%7B+%5Cinfty%20%7Da_k%5Clambda_n%5Ek%29P%5E%7B-1%7D)
![\small =Pdiag(f(\lambda_1),f(\lambda_2),...,f(\lambda_n))P^{-1}](https://latex.csdn.net/eq?%5Cdpi%7B120%7D%20%5Csmall%20%3DPdiag%28f%28%5Clambda_1%29%2Cf%28%5Clambda_2%29%2C...%2Cf%28%5Clambda_n%29%29P%5E%7B-1%7D)
同理可证:![\small f(At)=Pdiag(f(\lambda_1t),f(\lambda_2t),...,f(\lambda_nt))P^{-1}](https://latex.csdn.net/eq?%5Cdpi%7B120%7D%20%5Csmall%20f%28At%29%3DPdiag%28f%28%5Clambda_1t%29%2Cf%28%5Clambda_2t%29%2C...%2Cf%28%5Clambda_nt%29%29P%5E%7B-1%7D)
【例6】已知 ,求 , 。
通过计算特征多项式和特征向量可得:
,![\small P^{-1}AP=\Lambda =\begin{bmatrix} -2 &0 &0 \\ 0& 1 & 0\\ 0& 0 &1 \end{bmatrix}](https://latex.csdn.net/eq?%5Cdpi%7B120%7D%20%5Csmall%20P%5E%7B-1%7DAP%3D%5CLambda%20%3D%5Cbegin%7Bbmatrix%7D%20-2%20%260%20%260%20%5C%5C%200%26%201%20%26%200%5C%5C%200%26%200%20%261%20%5Cend%7Bbmatrix%7D)
故 ![\small e^{At}=P\begin{bmatrix} e^{-2t} & & \\ & e^t & \\ & & e^t \end{bmatrix}P^{-1}=\begin{bmatrix} 2e^t-e^{-2t} & 2e^t-2e^{-2t} &0 \\ e^{-2t}-e^t & 2e^{-2t}-e^t& 0\\ e^{-2t}-e^t& 2e^{-2t}-2e^t & e^t \end{bmatrix}](https://latex.csdn.net/eq?%5Cdpi%7B120%7D%20%5Csmall%20e%5E%7BAt%7D%3DP%5Cbegin%7Bbmatrix%7D%20e%5E%7B-2t%7D%20%26%20%26%20%5C%5C%20%26%20e%5Et%20%26%20%5C%5C%20%26%20%26%20e%5Et%20%5Cend%7Bbmatrix%7DP%5E%7B-1%7D%3D%5Cbegin%7Bbmatrix%7D%202e%5Et-e%5E%7B-2t%7D%20%26%202e%5Et-2e%5E%7B-2t%7D%20%260%20%5C%5C%20e%5E%7B-2t%7D-e%5Et%20%26%202e%5E%7B-2t%7D-e%5Et%26%200%5C%5C%20e%5E%7B-2t%7D-e%5Et%26%202e%5E%7B-2t%7D-2e%5Et%20%26%20e%5Et%20%5Cend%7Bbmatrix%7D)
![\small cosA=P\begin{bmatrix} cos(-2) & & \\ & cos1 & \\ & & cos1 \end{bmatrix}P^{-1}=\begin{bmatrix} 2cos1-cos2 & 2cos1-2cos2&0\\ cos2-cos1 &2cos2-cos1 &0 \\ cos2-cos1 &2cos2-2cos1 & cos1 \end{bmatrix}](https://latex.csdn.net/eq?%5Cdpi%7B120%7D%20%5Csmall%20cosA%3DP%5Cbegin%7Bbmatrix%7D%20cos%28-2%29%20%26%20%26%20%5C%5C%20%26%20cos1%20%26%20%5C%5C%20%26%20%26%20cos1%20%5Cend%7Bbmatrix%7DP%5E%7B-1%7D%3D%5Cbegin%7Bbmatrix%7D%202cos1-cos2%20%26%202cos1-2cos2%260%5C%5C%20cos2-cos1%20%262cos2-cos1%20%260%20%5C%5C%20cos2-cos1%20%262cos2-2cos1%20%26%20cos1%20%5Cend%7Bbmatrix%7D)
方法三:利用Jordan标准形:设 ,则存在 ,使得
![\small P^{-1}AP=J=\begin{bmatrix} J_{r_1}(\lambda_1) & & & \\ &J_{r_2}(\lambda_2) & & \\ & & ...& \\ & & & J_{r_s}(\lambda_s) \end{bmatrix}](https://latex.csdn.net/eq?%5Cdpi%7B120%7D%20%5Csmall%20P%5E%7B-1%7DAP%3DJ%3D%5Cbegin%7Bbmatrix%7D%20J_%7Br_1%7D%28%5Clambda_1%29%20%26%20%26%20%26%20%5C%5C%20%26J_%7Br_2%7D%28%5Clambda_2%29%20%26%20%26%20%5C%5C%20%26%20%26%20...%26%20%5C%5C%20%26%20%26%20%26%20J_%7Br_s%7D%28%5Clambda_s%29%20%5Cend%7Bbmatrix%7D)
其中 是特征值 对应的Jordan块。
![\small f(At)=\sum_{k=0}^{+\infty }a_k(At)^k=\sum_{k=0}^{+\infty }a_kt^k(PJP^{-1})^k=P(\sum_{k=0}^{+\infty }a_kt^kJ^k)P^{-1}](https://latex.csdn.net/eq?%5Cdpi%7B120%7D%20%5Csmall%20f%28At%29%3D%5Csum_%7Bk%3D0%7D%5E%7B+%5Cinfty%20%7Da_k%28At%29%5Ek%3D%5Csum_%7Bk%3D0%7D%5E%7B+%5Cinfty%20%7Da_kt%5Ek%28PJP%5E%7B-1%7D%29%5Ek%3DP%28%5Csum_%7Bk%3D0%7D%5E%7B+%5Cinfty%20%7Da_kt%5EkJ%5Ek%29P%5E%7B-1%7D)
![\small =Pdiag(f(J_{r_1}(\lambda_1)t),f(J_{r_2}(\lambda_2)t),...,f(J_{r_s}(\lambda_s)t))P^{-1}](https://latex.csdn.net/eq?%5Cdpi%7B120%7D%20%5Csmall%20%3DPdiag%28f%28J_%7Br_1%7D%28%5Clambda_1%29t%29%2Cf%28J_%7Br_2%7D%28%5Clambda_2%29t%29%2C...%2Cf%28J_%7Br_s%7D%28%5Clambda_s%29t%29%29P%5E%7B-1%7D)
![\small f(J_{r_i}(\lambda_i)t\begin{bmatrix} f(\lambda)& \frac{t}{1!} f'(\lambda)&... &\frac{t^{r_i-1}}{(r_i-1)!} f^{(r_i-1)}(\lambda) \\ & f(\lambda) &... &... \\ & & ...& \frac{t}{1!} f'(\lambda) \\ & & & f(\lambda) \end{bmatrix}](https://latex.csdn.net/eq?%5Cdpi%7B120%7D%20%5Csmall%20f%28J_%7Br_i%7D%28%5Clambda_i%29t%5Cbegin%7Bbmatrix%7D%20f%28%5Clambda%29%26%20%5Cfrac%7Bt%7D%7B1%21%7D%20f%27%28%5Clambda%29%26...%20%26%5Cfrac%7Bt%5E%7Br_i-1%7D%7D%7B%28r_i-1%29%21%7D%20f%5E%7B%28r_i-1%29%7D%28%5Clambda%29%20%5C%5C%20%26%20f%28%5Clambda%29%20%26...%20%26...%20%5C%5C%20%26%20%26%20...%26%20%5Cfrac%7Bt%7D%7B1%21%7D%20f%27%28%5Clambda%29%20%5C%5C%20%26%20%26%20%26%20f%28%5Clambda%29%20%5Cend%7Bbmatrix%7D)
