PA=LU

$PA=LU$

对矩阵$A$做$LU$分解（考虑行交换）
$A = \begin{bmatrix} a_{11} & a_{12} & a_{13} & a_{14} \\ a_{21} & a_{22} & a_{23} & a_{24} \\ a_{31} & a_{32} & a_{33} & a_{34} \\ a_{41} & a_{42} & a_{43} & a_{44} \end{bmatrix}$
第一步，将矩阵$A$的第一列元素$a_{11},a_{21},a_{31},a_{41}$中绝对值最大值所在行与第$1$行交换，即构造行交换矩阵$P_{1}$，用行交换矩阵$P_{1}$左乘矩阵$A$完成上述第一次行交换。如果$a_{41}$的绝对值最大，则将第$4$行和第$1$行交换，对应的行交换矩阵就是将$4$阶单位矩阵的第$1$行和第$4$行交换后得到的矩阵
$P = \left[ \begin{matrix} 0 & 0 & 0 & 1 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 1 & 0 & 0 & 0 \end{matrix} \right]$
用矩阵$P$左乘矩阵$A$相当于交换矩阵$A$的第$1$行和第$4$行
$\begin{aligned} PA &= \left[ \begin{matrix} 0 & 0 & 0 & 1 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 1 & 0 & 0 & 0 \end{matrix} \right] \begin{bmatrix} a_{11} & a_{12} & a_{13} & a_{14} \\ a_{21} & a_{22} & a_{23} & a_{24} \\ a_{31} & a_{32} & a_{33} & a_{34} \\ a_{41} & a_{42} & a_{43} & a_{44} \end{bmatrix}\\ &= \begin{bmatrix} a_{41} & a_{42} & a_{43} & a_{44} \\ a_{21} & a_{22} & a_{23} & a_{24} \\ a_{31} & a_{32} & a_{33} & a_{34} \\ a_{11} & a_{12} & a_{13} & a_{14} \end{bmatrix} \end{aligned}$
一般的，将第一次行交换后的矩阵记为
$P_{1}A = \begin{bmatrix} b_{11} & b_{12} & b_{13} & b_{14} \\ b_{21} & b_{22} & b_{23} & b_{24} \\ b_{31} & b_{32} & b_{33} & b_{34} \\ b_{41} & b_{42} & b_{43} & b_{44} \end{bmatrix}$
第二步，构造矩阵$M_{1}$
$M_{1} = \left[ \begin{matrix} 1 & 0 & 0 & 0 \\ -l_{21} & 1 & 0 & 0 \\ -l_{31} & 0 & 1 & 0 \\ -l_{41} & 0 & 0 & 1 \end{matrix} \right]$
其中$\displaystyle l_{i1} =\frac{b_{i1}}{b_{11}}, i = 2, 3, 4.$ 用矩阵$M_{1}$左乘$(P_{1}A)$
$M_{1}P_{1}A = \left[ \begin{matrix} b_{11} & b_{12} & b_{13} & b_{14} \\ 0 & b'_{22} & b'_{23} & b'_{24} \\ 0 & b'_{32} & b'_{33} & b'_{34} \\ 0 & b'_{42} & b'_{43} & b'_{44} \end{matrix} \right]$
第三步，将矩阵$(M_{1}P_{1}A)$的第二列主对角线以下的元素$b'_{22},b'_{32},b'_{42}$中绝对值最大值所在行与第$2$行交换，即构造行交换矩阵$P_{2}$，用行交换矩阵$P_{2}$左乘矩阵$(M_{1}P_{1}A)$完成第二次行交换。记第二次行交换后的矩阵为
$P_{2}M_{1}P_{1}A = \begin{bmatrix} b_{11} & b_{12} & b_{13} & b_{14} \\ 0 & c_{22} & c_{23} & c_{24} \\ 0 & c_{32} & c_{33} & c_{34} \\ 0 & c_{42} & c_{43} & c_{44} \end{bmatrix}$
第四步，构造矩阵$M_{2}$
$M_{2} = \left[ \begin{matrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & -l_{32} & 1 & 0 \\ 0 & -l_{42} & 0 & 1 \end{matrix} \right]$
其中$\displaystyle l_{i2} =\frac{c_{i2}}{c_{22}}, i = 3, 4.$ 用矩阵$M_{2}$左乘$(P_{2}M_{1}P_{1}A)$
