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矩阵代数(四)- 分块矩阵

矩阵代数(四)- 分块矩阵

作者: mHubery | 来源:发表于2019-03-06 21:01 被阅读0次

    小结

    1. 分块矩阵
    2. 分块矩阵运算
    3. 分块矩阵的逆

    分块矩阵

    矩阵\boldsymbol{A} = \left[\begin{array}{ccc|cc|c} 3 & 0 & -1 & 5 & 9 & -2 \\ -5 & 2 & 4 & 0 & -3 & 1 \\ \hline -8 & -6 & 3 & 1 & 7 & -4 \end{array}\right],也可写成2 \times 3分块矩阵\boldsymbol{A}=\begin{bmatrix}\boldsymbol{A_1\!_1} & \boldsymbol{A_1\!_2} & \boldsymbol{A_1\!_3} \\ \boldsymbol{A_2\!_1} & \boldsymbol{A_2\!_2} & \boldsymbol{A_2\!_3}\end{bmatrix}的形状,它的元素是分块(子矩阵)
    \begin{aligned} &\boldsymbol{A_1\!_1} =\begin{bmatrix} 3 & 0 & -1 \\ -5 & 2 & 4\end{bmatrix} &&\boldsymbol{A_1\!_2} =\begin{bmatrix} 5 & 9 \\ 0 & -3\end{bmatrix} &&\boldsymbol{A_1\!_3} =\begin{bmatrix} -2 \\ 1 \end{bmatrix} \\ &\boldsymbol{A_1\!_2} =\begin{bmatrix} 5 & 9 \\ 0 & -3\end{bmatrix} &&\boldsymbol{A_2\!_2} =\begin{bmatrix}1 \\ 7 \end{bmatrix} &&\boldsymbol{A_2\!_3} =\begin{bmatrix} -4 \end{bmatrix}\end{aligned}

    加法与标量乘法

    若矩阵\boldsymbol{A}\boldsymbol{B}有相同维数且以相同方式分块,则自然有矩阵的和\boldsymbol{A} + \boldsymbol{B}也以同样方式分块。这时\boldsymbol{A} + \boldsymbol{B}的每一个分块恰好是\boldsymbol{A}\boldsymbol{B}对应分块的(矩阵)和。分块矩阵乘以一个标量也可以逐块计算。

    分块矩阵的乘法

    \boldsymbol{A} = \left[\begin{array}{ccc|cc} 2 & -3 & 1 & 0 & -4 \\ 1 & 5 & -2 & 3 & 1 \\ \hline 0 & -4 & -2 & 7 & -1 \end{array}\right]=\begin{bmatrix}\boldsymbol{A_1\!_1} & \boldsymbol{A_1\!_2} \\ \boldsymbol{A_2\!_1} & \boldsymbol{A_2\!_2}\end{bmatrix}\boldsymbol{B} =\begin{bmatrix}6 & 4 \\ -2 & 1\\ -3 & 7\\ \hline -1 & 3 \\ 5 & 2\end{bmatrix}=\begin{bmatrix}\boldsymbol{B_1} & \boldsymbol{B_2}\end{bmatrix}
    \boldsymbol{A}的5列被分成3列一组和2列一组。\boldsymbol{B}的5行按同样方法分块---被分成3行一组和2行一组。我们称\boldsymbol{A}\boldsymbol{B}的分块是与分块乘法相一致的。\boldsymbol{AB}的乘积可以被写成\boldsymbol{AB}=\begin{bmatrix}\boldsymbol{A_1\!_1} & \boldsymbol{A_1\!_2} \\ \boldsymbol{A_2\!_1} & \boldsymbol{A_2\!_2}\end{bmatrix}\begin{bmatrix}\boldsymbol{B_1} &\boldsymbol{B_2}\end{bmatrix}=\begin{bmatrix}\boldsymbol{A_1\!_1}\boldsymbol{B_1} + \boldsymbol{A_1\!_2}\boldsymbol{B_2} \\ \boldsymbol{A_2\!_1}\boldsymbol{B_1} + \boldsymbol{A_2\!_2}\boldsymbol{B_2}\end{bmatrix}
    \boldsymbol{A_1\!_1}\boldsymbol{B_1}=\begin{bmatrix} 2 & -3 & 1 \\ 1 & 5 & -2\end{bmatrix}\begin{bmatrix}6 & 4 \\ -2 & 1\\ -3 & 7\end{bmatrix}=\begin{bmatrix}15 & 12 \\ 2 & -5\end{bmatrix}
    \boldsymbol{A_1\!_2}\boldsymbol{B_2}=\begin{bmatrix} 0 & -4 \\ 3 & -1\end{bmatrix}\begin{bmatrix}-1 & 3 \\ 5 & 2\end{bmatrix}=\begin{bmatrix}-20 & -8 \\ -8 & 7\end{bmatrix}
    \boldsymbol{A_1\!_1}\boldsymbol{B_1} + \boldsymbol{A_1\!_2}\boldsymbol{B_2}=\begin{bmatrix}15 & 12 \\ 2 & -5\end{bmatrix} + \begin{bmatrix}-20 & -8 \\ -8 & 7\end{bmatrix}=\begin{bmatrix}-5 & 4 \\ -6 & 2\end{bmatrix}