设 , 是 的 个特征值,则矩阵函数 的特征值为:
![\small f(\lambda_1),f(\lambda_2),...,f(\lambda_n)](https://latex.csdn.net/eq?%5Cdpi%7B120%7D%20%5Csmall%20f%28%5Clambda_1%29%2Cf%28%5Clambda_2%29%2C...%2Cf%28%5Clambda_n%29)
【例7】已知 ,求 。
通过计算特征多项式和特征向量可得:
![\small P=\begin{bmatrix} 1 &0 &0 \\ -1& -1 & 1\\ 2 & 1 &0 \end{bmatrix},P^{-1}AP=J=\begin{bmatrix} 1 &1 &0 \\ 0 & 1 & 0\\ 0 & 0 & 2 \end{bmatrix}](https://latex.csdn.net/eq?%5Cdpi%7B120%7D%20%5Csmall%20P%3D%5Cbegin%7Bbmatrix%7D%201%20%260%20%260%20%5C%5C%20-1%26%20-1%20%26%201%5C%5C%202%20%26%201%20%260%20%5Cend%7Bbmatrix%7D%2CP%5E%7B-1%7DAP%3DJ%3D%5Cbegin%7Bbmatrix%7D%201%20%261%20%260%20%5C%5C%200%20%26%201%20%26%200%5C%5C%200%20%26%200%20%26%202%20%5Cend%7Bbmatrix%7D)
故 ![\small e^{A}=P\begin{bmatrix} e &e &0 \\ 0& e& 0\\ 0&0 &e^2 \end{bmatrix}P^{-1}=\begin{bmatrix} -e &0 &e \\ 3e-e^2 & e^2 & -2e+e^2\\ -4e &0 &3e \end{bmatrix}](https://latex.csdn.net/eq?%5Cdpi%7B120%7D%20%5Csmall%20e%5E%7BA%7D%3DP%5Cbegin%7Bbmatrix%7D%20e%20%26e%20%260%20%5C%5C%200%26%20e%26%200%5C%5C%200%260%20%26e%5E2%20%5Cend%7Bbmatrix%7DP%5E%7B-1%7D%3D%5Cbegin%7Bbmatrix%7D%20-e%20%260%20%26e%20%5C%5C%203e-e%5E2%20%26%20e%5E2%20%26%20-2e+e%5E2%5C%5C%20-4e%20%260%20%263e%20%5Cend%7Bbmatrix%7D)
![\small sin(At)=P\begin{bmatrix} sint & tcost &0 \\ 0&sint & 0\\ 0 & 0 & sin2t \end{bmatrix}P^{-1}](https://latex.csdn.net/eq?%5Cdpi%7B120%7D%20%5Csmall%20sin%28At%29%3DP%5Cbegin%7Bbmatrix%7D%20sint%20%26%20tcost%20%260%20%5C%5C%200%26sint%20%26%200%5C%5C%200%20%26%200%20%26%20sin2t%20%5Cend%7Bbmatrix%7DP%5E%7B-1%7D)
方法四:待定系数法:用待定系数法求矩阵函数 或 的步骤如下:
(1)求矩阵 的特征多项式。
(2)设 ,根据 或 , ,列方程组求解 。
(3)计算 或 =![\small b_{n-1}(t)A^{n-1}+...+b_1(t)A+b_0(t)I](https://latex.csdn.net/eq?%5Cdpi%7B120%7D%20%5Csmall%20b_%7Bn-1%7D%28t%29A%5E%7Bn-1%7D+...+b_1%28t%29A+b_0%28t%29I)
【例8】已知 ,求 。
可求得 ,设 ![\small r(\lambda)=b_2\lambda^2+b_1\lambda+b_0](https://latex.csdn.net/eq?%5Cdpi%7B120%7D%20%5Csmall%20r%28%5Clambda%29%3Db_2%5Clambda%5E2+b_1%5Clambda+b_0)
则由 ![\small \left\{\begin{matrix} r(1)=b_2+b_1+b_0=e^t\\ r'(1)=2b_2+b_1=te^t\\ r(2)=4b_2+2b_1+b_0=e^{2t} \end{matrix}\right.\Rightarrow \left\{\begin{matrix} b_2=e^{2t}-e^t-te^t\\ b_1=-2e^{2t}+2e^t+3te^t\\ b_0=e^{2t}-2te^t \end{matrix}\right.](https://latex.csdn.net/eq?%5Cdpi%7B120%7D%20%5Csmall%20%5Cleft%5C%7B%5Cbegin%7Bmatrix%7D%20r%281%29%3Db_2+b_1+b_0%3De%5Et%5C%5C%20r%27%281%29%3D2b_2+b_1%3Dte%5Et%5C%5C%20r%282%29%3D4b_2+2b_1+b_0%3De%5E%7B2t%7D%20%5Cend%7Bmatrix%7D%5Cright.%5CRightarrow%20%5Cleft%5C%7B%5Cbegin%7Bmatrix%7D%20b_2%3De%5E%7B2t%7D-e%5Et-te%5Et%5C%5C%20b_1%3D-2e%5E%7B2t%7D+2e%5Et+3te%5Et%5C%5C%20b_0%3De%5E%7B2t%7D-2te%5Et%20%5Cend%7Bmatrix%7D%5Cright.)
![\small e^{At}=b_2A^2+b_1A+b_0I=\begin{bmatrix} e^{2t}-2te^t & 0&te^t \\ -e^{2t}+e^t+2te^t& e^{2t}&e^{2t}-e^t-te^t \\ -4te^t &0 & 2te^t+e^t \end{bmatrix}](https://latex.csdn.net/eq?%5Cdpi%7B120%7D%20%5Csmall%20e%5E%7BAt%7D%3Db_2A%5E2+b_1A+b_0I%3D%5Cbegin%7Bbmatrix%7D%20e%5E%7B2t%7D-2te%5Et%20%26%200%26te%5Et%20%5C%5C%20-e%5E%7B2t%7D+e%5Et+2te%5Et%26%20e%5E%7B2t%7D%26e%5E%7B2t%7D-e%5Et-te%5Et%20%5C%5C%20-4te%5Et%20%260%20%26%202te%5Et+e%5Et%20%5Cend%7Bbmatrix%7D)
同理 ![\small \left\{\begin{matrix} r(1)=b_2+b_1+b_0=cos1\\ r'(1)=2b_2+b_1=-sin1\\ r(2)=4b_2+2b_1+b_0=cos2 \end{matrix}\right.\Rightarrow \left\{\begin{matrix} b_2=sin1-cos1+cos2\\ b_1=-3sin1+2cos1-2cos2\\ b_0=2sin1+cos2 \end{matrix}\right.](https://latex.csdn.net/eq?%5Cdpi%7B120%7D%20%5Csmall%20%5Cleft%5C%7B%5Cbegin%7Bmatrix%7D%20r%281%29%3Db_2+b_1+b_0%3Dcos1%5C%5C%20r%27%281%29%3D2b_2+b_1%3D-sin1%5C%5C%20r%282%29%3D4b_2+2b_1+b_0%3Dcos2%20%5Cend%7Bmatrix%7D%5Cright.%5CRightarrow%20%5Cleft%5C%7B%5Cbegin%7Bmatrix%7D%20b_2%3Dsin1-cos1+cos2%5C%5C%20b_1%3D-3sin1+2cos1-2cos2%5C%5C%20b_0%3D2sin1+cos2%20%5Cend%7Bmatrix%7D%5Cright.)
