3.3. 矩阵函数

3.3.1. 概念

3.3.2. 求解方法

  • 方法1: 待定系数法;

  • 方法2: 数项级数求和法, 写成矩阵级数求和形式, 计算 \({\bm A}^k\) , 代入求解;

  • 方法3: 对角矩阵法, 根据 \({\bm A} = {\bm P}{\bm \Lambda}{\bm P}^{-1}\) , 有 \({\bm A}^k = {\bm P}{\bm \Lambda}^k{\bm P}^{-1}\) 可知 \(f({\bm A}) = {\bm P}f({\bm \Lambda}){\bm P}^{-1}\) , 其中 \(f({\bm \Lambda}) = (f(\Lambda_{ii})) = (f(\lambda_i))\) 据此可求;

  • 方法4: Jordan标准型法, 对于不能化为对角阵的矩阵, 采用此方法, 注意此时 \(f({\bm \Lambda}) {\neq} (f(\Lambda_{ii}))\)

待定系数法

数项级数求和法

对角矩阵法

根据 \({\bm A} = {\bm P}{\bm \Lambda}{\bm P}^{-1}\) , 有 \({\bm A}^k = {\bm P}{\bm \Lambda}^k{\bm P}^{-1}\) 可知 \(f({\bm A}) = {\bm P}f({\bm \Lambda}){\bm P}^{-1}\) , 其中 \(f({\bm \Lambda}) = (f(\Lambda_{ii})) = (f(\lambda_i))\)

Jordan标准型法

对于不能化为对角阵的, 采用Jordan标准形法来求

\[f({\bm{A}}) = \sum\limits_{k = 0}^\infty {{c_k}{{\bm{A}}^k}} = \sum\limits_{k = 0}^\infty {{c_k}{\bm{P}}{{\bm{J}}^k}{{\bm{P}}^{ - 1}}} = {\bm{P}}\left( {\sum\limits_{k = 0}^\infty {{c_k}{{\bm{J}}^k}} } \right){{\bm{P}}^{ - 1}} \]

即

\[f({\bm{A}}) = {\bm{P}}\left[ {\begin{array}{ccccc} {f({{\bm{J}}_1})}&{}&{}\\ {}& \ddots &{}\\ {}&{}&{f({{\bm{J}}_s})} \end{array}} \right]{{\bm{P}}^{ - 1}} \]

其中

\[f({{\bm{J}}_i}) = \left[ {\begin{array}{ccccc} {f({\lambda _i})}&{\frac{{{f^{(1)}}({\lambda _i})}}{{1!}}}&{\frac{{{f^{(2)}}({\lambda _i})}}{{2!}}}& \cdots &{\frac{{{f^{({m_i} - 1)}}({\lambda _i})}}{{({m_i} - 1)!}}}\\ {}&{f({\lambda _i})}&{\frac{{{f^{(1)}}({\lambda _i})}}{{1!}}}& \ddots & \vdots \\ {}&{}& \ddots & \ddots &{\frac{{{f^{(2)}}({\lambda _i})}}{{2!}}}\\ {}&{}&{}&{f({\lambda _i})}&{\frac{{{f^{(1)}}({\lambda _i})}}{{1!}}}\\ {}&{}&{}&{}&{f({\lambda _i})} \end{array}} \right] \]

其中, \(m_i\) 为Jordan块 \({\bm J}_i\) 的阶数, 也是特征值 \(\lambda_i\) 的重数.

提示

只需要求出第一行, 斜向右下顺着写即可.

3.3.3. 总结

  • \(e^z = 1 + \frac{z}{1!} + \frac{z^2}{2!} + \frac{z^3}{3!} + \cdots\)

  • \({\rm{cos}}z = 1 - \frac{z^2}{2!} + \frac{z^4}{4!} - \cdots\)

  • \({\rm{sin}}z = \frac{z}{1!} - \frac{z^3}{3!} + \frac{z^5}{5!} - \cdots\)

  • \(\frac{1}{1-z} = \sum_{n=0}^{\infty}z^n, |z|<1\)

  • \(e^{jz} = {\rm cos}z + j{\rm sin}z\)

  • \({\rm cos}z = \frac{e^{jz} + e^{-jz}}{2}\)

  • \({\rm sin}z = \frac{e^{jz} - e^{-jz}}{2}\)

  • \({\rm cos}(-z) = {\rm cos}z\)

  • \({\rm sin}(-z) = -{\rm sin}z\)

  • \(e^{\bm A} = 1 + \frac{{\bm A}}{1!} + \frac{{\bm A}^2}{2!} + \frac{{\bm A}^3}{3!} + \cdots\)

  • \({\rm{cos}}{\bm A} = 1 - \frac{{\bm A}^2}{2!} + \frac{{\bm A}^4}{4!} - \cdots\)

  • \({\rm{sin}}{\bm A} = \frac{{\bm A}}{1!} - \frac{{\bm A}^3}{3!} + \frac{{\bm A}^5}{5!} - \cdots\)

  • \({({\bm I} - {\bm A})^{-1}} = \sum_{n=0}^{\infty}{\bm A}^n\)

  • \(e^{j{\bm A}} = {\rm cos}{\bm A} + j{\rm sin}{\bm A}\)

  • \({\rm cos}{\bm A} = \frac{e^{j{\bm A}} + e^{-j{\bm A}}}{2}\)

  • \({\rm sin}{\bm A} = \frac{e^{j{\bm A}} - e^{-j{\bm A}}}{2}\)

  • \({\rm cos}(-{\bm A}) = {\rm cos}{\bm A}\)

  • \({\rm sin}(-{\bm A}) = -{\rm sin}{\bm A}\)

  • 当 \({\bm{AB}} {\neq} {\bm{BA}}\) 时, \(e^{\bm A}e^{\bm B} {\neq} e^{\bm B}e^{\bm A} {\neq} e^{\bm{AB}}\)

  • 当 \({\bm{AB}} = {\bm{BA}}\) 时, \(e^{\bm A}e^{\bm B} = e^{\bm B}e^{\bm A} = e^{\bm{AB}}\)