$M_{2}P_{2}M_{1}P_{1}A = \left[ \begin{matrix} b_{11} & b_{12} & b_{13} & b_{14} \\ 0 & c_{22} & c_{23} & c_{24} \\ 0 & 0 & c'_{33} & c'_{34} \\ 0 & 0 & c'_{43} & c'_{44} \end{matrix} \right]$
第五步，将矩阵$(M_{2}P_{2}M_{1}P_{1}A)$的第三列主对角线以下的元素$c'_{33},c'_{43}$中绝对值最大值所在行与第$3$行交换，即构造行交换矩阵$P_{3}$，用行交换矩阵$P_{3}$左乘矩阵$(M_{2}P_{2}M_{1}P_{1}A)$完成第三次行交换。记第三次行交换后的矩阵为
$P_{3}M_{2}P_{2}M_{1}P_{1}A = \begin{bmatrix} b_{11} & b_{12} & b_{13} & b_{14} \\ 0 & c_{22} & c_{23} & c_{24} \\ 0 & 0 & d_{33} & d_{34} \\ 0 & 0 & d_{43} & d_{44} \end{bmatrix}$
第六步，构造矩阵$M_{3}$
$M_{3} = \left[ \begin{matrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & -l_{43} & 1 \end{matrix} \right]$
其中$\displaystyle l_{43} =\frac{d_{43}}{d_{33}}.$ 用矩阵$M_{3}$左乘矩阵$(P_{3}M_{2}P_{2}M_{1}P_{1}A)$
$\begin{aligned} U &= M_{3}P_{3}M_{2}P_{2}M_{1}P_{1}A \\ &= \left[ \begin{matrix} b_{11} & b_{12} & b_{13} & b_{14} \\ 0 & c_{22} & c_{23} & c_{24} \\ 0 & 0 & d_{33} & d_{34} \\ 0 & 0 & 0 & d'_{44} \end{matrix} \right] \end{aligned}$
用矩阵$(M_{3}P_{3}M_{2}P_{2}M_{1}P_{1})$的逆矩阵$(M_{3}P_{3}M_{2}P_{2}M_{1}P_{1})^{-1}$左乘上式
$\begin{aligned} A &= (M_{3}P_{3}M_{2}P_{2}M_{1}P_{1})^{-1}U \\ &= P_{1}^{-1}M_{1}^{-1}P_{2}^{-1}M_{2}^{-1}P_{3}^{-1}M_{3}^{-1}U \end{aligned}$
再用矩阵$(P_{3}P_{2}P_{1})$左乘上式
$\begin{aligned} & P_{3}P_{2}P_{1}A \\ =& P_{3}P_{2}P_{1}P_{1}^{-1}M_{1}^{-1}P_{2}^{-1}M_{2}^{-1}P_{3}^{-1}M_{3}^{-1}U \\ =&P_{3}P_{2}M_{1}^{-1}P_{2}^{-1}P_{3}^{-1}P_{3}M_{2}^{-1}P_{3}^{-1}M_{3}^{-1}U \\ =&(P_{3}P_{2}M_{1}^{-1}P_{2}^{-1}P_{3}^{-1})(P_{3}M_{2}^{-1}P_{3}^{-1})(M_{3}^{-1})U \\ =&L_{1}L_{2}L_{3}U \end{aligned}$
其中
$M_{1}^{-1} = \left[ \begin{matrix} 1 & 0 & 0 & 0 \\ l_{21} & 1 & 0 & 0 \\ l_{31} & 0 & 1 & 0 \\ l_{41} & 0 & 0 & 1 \end{matrix} \right]$
$M_{2}^{-1} = \left[ \begin{matrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & l_{32} & 1 & 0 \\ 0 & l_{42} & 0 & 1 \end{matrix} \right]$
$M_{3}^{-1} = \left[ \begin{matrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & l_{43} & 1 \end{matrix} \right]$
$\begin{aligned} P_{1}^{-1} &=P_{1} \\ P_{2}^{-1} &=P_{2} \\ P_{3}^{-1} &=P_{3} \end{aligned}$
记
$\begin{aligned} L_{1} &=P_{3}P_{2}M_{1}^{-1}P_{2}^{-1}P_{3}^{-1} \\ L_{2}&=P_{3}M_{2}^{-1}P_{3}^{-1} \\ L_{3}&=M_{3}^{-1} \\ L &=L_{1}L_{2}L_{3} \\ P&=P_{3}P_{2}P_{1} \end{aligned}$
最后我们得到
$\boxed{PA=LU}$

来源：CSDN

作者：无所不能的Matlab

链接：https://blog.csdn.net/weixin_44969779/article/details/104117147