![\small cosA=\begin{bmatrix} 2sin1+cos2& 0& -sin1\\ -cos2+cos1-2sin1& cos2 &cos2-cos1+sin1 \\ 4sin1 & 0 & -2sin1+cos1 \end{bmatrix}](https://latex.csdn.net/eq?%5Cdpi%7B120%7D%20%5Csmall%20cosA%3D%5Cbegin%7Bbmatrix%7D%202sin1+cos2%26%200%26%20-sin1%5C%5C%20-cos2+cos1-2sin1%26%20cos2%20%26cos2-cos1+sin1%20%5C%5C%204sin1%20%26%200%20%26%20-2sin1+cos1%20%5Cend%7Bbmatrix%7D)
如果求得次数比矩阵 特征多项式低的零化多项式——最小多项式
则计算矩阵函数就更容易一些。
【例9】已知 ,求 。
可求得 的Jordan标准形为 ![\small J=\begin{bmatrix} 2 & 0 &0 \\ 0 & 2 &1 \\ 0&0 & 2 \end{bmatrix}](https://latex.csdn.net/eq?%5Cdpi%7B120%7D%20%5Csmall%20J%3D%5Cbegin%7Bbmatrix%7D%202%20%26%200%20%260%20%5C%5C%200%20%26%202%20%261%20%5C%5C%200%260%20%26%202%20%5Cend%7Bbmatrix%7D)
于是 的最小多项式为 ![\small m_A(\lambda)=(\lambda-2)^2](https://latex.csdn.net/eq?%5Cdpi%7B120%7D%20%5Csmall%20m_A%28%5Clambda%29%3D%28%5Clambda-2%29%5E2)
设 ![\small r(\lambda)=b_1\lambda+b_0](https://latex.csdn.net/eq?%5Cdpi%7B120%7D%20%5Csmall%20r%28%5Clambda%29%3Db_1%5Clambda+b_0)
由 ![\small \left\{\begin{matrix} r(2)=2b_1+b_0=e^{2t}\\ r'(2)=b_1=te^{2t} \end{matrix}\right.\Rightarrow \left\{\begin{matrix} b_1=te^{2t}\\ b_0=(1-2t)e^{2t} \end{matrix}\right.](https://latex.csdn.net/eq?%5Cdpi%7B120%7D%20%5Csmall%20%5Cleft%5C%7B%5Cbegin%7Bmatrix%7D%20r%282%29%3D2b_1+b_0%3De%5E%7B2t%7D%5C%5C%20r%27%282%29%3Db_1%3Dte%5E%7B2t%7D%20%5Cend%7Bmatrix%7D%5Cright.%5CRightarrow%20%5Cleft%5C%7B%5Cbegin%7Bmatrix%7D%20b_1%3Dte%5E%7B2t%7D%5C%5C%20b_0%3D%281-2t%29e%5E%7B2t%7D%20%5Cend%7Bmatrix%7D%5Cright.)
于是 ![\small e^{At}=b_1A+b_0I=e^{2t}\begin{bmatrix} 1+t & t &-t \\ -2t &1-2t &2t \\ -t &-t &1+t \end{bmatrix}](https://latex.csdn.net/eq?%5Cdpi%7B120%7D%20%5Csmall%20e%5E%7BAt%7D%3Db_1A+b_0I%3De%5E%7B2t%7D%5Cbegin%7Bbmatrix%7D%201+t%20%26%20t%20%26-t%20%5C%5C%20-2t%20%261-2t%20%262t%20%5C%5C%20-t%20%26-t%20%261+t%20%5Cend%7Bbmatrix%7D)
又由于 ![\small \left\{\begin{matrix} r(2)=2b_1+b_0=sin2\\ r'(2)=b_1=cos2 \end{matrix}\right.\Rightarrow \left\{\begin{matrix} b_1=cos2\\ b_0=sin2-2cos2 \end{matrix}\right.](https://latex.csdn.net/eq?%5Cdpi%7B120%7D%20%5Csmall%20%5Cleft%5C%7B%5Cbegin%7Bmatrix%7D%20r%282%29%3D2b_1+b_0%3Dsin2%5C%5C%20r%27%282%29%3Db_1%3Dcos2%20%5Cend%7Bmatrix%7D%5Cright.%5CRightarrow%20%5Cleft%5C%7B%5Cbegin%7Bmatrix%7D%20b_1%3Dcos2%5C%5C%20b_0%3Dsin2-2cos2%20%5Cend%7Bmatrix%7D%5Cright.)
故 ![\small sinA=b_1A+b_0I=\begin{bmatrix} sin2+cos2 & cos2 & -cos2\\ -2cos2 & sin2-2cos2 &2cos2 \\ -cos2 &-cos2 & sin2+cos2 \end{bmatrix}](https://latex.csdn.net/eq?%5Cdpi%7B120%7D%20%5Csmall%20sinA%3Db_1A+b_0I%3D%5Cbegin%7Bbmatrix%7D%20sin2+cos2%20%26%20cos2%20%26%20-cos2%5C%5C%20-2cos2%20%26%20sin2-2cos2%20%262cos2%20%5C%5C%20-cos2%20%26-cos2%20%26%20sin2+cos2%20%5Cend%7Bbmatrix%7D)
3.3,矩阵函数的性质
常见的矩阵函数有 ,它们有些性质与普通的指数函数和三角函数相同;有些性质与一般指数函数和三角函数不同,这是由于矩阵乘法不满足交换律。
对任意 ,总有:
(1) ,![\small cos(-A)=cosA](https://latex.csdn.net/eq?%5Cdpi%7B120%7D%20%5Csmall%20cos%28-A%29%3DcosA)
(2) , ,![\small sinA=\frac{1}{2i}(e^{iA}-e^{-iA})](https://latex.csdn.net/eq?%5Cdpi%7B120%7D%20%5Csmall%20sinA%3D%5Cfrac%7B1%7D%7B2i%7D%28e%5E%7BiA%7D-e%5E%7B-iA%7D%29)
对任意 ,且 ,则:
(1)![\small e^{A+B}=e^Ae^B=e^Be^A](https://latex.csdn.net/eq?%5Cdpi%7B120%7D%20%5Csmall%20e%5E%7BA+B%7D%3De%5EAe%5EB%3De%5EBe%5EA)
(2)![\small sin(A+B)=sinAcosB+cosAsinB;cos(A+B)=cosAcosB-sinAsinB](https://latex.csdn.net/eq?%5Cdpi%7B120%7D%20%5Csmall%20sin%28A+B%29%3DsinAcosB+cosAsinB%3Bcos%28A+B%29%3DcosAcosB-sinAsinB)
(3)
对任意 ,则有:
(1)![\small dete^A=e^{trA};(e^A)^{-1}=e^{-A}](https://latex.csdn.net/eq?%5Cdpi%7B120%7D%20%5Csmall%20dete%5EA%3De%5E%7BtrA%7D%3B%28e%5EA%29%5E%7B-1%7D%3De%5E%7B-A%7D)
4,矩阵的微分和积分
4.1,函数矩阵的微分和积分
设以变量 的函数为元素的矩阵 称为函数矩阵,其中 都是变量 的函数。
(1)若 ,则称 是定义在 上的。
(2)又若每个 在 上连续、可微、可积,则称 在 上连续、可微、可积的。
(3)当 可微时,规定其导数为:![\small \frac{d}{dt}A(t)=(\frac{d}{dt}a_{ij}(t))_{m\times n}](https://latex.csdn.net/eq?%5Cdpi%7B120%7D%20%5Csmall%20%5Cfrac%7Bd%7D%7Bdt%7DA%28t%29%3D%28%5Cfrac%7Bd%7D%7Bdt%7Da_%7Bij%7D%28t%29%29_%7Bm%5Ctimes%20n%7D)
(4)当 在 上可积时,规定 在 上积分为:![\small \int_{a}^{b}A(t)dt=(\int_{a}^{b}a_{ij}(t)dt)_{m\times n}](https://latex.csdn.net/eq?%5Cdpi%7B120%7D%20%5Csmall%20%5Cint_%7Ba%7D%5E%7Bb%7DA%28t%29dt%3D%28%5Cint_%7Ba%7D%5E%7Bb%7Da_%7Bij%7D%28t%29dt%29_%7Bm%5Ctimes%20n%7D)
【例10】求函数矩阵 的导数