    \boldsymbol{A}=\begin{bmatrix}-3 & 1 & 2 \\ 1 & -4 & 5\end{bmatrix}\boldsymbol{B}=\begin{bmatrix}a & b \\ c & d\\ e & f\end{bmatrix}。验证\boldsymbol{AB}=col_1(\boldsymbol{A})row_1(\boldsymbol{B}) + col_2(\boldsymbol{A})row_2(\boldsymbol{B}) + col_3(\boldsymbol{A})row_3(\boldsymbol{B})
    \begin{aligned} &col_1(\boldsymbol{A})row_1(\boldsymbol{B})=\begin{bmatrix}-3 \\ 1\end{bmatrix}\begin{bmatrix}a & b\end{bmatrix}=\begin{bmatrix}-3a & -3b \\ a & b\end{bmatrix} \\ &col_2(\boldsymbol{A})row_2(\boldsymbol{B})=\begin{bmatrix}1 \\ -4\end{bmatrix}\begin{bmatrix}c & d\end{bmatrix}=\begin{bmatrix}c & d \\ -4c & -4d\end{bmatrix} \\ &col_3(\boldsymbol{A})row_3(\boldsymbol{B})=\begin{bmatrix}2 \\ 5\end{bmatrix}\begin{bmatrix}e & f\end{bmatrix}=\begin{bmatrix}2e & 2f \\ 5e & 5f\end{bmatrix} \end{aligned}
    于是\sum_{k=1}^3{col_k(\boldsymbol{A})row_k(\boldsymbol{B})}=\begin{bmatrix}-3a + c + 2e & -3b + d + 2f \\ a - 4c + 5e & b - 4d + 5f\end{bmatrix}
    这个矩阵恰好就是\boldsymbol{AB}

    定理10\boldsymbol{AB}的列行展开)
    \boldsymbol{A}m \times n矩阵,\boldsymbol{B}n \times p矩阵,则\begin{aligned} \boldsymbol{AB}&=\begin{bmatrix}col_1(\boldsymbol{A}) & \cdots & col_n(\boldsymbol{A})\end{bmatrix} \begin{bmatrix}row_1(\boldsymbol{B}) \\ \vdots \\ row_n(\boldsymbol{B})\end{bmatrix} \\ &=col_1(\boldsymbol{A})row_1(\boldsymbol{B}) + \cdots + col_n(\boldsymbol{A})row_n(\boldsymbol{B})\end{aligned}

    分块矩阵的逆

    形如\boldsymbol{A}=\begin{bmatrix} \boldsymbol{A_1\!_1} &\boldsymbol{A_1\!_2} \\ \boldsymbol{0} & \boldsymbol{A_2\!_2}\end{bmatrix}的矩阵称为分块上三角矩阵。设\boldsymbol{A_1\!_1}p \times p矩阵,\boldsymbol{A_2\!_2}q \times q矩阵。求\boldsymbol{A}^{-1}的表达式。
    解:用\boldsymbol{B}表示\boldsymbol{A}^{-1}且把它分块,使得\begin{bmatrix}\boldsymbol{A_1\!_1} &\boldsymbol{A_1\!_2} \\ \boldsymbol{0} & \boldsymbol{A_2\!_2}\end{bmatrix}\begin{bmatrix} \boldsymbol{B_1\!_1} &\boldsymbol{B_1\!_2} \\ \boldsymbol{B_2\!_1} & \boldsymbol{B_2\!_2}\end{bmatrix}=\begin{bmatrix}\boldsymbol{I_p}& \boldsymbol{0} \\ \boldsymbol{0} & \boldsymbol{I_q}\end{bmatrix}
    则有:\begin{aligned} \boldsymbol{A_1\!_1}\boldsymbol{B_1\!_1}+\boldsymbol{A_1\!_2}\boldsymbol{B_2\!_1} &= \boldsymbol{I_p}\\ \boldsymbol{A_1\!_1}\boldsymbol{B_1\!_2}+\boldsymbol{A_1\!_2}\boldsymbol{B_2\!_2} &= \boldsymbol{0}\\ \boldsymbol{A_2\!_2}\boldsymbol{B_2\!_1} &= \boldsymbol{0}\\ \boldsymbol{A_2\!_2}\boldsymbol{B_2\!_2} &= \boldsymbol{I_q} \end{aligned}
    \boldsymbol{A}可逆,且\boldsymbol{A_2\!_2}所在行其余列全为\boldsymbol{0},可知\boldsymbol{A_2\!_2}必有q个主元位置,故\boldsymbol{B_2\!_2} = \boldsymbol{A_2\!_2}^{-1}
    \boldsymbol{B_2\!_1}=\boldsymbol{A_2\!_2}^{-1}\boldsymbol{0}=\boldsymbol{0}
    于是\boldsymbol{A_1\!_1}\boldsymbol{B_1\!_1}+\boldsymbol{A_1\!_2}\boldsymbol{B_2\!_1}=\boldsymbol{A_1\!_1}\boldsymbol{B_1\!_1}=\boldsymbol{I_p}。同理\boldsymbol{B_1\!_1}=\boldsymbol{A_1\!_1}^{-1}.
    最后\boldsymbol{A_1\!_1}\boldsymbol{B_1\!_2}=-\boldsymbol{A_1\!_2}\boldsymbol{B_2\!_2}=-\boldsymbol{A_1\!_2}\boldsymbol{A_2\!_2}^{-1}
    \boldsymbol{B_1\!_2}=-\boldsymbol{A_1\!_1}^{-1}\boldsymbol{A_1\!_2}\boldsymbol{A_2\!_2}^{-1}
    最终得:\boldsymbol{A}^{-1}=\boldsymbol{B}=\begin{bmatrix}\boldsymbol{A_1\!_1}^{-1} & -\boldsymbol{A_1\!_1}^{-1}\boldsymbol{A_1\!_2}\boldsymbol{A_2\!_2}^{-1} \\ \boldsymbol{0} & \boldsymbol{A_2\!_2}^{-1}\end{bmatrix}

    分块对角矩阵是一个分块矩阵,除了主对角上各分块外,其余全是零分块。这样的一个矩阵是可逆的当且仅当主对角线上各分块都是可逆的。

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