![\small \frac{d}{dt}A(t)=(\frac{d}{dt}a_{ij}(t))=\begin{bmatrix} cost & -sint & 1\\ 2^tln2 & e^t &2t \\ 0 & 0 & 3t^2 \end{bmatrix}](https://latex.csdn.net/eq?%5Cdpi%7B120%7D%20%5Csmall%20%5Cfrac%7Bd%7D%7Bdt%7DA%28t%29%3D%28%5Cfrac%7Bd%7D%7Bdt%7Da_%7Bij%7D%28t%29%29%3D%5Cbegin%7Bbmatrix%7D%20cost%20%26%20-sint%20%26%201%5C%5C%202%5Etln2%20%26%20e%5Et%20%262t%20%5C%5C%200%20%26%200%20%26%203t%5E2%20%5Cend%7Bbmatrix%7D)
设 与 为适当阶的可微矩阵,则:
(1)![\small \frac{d}{dt}(A(t)+B(t))=\frac{d}{dt}(A(t))+\frac{d}{dt}(B(t))](https://latex.csdn.net/eq?%5Cdpi%7B120%7D%20%5Csmall%20%5Cfrac%7Bd%7D%7Bdt%7D%28A%28t%29+B%28t%29%29%3D%5Cfrac%7Bd%7D%7Bdt%7D%28A%28t%29%29+%5Cfrac%7Bd%7D%7Bdt%7D%28B%28t%29%29)
(2)当 是可微函数时,有 ![\small \frac{d}{dt}(\lambda(t)A(t))=(\frac{d}{dt}\lambda(t))A(t)+\lambda(t)\frac{d}{dt}(A(t))](https://latex.csdn.net/eq?%5Cdpi%7B120%7D%20%5Csmall%20%5Cfrac%7Bd%7D%7Bdt%7D%28%5Clambda%28t%29A%28t%29%29%3D%28%5Cfrac%7Bd%7D%7Bdt%7D%5Clambda%28t%29%29A%28t%29+%5Clambda%28t%29%5Cfrac%7Bd%7D%7Bdt%7D%28A%28t%29%29)
(3)![\small \frac{d}{dt}(A(t)B(t))=\frac{d}{dt}(A(t))B(t)+A(t)\frac{d}{dt}(B(t))](https://latex.csdn.net/eq?%5Cdpi%7B120%7D%20%5Csmall%20%5Cfrac%7Bd%7D%7Bdt%7D%28A%28t%29B%28t%29%29%3D%5Cfrac%7Bd%7D%7Bdt%7D%28A%28t%29%29B%28t%29+A%28t%29%5Cfrac%7Bd%7D%7Bdt%7D%28B%28t%29%29)
(4)当 关于 可微时,有 ![\small \frac{d}{dt}(A(u))=f'(t)\frac{d}{du}(A(u))](https://latex.csdn.net/eq?%5Cdpi%7B120%7D%20%5Csmall%20%5Cfrac%7Bd%7D%7Bdt%7D%28A%28u%29%29%3Df%27%28t%29%5Cfrac%7Bd%7D%7Bdu%7D%28A%28u%29%29)
(5)当 是可微矩阵时,有 ![\small \frac{d}{dt}(A^{-1}(t))=-A^{-1}(t)\frac{d}{dt}(A(t))A^{-1}(t)](https://latex.csdn.net/eq?%5Cdpi%7B120%7D%20%5Csmall%20%5Cfrac%7Bd%7D%7Bdt%7D%28A%5E%7B-1%7D%28t%29%29%3D-A%5E%7B-1%7D%28t%29%5Cfrac%7Bd%7D%7Bdt%7D%28A%28t%29%29A%5E%7B-1%7D%28t%29)
(6)由于 仍是函数矩阵,如果它仍是可导的,即可以定义其二阶导数。
(7)函数矩阵的高阶导数:![\small \frac{d^k}{dt^k}A(t)=\frac{d}{dt}\left ( \frac{d^{k-1}}{dt^{k-1}}A(t) \right )](https://latex.csdn.net/eq?%5Cdpi%7B120%7D%20%5Csmall%20%5Cfrac%7Bd%5Ek%7D%7Bdt%5Ek%7DA%28t%29%3D%5Cfrac%7Bd%7D%7Bdt%7D%5Cleft%20%28%20%5Cfrac%7Bd%5E%7Bk-1%7D%7D%7Bdt%5E%7Bk-1%7D%7DA%28t%29%20%5Cright%20%29)
设 ,则有(利用逐项求导的方法)
(1)![\small \frac{d}{dt}(e^{At})=Ae^{At}=e^{At}A](https://latex.csdn.net/eq?%5Cdpi%7B120%7D%20%5Csmall%20%5Cfrac%7Bd%7D%7Bdt%7D%28e%5E%7BAt%7D%29%3DAe%5E%7BAt%7D%3De%5E%7BAt%7DA)
(2)![\small \frac{d}{dt}(sin(At))=Acos(At)=cos(At)A](https://latex.csdn.net/eq?%5Cdpi%7B120%7D%20%5Csmall%20%5Cfrac%7Bd%7D%7Bdt%7D%28sin%28At%29%29%3DAcos%28At%29%3Dcos%28At%29A)
(3) ![\small \frac{d}{dt}(cos(At))=-Asin(At)=-sin(At)A](https://latex.csdn.net/eq?%5Cdpi%7B120%7D%20%5Csmall%20%5Cfrac%7Bd%7D%7Bdt%7D%28cos%28At%29%29%3D-Asin%28At%29%3D-sin%28At%29A)
设 与 是区间 上适当阶的可积矩阵, 是适当阶的常数矩阵, ,则:
(1)![\small \int_{a}^{b}(A(t)+B(t))dt=\int_{a}^{b}A(t)dt+\int_{a}^{b}B(t)dt](https://latex.csdn.net/eq?%5Cdpi%7B120%7D%20%5Csmall%20%5Cint_%7Ba%7D%5E%7Bb%7D%28A%28t%29+B%28t%29%29dt%3D%5Cint_%7Ba%7D%5E%7Bb%7DA%28t%29dt+%5Cint_%7Ba%7D%5E%7Bb%7DB%28t%29dt)
(2)![\small \int_{a}^{b}(\lambda A(t))dt=\lambda\int_{a}^{b} A(t)dt](https://latex.csdn.net/eq?%5Cdpi%7B120%7D%20%5Csmall%20%5Cint_%7Ba%7D%5E%7Bb%7D%28%5Clambda%20A%28t%29%29dt%3D%5Clambda%5Cint_%7Ba%7D%5E%7Bb%7D%20A%28t%29dt)
(3) ,![\small \int_{a}^{b}(AB(t))dt=A(\int_{a}^{b} B(t)dt)](https://latex.csdn.net/eq?%5Cdpi%7B120%7D%20%5Csmall%20%5Cint_%7Ba%7D%5E%7Bb%7D%28AB%28t%29%29dt%3DA%28%5Cint_%7Ba%7D%5E%7Bb%7D%20B%28t%29dt%29)
(4)若 在区间 上连续,则对任意 ,有 ![\small \frac{d}{dt}\int_{a}^{t}A(\tau )d\tau =A(t)](https://latex.csdn.net/eq?%5Cdpi%7B120%7D%20%5Csmall%20%5Cfrac%7Bd%7D%7Bdt%7D%5Cint_%7Ba%7D%5E%7Bt%7DA%28%5Ctau%20%29d%5Ctau%20%3DA%28t%29)
(5)若 在区间 上连续可微时,有 ![\small \int_{a}^{b}A'(t)dt=A(b)-A(a)](https://latex.csdn.net/eq?%5Cdpi%7B120%7D%20%5Csmall%20%5Cint_%7Ba%7D%5E%7Bb%7DA%27%28t%29dt%3DA%28b%29-A%28a%29)
4.2,数量矩阵对矩阵变量的导数
设 是以矩阵 为自变量的 元函数,且 都存在,规定 对矩阵变量 的导数为 为:
![\small \frac{df}{dX}=(\frac{\partial f}{\partial x_{ij}})_{m\times n}=\begin{bmatrix} \frac{\partial f}{\partial x_{11}} &... &\frac{\partial f}{\partial x_{1n}} \\ ... & & ...\\ \frac{\partial f}{\partial x_{m1}} & ...& \frac{\partial f}{\partial x_{mn}} \end{bmatrix}](https://latex.csdn.net/eq?%5Cdpi%7B120%7D%20%5Csmall%20%5Cfrac%7Bdf%7D%7BdX%7D%3D%28%5Cfrac%7B%5Cpartial%20f%7D%7B%5Cpartial%20x_%7Bij%7D%7D%29_%7Bm%5Ctimes%20n%7D%3D%5Cbegin%7Bbmatrix%7D%20%5Cfrac%7B%5Cpartial%20f%7D%7B%5Cpartial%20x_%7B11%7D%7D%20%26...%20%26%5Cfrac%7B%5Cpartial%20f%7D%7B%5Cpartial%20x_%7B1n%7D%7D%20%5C%5C%20...%20%26%20%26%20...%5C%5C%20%5Cfrac%7B%5Cpartial%20f%7D%7B%5Cpartial%20x_%7Bm1%7D%7D%20%26%20...%26%20%5Cfrac%7B%5Cpartial%20f%7D%7B%5Cpartial%20x_%7Bmn%7D%7D%20%5Cend%7Bbmatrix%7D)
特别的,以 为自变量的函数 的导数:
![\small \frac{df}{dx}=\left [ \frac{df}{d\xi _1},\frac{df}{d\xi _2},...,\frac{df}{d\xi _n} \right ]^T](https://latex.csdn.net/eq?%5Cdpi%7B120%7D%20%5Csmall%20%5Cfrac%7Bdf%7D%7Bdx%7D%3D%5Cleft%20%5B%20%5Cfrac%7Bdf%7D%7Bd%5Cxi%20_1%7D%2C%5Cfrac%7Bdf%7D%7Bd%5Cxi%20_2%7D%2C...%2C%5Cfrac%7Bdf%7D%7Bd%5Cxi%20_n%7D%20%5Cright%20%5D%5ET)
称为数量函数对向量变量的导数,记为 。
设 是给定的向量, 为向量变量,且 ,求 。
解:因为 ,![\small \frac{\partial f}{\partial \xi_j}=a_j(j=1,2,...,n)](https://latex.csdn.net/eq?%5Cdpi%7B120%7D%20%5Csmall%20%5Cfrac%7B%5Cpartial%20f%7D%7B%5Cpartial%20%5Cxi_j%7D%3Da_j%28j%3D1%2C2%2C...%2Cn%29)
所以 ![\small \frac{df}{dx}=\left [\frac{\partial f}{\partial \xi_1} ,\frac{\partial f}{\partial \xi_2} ,...,\frac{\partial f}{\partial \xi_n} \right ]^T=\left ( a_1,a_2,...,a_n \right )^T=\alpha](https://latex.csdn.net/eq?%5Cdpi%7B120%7D%20%5Csmall%20%5Cfrac%7Bdf%7D%7Bdx%7D%3D%5Cleft%20%5B%5Cfrac%7B%5Cpartial%20f%7D%7B%5Cpartial%20%5Cxi_1%7D%20%2C%5Cfrac%7B%5Cpartial%20f%7D%7B%5Cpartial%20%5Cxi_2%7D%20%2C...%2C%5Cfrac%7B%5Cpartial%20f%7D%7B%5Cpartial%20%5Cxi_n%7D%20%5Cright%20%5D%5ET%3D%5Cleft%20%28%20a_1%2Ca_2%2C...%2Ca_n%20%5Cright%20%29%5ET%3D%5Calpha)
设 是给定的矩阵, 是矩阵变量,且 ,求 。
解:因为 ![\small AX=(\sum_{k=1}^na_{ik}x_{kj})_{m\times m}](https://latex.csdn.net/eq?%5Cdpi%7B120%7D%20%5Csmall%20AX%3D%28%5Csum_%7Bk%3D1%7D%5Ena_%7Bik%7Dx_%7Bkj%7D%29_%7Bm%5Ctimes%20m%7D)
所以 ![\small f(X)=tr(AX)=\sum_{s=1}^m\sum_{k=1}^na_{sk}x_{ks}](https://latex.csdn.net/eq?%5Cdpi%7B120%7D%20%5Csmall%20f%28X%29%3Dtr%28AX%29%3D%5Csum_%7Bs%3D1%7D%5Em%5Csum_%7Bk%3D1%7D%5Ena_%7Bsk%7Dx_%7Bks%7D)
而 ![\small \frac{\partial f}{\partial x_{ij}}=a_{ij}(i=1,2,...,n;j=1,2,...,n)](https://latex.csdn.net/eq?%5Cdpi%7B120%7D%20%5Csmall%20%5Cfrac%7B%5Cpartial%20f%7D%7B%5Cpartial%20x_%7Bij%7D%7D%3Da_%7Bij%7D%28i%3D1%2C2%2C...%2Cn%3Bj%3D1%2C2%2C...%2Cn%29)
故 ![\small \frac{df}{dX}=\left ( \frac{\partial f}{\partial x_{ij}}\right )=(a_{ji})_{n\times m}=A^T](https://latex.csdn.net/eq?%5Cdpi%7B120%7D%20%5Csmall%20%5Cfrac%7Bdf%7D%7BdX%7D%3D%5Cleft%20%28%20%5Cfrac%7B%5Cpartial%20f%7D%7B%5Cpartial%20x_%7Bij%7D%7D%5Cright%20%29%3D%28a_%7Bji%7D%29_%7Bn%5Ctimes%20m%7D%3DA%5ET)
设 是给定的矩阵, 是向量变量,且 ,求 。
解:因为 ![\small f(x)=x^TAx=\sum_{s=1}^n\xi_s\left ( \sum_{k=1}^na_{sk}\, \xi_k \right )](https://latex.csdn.net/eq?%5Cdpi%7B120%7D%20%5Csmall%20f%28x%29%3Dx%5ETAx%3D%5Csum_%7Bs%3D1%7D%5En%5Cxi_s%5Cleft%20%28%20%5Csum_%7Bk%3D1%7D%5Ena_%7Bsk%7D%5C%2C%20%5Cxi_k%20%5Cright%20%29)
所以 ![\small \frac{\partial f}{\partial \xi_j}=\sum_{s=1}^na_{sj}\, \xi_s+\sum_{k=1}^na_{jk}\, \xi_k(j=1,2,...,n)](https://latex.csdn.net/eq?%5Cdpi%7B120%7D%20%5Csmall%20%5Cfrac%7B%5Cpartial%20f%7D%7B%5Cpartial%20%5Cxi_j%7D%3D%5Csum_%7Bs%3D1%7D%5Ena_%7Bsj%7D%5C%2C%20%5Cxi_s+%5Csum_%7Bk%3D1%7D%5Ena_%7Bjk%7D%5C%2C%20%5Cxi_k%28j%3D1%2C2%2C...%2Cn%29)
所以 ![\small \frac{df}{dx}=\begin{bmatrix} \frac{\partial f}{\partial \xi_1}\\ ...\\ \frac{\partial f}{\partial \xi_n} \end{bmatrix}=\begin{bmatrix} \sum_{s=1}^na_{s1}\, \xi_s+\sum_{k=1}^na_{1k}\, \xi_k\\ ...\\ \sum_{s=1}^na_{sn}\, \xi_s+\sum_{k=1}^na_{nk}\, \xi_k \end{bmatrix}=(A^T+A)x](https://latex.csdn.net/eq?%5Cdpi%7B120%7D%20%5Csmall%20%5Cfrac%7Bdf%7D%7Bdx%7D%3D%5Cbegin%7Bbmatrix%7D%20%5Cfrac%7B%5Cpartial%20f%7D%7B%5Cpartial%20%5Cxi_1%7D%5C%5C%20...%5C%5C%20%5Cfrac%7B%5Cpartial%20f%7D%7B%5Cpartial%20%5Cxi_n%7D%20%5Cend%7Bbmatrix%7D%3D%5Cbegin%7Bbmatrix%7D%20%5Csum_%7Bs%3D1%7D%5Ena_%7Bs1%7D%5C%2C%20%5Cxi_s+%5Csum_%7Bk%3D1%7D%5Ena_%7B1k%7D%5C%2C%20%5Cxi_k%5C%5C%20...%5C%5C%20%5Csum_%7Bs%3D1%7D%5Ena_%7Bsn%7D%5C%2C%20%5Cxi_s+%5Csum_%7Bk%3D1%7D%5Ena_%7Bnk%7D%5C%2C%20%5Cxi_k%20%5Cend%7Bbmatrix%7D%3D%28A%5ET+A%29x)
4.3,矩阵值函数对矩阵变量的导数
设矩阵 的元素 都是以矩阵变量 的函数,称 为矩阵值函数,规定 对矩阵变量 的导数为 。
,其中 ![\small \frac{\partial F}{\partial x_{ij}}=\begin{bmatrix} \frac{\partial f_{11}}{\partial x_{ij}}&... & \frac{\partial f_{1t}}{\partial x_{ij}} \\ ...& &... \\ \frac{\partial f_{s1}}{\partial x_{ij}} &... & \frac{\partial f_{st}}{\partial x_{ij}} \end{bmatrix}](https://latex.csdn.net/eq?%5Cdpi%7B120%7D%20%5Csmall%20%5Cfrac%7B%5Cpartial%20F%7D%7B%5Cpartial%20x_%7Bij%7D%7D%3D%5Cbegin%7Bbmatrix%7D%20%5Cfrac%7B%5Cpartial%20f_%7B11%7D%7D%7B%5Cpartial%20x_%7Bij%7D%7D%26...%20%26%20%5Cfrac%7B%5Cpartial%20f_%7B1t%7D%7D%7B%5Cpartial%20x_%7Bij%7D%7D%20%5C%5C%20...%26%20%26...%20%5C%5C%20%5Cfrac%7B%5Cpartial%20f_%7Bs1%7D%7D%7B%5Cpartial%20x_%7Bij%7D%7D%20%26...%20%26%20%5Cfrac%7B%5Cpartial%20f_%7Bst%7D%7D%7B%5Cpartial%20x_%7Bij%7D%7D%20%5Cend%7Bbmatrix%7D)
其结果为 的矩阵。
设 为向量变量,求 和 。
由定义得,![\small \frac{dx}{dx^T}=\left ( \frac{\partial x}{\partial \xi_1},\frac{\partial x}{\partial \xi_2},...,\frac{\partial x}{\partial \xi_n} \right )=I_n](https://latex.csdn.net/eq?%5Cdpi%7B120%7D%20%5Csmall%20%5Cfrac%7Bdx%7D%7Bdx%5ET%7D%3D%5Cleft%20%28%20%5Cfrac%7B%5Cpartial%20x%7D%7B%5Cpartial%20%5Cxi_1%7D%2C%5Cfrac%7B%5Cpartial%20x%7D%7B%5Cpartial%20%5Cxi_2%7D%2C...%2C%5Cfrac%7B%5Cpartial%20x%7D%7B%5Cpartial%20%5Cxi_n%7D%20%5Cright%20%29%3DI_n)
类似由定义得,![\small \frac{dx^T}{dx}=\begin{bmatrix} \frac{\partial x^T}{\partial \xi_1}\\ ...\\ \frac{\partial x^T}{\partial \xi_n} \end{bmatrix}=I_n](https://latex.csdn.net/eq?%5Cdpi%7B120%7D%20%5Csmall%20%5Cfrac%7Bdx%5ET%7D%7Bdx%7D%3D%5Cbegin%7Bbmatrix%7D%20%5Cfrac%7B%5Cpartial%20x%5ET%7D%7B%5Cpartial%20%5Cxi_1%7D%5C%5C%20...%5C%5C%20%5Cfrac%7B%5Cpartial%20x%5ET%7D%7B%5Cpartial%20%5Cxi_n%7D%20%5Cend%7Bbmatrix%7D%3DI_n)
5,矩阵分析应用举例
5.1,一阶常系数线性微分方程组
有常系数微分方程组
![\small \left\{\begin{matrix} \frac{dx_1(t)}{dt}=a_{11}x_1(t)+a_{12}x_2(t)+...+a_{1n}x_n(t)+f_1(t)\\ \frac{dx_2(t)}{dt}=a_{21}x_1(t)+a_{22}x_2(t)+...+a_{2n}x_n(t)+f_2(t)\\ ...\\ \frac{dx_n(t)}{dt}=a_{n1}x_1(t)+a_{n2}x_2(t)+...+a_{nn}x_n(t)+f_n(t) \end{matrix}\right.](https://latex.csdn.net/eq?%5Cdpi%7B120%7D%20%5Csmall%20%5Cleft%5C%7B%5Cbegin%7Bmatrix%7D%20%5Cfrac%7Bdx_1%28t%29%7D%7Bdt%7D%3Da_%7B11%7Dx_1%28t%29+a_%7B12%7Dx_2%28t%29+...+a_%7B1n%7Dx_n%28t%29+f_1%28t%29%5C%5C%20%5Cfrac%7Bdx_2%28t%29%7D%7Bdt%7D%3Da_%7B21%7Dx_1%28t%29+a_%7B22%7Dx_2%28t%29+...+a_%7B2n%7Dx_n%28t%29+f_2%28t%29%5C%5C%20...%5C%5C%20%5Cfrac%7Bdx_n%28t%29%7D%7Bdt%7D%3Da_%7Bn1%7Dx_1%28t%29+a_%7Bn2%7Dx_2%28t%29+...+a_%7Bnn%7Dx_n%28t%29+f_n%28t%29%20%5Cend%7Bmatrix%7D%5Cright.)
满足初始条件 的解。
, , ,![\small f(t)=\begin{bmatrix} f_1(t)\\ f_2(t)\\ ...\\ f_n(t) \end{bmatrix}](https://latex.csdn.net/eq?%5Cdpi%7B120%7D%20%5Csmall%20f%28t%29%3D%5Cbegin%7Bbmatrix%7D%20f_1%28t%29%5C%5C%20f_2%28t%29%5C%5C%20...%5C%5C%20f_n%28t%29%20%5Cend%7Bbmatrix%7D)
则上述微分方程组可写为:
![\small \left\{\begin{matrix} \frac{dx(t)}{dt}=Ax(t)+f(t)\\ x(t_0)=c \end{matrix}\right.](https://latex.csdn.net/eq?%5Cdpi%7B120%7D%20%5Csmall%20%5Cleft%5C%7B%5Cbegin%7Bmatrix%7D%20%5Cfrac%7Bdx%28t%29%7D%7Bdt%7D%3DAx%28t%29+f%28t%29%5C%5C%20x%28t_0%29%3Dc%20%5Cend%7Bmatrix%7D%5Cright.)
通过常数变异法来求解
因为 ![\small \frac{d}{dt}(e^{-At}x(t))=e^{-At}(-A)x(t)+e^{-At}\frac{dx(t)}{dt}=e^{-At}f(t)](https://latex.csdn.net/eq?%5Cdpi%7B120%7D%20%5Csmall%20%5Cfrac%7Bd%7D%7Bdt%7D%28e%5E%7B-At%7Dx%28t%29%29%3De%5E%7B-At%7D%28-A%29x%28t%29+e%5E%7B-At%7D%5Cfrac%7Bdx%28t%29%7D%7Bdt%7D%3De%5E%7B-At%7Df%28t%29)
在 上积分,得
![\small \int_{t_0}^{t}\frac{d}{d\tau }(e^{-A\tau }x(\tau))d \tau =\int_{t_0}^{t}e^{-A\tau}f(\tau)d\tau](https://latex.csdn.net/eq?%5Cdpi%7B120%7D%20%5Csmall%20%5Cint_%7Bt_0%7D%5E%7Bt%7D%5Cfrac%7Bd%7D%7Bd%5Ctau%20%7D%28e%5E%7B-A%5Ctau%20%7Dx%28%5Ctau%29%29d%20%5Ctau%20%3D%5Cint_%7Bt_0%7D%5E%7Bt%7De%5E%7B-A%5Ctau%7Df%28%5Ctau%29d%5Ctau)
于是微粉刺方程组的解为:![\small x(t)=e^{A(t-t_0)}c+e^{At}\int_{t_0}^{t}e^{-A\tau }f(\tau )d\tau](https://latex.csdn.net/eq?%5Cdpi%7B120%7D%20%5Csmall%20x%28t%29%3De%5E%7BA%28t-t_0%29%7Dc+e%5E%7BAt%7D%5Cint_%7Bt_0%7D%5E%7Bt%7De%5E%7B-A%5Ctau%20%7Df%28%5Ctau%20%29d%5Ctau)
【例11】求解微分方程组初值问题。
![\small \left\{\begin{matrix} \frac{dx_1(t)}{dt}=-x_1(t)+x_3(t)+1\\ \frac{dx_2(t)}{dt}=-x_1(t)+2x_2(t)-1\\ \frac{dx_3(t)}{dt}=-4x_1(t)+3x_3(t)+2\\ x_(0)=1,x_2(0)=0,x_3(0)=1 \end{matrix}\right.](https://latex.csdn.net/eq?%5Cdpi%7B120%7D%20%5Csmall%20%5Cleft%5C%7B%5Cbegin%7Bmatrix%7D%20%5Cfrac%7Bdx_1%28t%29%7D%7Bdt%7D%3D-x_1%28t%29+x_3%28t%29+1%5C%5C%20%5Cfrac%7Bdx_2%28t%29%7D%7Bdt%7D%3D-x_1%28t%29+2x_2%28t%29-1%5C%5C%20%5Cfrac%7Bdx_3%28t%29%7D%7Bdt%7D%3D-4x_1%28t%29+3x_3%28t%29+2%5C%5C%20x_%280%29%3D1%2Cx_2%280%29%3D0%2Cx_3%280%29%3D1%20%5Cend%7Bmatrix%7D%5Cright.)
记 , , ,![\small f(t)=\begin{bmatrix} 1\\ -1\\ 2 \end{bmatrix}](https://latex.csdn.net/eq?%5Cdpi%7B120%7D%20%5Csmall%20f%28t%29%3D%5Cbegin%7Bbmatrix%7D%201%5C%5C%20-1%5C%5C%202%20%5Cend%7Bbmatrix%7D)
则微分方程组能写成矩阵的形式:
![\small \left\{\begin{matrix} \frac{dx(t)}{dt}=Ax(t)+f(t)\\ x(t_0)=c \end{matrix}\right.](https://latex.csdn.net/eq?%5Cdpi%7B120%7D%20%5Csmall%20%5Cleft%5C%7B%5Cbegin%7Bmatrix%7D%20%5Cfrac%7Bdx%28t%29%7D%7Bdt%7D%3DAx%28t%29+f%28t%29%5C%5C%20x%28t_0%29%3Dc%20%5Cend%7Bmatrix%7D%5Cright.)
可求得:![\small e^{At}=\begin{bmatrix} e^{2t}-2te^t & 0&te^t \\ -e^{2t}+e^t+2te^t& e^{2t}&e^{2t}-e^t-te^t \\ -4te^t &0 & 2te^t+e^t \end{bmatrix}](https://latex.csdn.net/eq?%5Cdpi%7B120%7D%20%5Csmall%20e%5E%7BAt%7D%3D%5Cbegin%7Bbmatrix%7D%20e%5E%7B2t%7D-2te%5Et%20%26%200%26te%5Et%20%5C%5C%20-e%5E%7B2t%7D+e%5Et+2te%5Et%26%20e%5E%7B2t%7D%26e%5E%7B2t%7D-e%5Et-te%5Et%20%5C%5C%20-4te%5Et%20%260%20%26%202te%5Et+e%5Et%20%5Cend%7Bbmatrix%7D)
因此微分方程的解为:![\small x(t)=e^{At}c+e^{At}\int_{0}^{t}e^{-A\tau }f(\tau )d\tau](https://latex.csdn.net/eq?%5Cdpi%7B120%7D%20%5Csmall%20x%28t%29%3De%5E%7BAt%7Dc+e%5E%7BAt%7D%5Cint_%7B0%7D%5E%7Bt%7De%5E%7B-A%5Ctau%20%7Df%28%5Ctau%20%29d%5Ctau)
,![\small \int_{0}^{t}e^{-A\tau }f(\tau)d\tau=\int_{0}^{t}\begin{bmatrix} e^{-\tau}\\ -e^{-\tau}\\ 2e^{-\tau} \end{bmatrix}d\tau=\begin{bmatrix} 1-e^{-t}\\ -1+e^{-t}\\ 2(1-e^{-t}) \end{bmatrix}](https://latex.csdn.net/eq?%5Cdpi%7B120%7D%20%5Csmall%20%5Cint_%7B0%7D%5E%7Bt%7De%5E%7B-A%5Ctau%20%7Df%28%5Ctau%29d%5Ctau%3D%5Cint_%7B0%7D%5E%7Bt%7D%5Cbegin%7Bbmatrix%7D%20e%5E%7B-%5Ctau%7D%5C%5C%20-e%5E%7B-%5Ctau%7D%5C%5C%202e%5E%7B-%5Ctau%7D%20%5Cend%7Bbmatrix%7Dd%5Ctau%3D%5Cbegin%7Bbmatrix%7D%201-e%5E%7B-t%7D%5C%5C%20-1+e%5E%7B-t%7D%5C%5C%202%281-e%5E%7B-t%7D%29%20%5Cend%7Bbmatrix%7D)
![\small e^{At}\int_{0}^{t}e^{-A\tau }f(\tau )d\tau=\begin{bmatrix} e^t-1\\ -e^t+1\\ 2e^t-2 \end{bmatrix}](https://latex.csdn.net/eq?%5Cdpi%7B120%7D%20%5Csmall%20e%5E%7BAt%7D%5Cint_%7B0%7D%5E%7Bt%7De%5E%7B-A%5Ctau%20%7Df%28%5Ctau%20%29d%5Ctau%3D%5Cbegin%7Bbmatrix%7D%20e%5Et-1%5C%5C%20-e%5Et+1%5C%5C%202e%5Et-2%20%5Cend%7Bbmatrix%7D)
故微分方程组的解为:![\small x(t)=\begin{bmatrix} x_1(t)\\ x_2(t)\\ x_3(t) \end{bmatrix}=\begin{bmatrix} e^t-te^t\\ te^t\\ e^t-2te^t \end{bmatrix}+\begin{bmatrix} e^t-1\\ -e^t+1\\ 2e^t-2 \end{bmatrix}=\begin{bmatrix} (2-t)e^t-1\\ (t-1)e^t+1\\ (3-2t)e^t-2 \end{bmatrix}](https://latex.csdn.net/eq?%5Cdpi%7B120%7D%20%5Csmall%20x%28t%29%3D%5Cbegin%7Bbmatrix%7D%20x_1%28t%29%5C%5C%20x_2%28t%29%5C%5C%20x_3%28t%29%20%5Cend%7Bbmatrix%7D%3D%5Cbegin%7Bbmatrix%7D%20e%5Et-te%5Et%5C%5C%20te%5Et%5C%5C%20e%5Et-2te%5Et%20%5Cend%7Bbmatrix%7D+%5Cbegin%7Bbmatrix%7D%20e%5Et-1%5C%5C%20-e%5Et+1%5C%5C%202e%5Et-2%20%5Cend%7Bbmatrix%7D%3D%5Cbegin%7Bbmatrix%7D%20%282-t%29e%5Et-1%5C%5C%20%28t-1%29e%5Et+1%5C%5C%20%283-2t%29e%5Et-2%20%5Cend%7Bbmatrix%7D)
5.2,Lyapunov矩阵方程AX+BX=F
在控制论和系统理论中,会遇到形如: 的矩阵方程,这个方程也称为 Lyapunov方程,关于这个方程的解有如下结果:
给定矩阵方程:
![\small AX+XB=F](https://latex.csdn.net/eq?%5Cdpi%7B120%7D%20%5Csmall%20AX+XB%3DF)
其中 ,如果 和 的所有特征值都具有负实部,则该方程组有唯一解:![\small X=-\int_{0}^{+\infty }e^{At}Fe^{Bt}dt](https://latex.csdn.net/eq?%5Cdpi%7B120%7D%20%5Csmall%20X%3D-%5Cint_%7B0%7D%5E%7B+%5Cinfty%20%7De%5E%7BAt%7DFe%5E%7BBt%7Ddt)
设 ,则微分方程:
![\small \left\{\begin{matrix} \frac{dX(t)}{dt}=AX(t)+X(t)B\\ X(0)=F \end{matrix}\right.](https://latex.csdn.net/eq?%5Cdpi%7B120%7D%20%5Csmall%20%5Cleft%5C%7B%5Cbegin%7Bmatrix%7D%20%5Cfrac%7BdX%28t%29%7D%7Bdt%7D%3DAX%28t%29+X%28t%29B%5C%5C%20X%280%29%3DF%20%5Cend%7Bmatrix%7D%5Cright.)
的解为:![\small X=e^{At}Fe^{Bt}](https://latex.csdn.net/eq?%5Cdpi%7B120%7D%20%5Csmall%20X%3De%5E%7BAt%7DFe%5E%7BBt%7D)
设 ,且 的所有特征值具有负实部,则矩阵方程:
![\small AX(t)+X(t)A^H=-F](https://latex.csdn.net/eq?%5Cdpi%7B120%7D%20%5Csmall%20AX%28t%29+X%28t%29A%5EH%3D-F)
的唯一解为:![\small X=\int_{0}^{+\infty }e^{At}Fe^{A^Ht}dt](https://latex.csdn.net/eq?%5Cdpi%7B120%7D%20%5Csmall%20X%3D%5Cint_%7B0%7D%5E%7B+%5Cinfty%20%7De%5E%7BAt%7DFe%5E%7BA%5EHt%7Ddt)
如果 是 正定矩阵,则 也是 正定矩阵。
5.3,最小二乘问题
在矩阵论实际应用中,最小二乘法是一种很重要的方法。因为实验有误差,数据有偏差,如何在纷乱复杂的环境中找到最优选择,最小二乘法就是解决该问题的一种重要方法。
【例12】现有 组数据: ,试将它们拟合为一条直线。
设这条直线为: ,由于各种实际原因使:
![y_i\ne kx_i+b(\exists i\in\left \{ 1,2,...,n \right \})](https://latex.csdn.net/eq?y_i%5Cne%20kx_i+b%28%5Cexists%20i%5Cin%5Cleft%20%5C%7B%201%2C2%2C...%2Cn%20%5Cright%20%5C%7D%29)
则可构造误差
![\varepsilon _i=y_i-(kx_i+b)(i=1,2,...,n)](https://latex.csdn.net/eq?%5Cvarepsilon%20_i%3Dy_i-%28kx_i+b%29%28i%3D1%2C2%2C...%2Cn%29)
为了不使得误差正负抵消,造成小的假象,进一步采取 平方和作为目标函数,即:
![\varepsilon =\sum_{i=1}^n\varepsilon _i^2=\sum_{i=1}^n[y_i-(kx_i+b)]^2](https://latex.csdn.net/eq?%5Cvarepsilon%20%3D%5Csum_%7Bi%3D1%7D%5En%5Cvarepsilon%20_i%5E2%3D%5Csum_%7Bi%3D1%7D%5En%5By_i-%28kx_i+b%29%5D%5E2)
希望求出最佳逼近对应的 和 使得目标函数为 ,也即最小二乘解。
令 ![\left\{\begin{matrix} \frac{\partial \varepsilon }{\partial k}=-2\sum_{i=1}^nx_i(y_i-kx_i-b)=0\\ \frac{\partial \varepsilon }{\partial b}=-2\sum_{i=1}^n(y_i-kx_i-b)=0 \end{matrix}\right.](https://latex.csdn.net/eq?%5Cleft%5C%7B%5Cbegin%7Bmatrix%7D%20%5Cfrac%7B%5Cpartial%20%5Cvarepsilon%20%7D%7B%5Cpartial%20k%7D%3D-2%5Csum_%7Bi%3D1%7D%5Enx_i%28y_i-kx_i-b%29%3D0%5C%5C%20%5Cfrac%7B%5Cpartial%20%5Cvarepsilon%20%7D%7B%5Cpartial%20b%7D%3D-2%5Csum_%7Bi%3D1%7D%5En%28y_i-kx_i-b%29%3D0%20%5Cend%7Bmatrix%7D%5Cright.)
也就是 和 的二元一次方程:
![\left\{\begin{matrix} (\sum_{i=1}^nx_i^2)k+(\sum_{i=1}^nx_i)b=\sum_{i=1}^nx_iy_i\\ (\sum_{i=1}^nx_i)k+nb=\sum_{i=1}^ny_i \end{matrix}\right.](https://latex.csdn.net/eq?%5Cleft%5C%7B%5Cbegin%7Bmatrix%7D%20%28%5Csum_%7Bi%3D1%7D%5Enx_i%5E2%29k+%28%5Csum_%7Bi%3D1%7D%5Enx_i%29b%3D%5Csum_%7Bi%3D1%7D%5Enx_iy_i%5C%5C%20%28%5Csum_%7Bi%3D1%7D%5Enx_i%29k+nb%3D%5Csum_%7Bi%3D1%7D%5Eny_i%20%5Cend%7Bmatrix%7D%5Cright.)
令 ![D=\begin{vmatrix} \sum_{i=1}^nx_i^2 &\sum_{i=1}^nx_i \\ \sum_{i=1}^nx_i & n \end{vmatrix},D_k=\begin{vmatrix} \sum_{i=1}^nx_iy_i &\sum_{i=1}^nx_i \\ \sum_{i=1}^ny_i & n \end{vmatrix},D_b=\begin{vmatrix} \sum_{i=1}^nx_i^2 &\sum_{i=1}^nx_i y_i\\ \sum_{i=1}^nx_i & \sum_{i=1}^ny_i \end{vmatrix}](https://latex.csdn.net/eq?D%3D%5Cbegin%7Bvmatrix%7D%20%5Csum_%7Bi%3D1%7D%5Enx_i%5E2%20%26%5Csum_%7Bi%3D1%7D%5Enx_i%20%5C%5C%20%5Csum_%7Bi%3D1%7D%5Enx_i%20%26%20n%20%5Cend%7Bvmatrix%7D%2CD_k%3D%5Cbegin%7Bvmatrix%7D%20%5Csum_%7Bi%3D1%7D%5Enx_iy_i%20%26%5Csum_%7Bi%3D1%7D%5Enx_i%20%5C%5C%20%5Csum_%7Bi%3D1%7D%5Eny_i%20%26%20n%20%5Cend%7Bvmatrix%7D%2CD_b%3D%5Cbegin%7Bvmatrix%7D%20%5Csum_%7Bi%3D1%7D%5Enx_i%5E2%20%26%5Csum_%7Bi%3D1%7D%5Enx_i%20y_i%5C%5C%20%5Csum_%7Bi%3D1%7D%5Enx_i%20%26%20%5Csum_%7Bi%3D1%7D%5Eny_i%20%5Cend%7Bvmatrix%7D)
最小二乘解为 ![k=\frac{D_k}{D},b=\frac{D_b}{D}](https://latex.csdn.net/eq?k%3D%5Cfrac%7BD_k%7D%7BD%7D%2Cb%3D%5Cfrac%7BD_b%7D%7BD%7D)
设 ,当线性方程组 无解时,则对任意 都有 。
此时希望找到这样的向量 ,它使得 取到 的最小值。
即: 。
称这个问题为最小二乘问题,称 为矛盾方程组 的最小二乘解。
设 ,若 是 的最小二乘解,则 是方程组
![A^TAx=A^Tb](https://latex.csdn.net/eq?A%5ETAx%3DA%5ETb)
的解,称上式为 的法方程组。
证明:由于
![f(x)=||Ax-b||_2^2=(Ax-b)^T(Ax-b)=x^TA^TAx-x^TA^TAx+b^Tb](https://latex.csdn.net/eq?f%28x%29%3D%7C%7CAx-b%7C%7C_2%5E2%3D%28Ax-b%29%5ET%28Ax-b%29%3Dx%5ETA%5ETAx-x%5ETA%5ETAx+b%5ETb)
若 为 的最小二乘解,则它应是 极小值点,从而
![0=\frac{df}{dx}|_{x_0}=2A^TAx_0-2A^Tb](https://latex.csdn.net/eq?0%3D%5Cfrac%7Bdf%7D%7Bdx%7D%7C_%7Bx_0%7D%3D2A%5ETAx_0-2A%5ETb)
推论1:若 和 都是矛盾方程 最小二乘解,则 或 。
推论2:若 非奇异,则矛盾方程 的最小二乘解为: ,且解唯一。
【例13】设 ,且 有解。
试求约束最小化问题: 的解满足的代数方程,
也就是求 在约束条件 的最小二乘解所满足的代数方程。
若 为 的极值点,则应有:
![\left\{\begin{matrix} \frac{\partial f}{\partial x}|_{(x_0,u_0)}=2A^TAx_0-2A^Tb+2B^Tu_0=0\\ \frac{\partial f}{\partial u}|_{(x_0,u_0)}=2(Bx_0-d)=0 \end{matrix}\right.](https://latex.csdn.net/eq?%5Cleft%5C%7B%5Cbegin%7Bmatrix%7D%20%5Cfrac%7B%5Cpartial%20f%7D%7B%5Cpartial%20x%7D%7C_%7B%28x_0%2Cu_0%29%7D%3D2A%5ETAx_0-2A%5ETb+2B%5ETu_0%3D0%5C%5C%20%5Cfrac%7B%5Cpartial%20f%7D%7B%5Cpartial%20u%7D%7C_%7B%28x_0%2Cu_0%29%7D%3D2%28Bx_0-d%29%3D0%20%5Cend%7Bmatrix%7D%5Cright.)
这说明极值点 应满足方程:
![\begin{bmatrix} A^TA &B^T \\ B & 0 \end{bmatrix}\begin{bmatrix} x\\ u \end{bmatrix}=\begin{bmatrix} A^Tb\\ d \end{bmatrix}](https://latex.csdn.net/eq?%5Cbegin%7Bbmatrix%7D%20A%5ETA%20%26B%5ET%20%5C%5C%20B%20%26%200%20%5Cend%7Bbmatrix%7D%5Cbegin%7Bbmatrix%7D%20x%5C%5C%20u%20%5Cend%7Bbmatrix%7D%3D%5Cbegin%7Bbmatrix%7D%20A%5ETb%5C%5C%20d%20%5Cend%7Bbmatrix%7D)
求解该方程组即可得到 在约束条件 下的最小二乘解 。
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