线性代数 : 矩阵的LU分解(可逆方阵)
矩阵的 L U LU LU分解
回顾矩阵的逆, A A A为可逆矩阵
A A − 1 = I AA^{-1} = I AA−1=I
A − 1 A = I A^{-1}A = I A−1A=I
可逆矩阵: A B AB AB乘积也是可逆矩阵
( A B ) ( A B ) − 1 = I (AB)(AB)^{-1} = I (AB)(AB)−1=I
A B B − 1 A − 1 = I ABB^{-1}A^{-1} = I ABB−1A−1=I
( A B ) − 1 = B − 1 A − 1 (AB)^{-1} = B^{-1}A^{-1} (AB)−1=B−1A−1
可逆矩阵:转置仍是可逆矩阵
A A − 1 = I AA^{-1} = I AA−1=I
( A A − 1 ) T = I T (AA^{-1})^T = I^T (AA−1)T=IT
( A − 1 ) T A T = I (A^{-1})^TA^T = I (A−1)TAT=I
可逆矩阵:逆的转置=转置的逆
( A − 1 ) T A T = I (A^{-1})^TA^T = I (A−1)TAT=I
( A − 1 ) T = ( A T ) − 1 (A^{-1})^T = (A^T)^{-1} (A−1)T=(AT)−1
A = L U A = LU A=LU
- L L L : 下三角矩阵
- U U U : 上三角矩阵
A → U : E A = U A \to U : EA = U A→U:EA=U
- E E E : 消元矩阵
L = ? L = ? L=?
- E − 1 E A = E − 1 U E^{-1}EA = E^{-1}U E−1EA=E−1U
- A = E − 1 U , L = E − 1 A = E^{-1}U, L = E^{-1} A=E−1U,L=E−1
A = L D U A = LDU A=LDU
- D D D : 主元矩阵
例1: 2 × 2 2 \times 2 2×2矩阵
A = [ 2 1 8 7 ] , E 2 , 1 A = U A = \left[\begin{matrix}2 & 1\\ 8 & 7\end{matrix}\right], E_{2,1}A = U A=[2817],E2,1A=U
[ 1 0 − 4 1 ] [ 2 1 8 7 ] = [ 2 1 0 3 ] \left[\begin{matrix}1 & 0 \\ -4 & 1\end{matrix}\right] \left[\begin{matrix}2 & 1 \\ 8 & 7\end{matrix}\right]= \left[\begin{matrix}2 & 1\\ 0 & 3\end{matrix}\right] [1−401][2817]=[2013]
E = [ 1 0 − 4 1 ] , U = [ 2 1 0 3 ] E = \left[\begin{matrix}1 & 0 \\ -4 & 1\end{matrix}\right], U = \left[\begin{matrix}2 & 1\\ 0 & 3\end{matrix}\right] E=[1−401],U=[2013]
L = E − 1 = [ 1 0 4 1 ] L = E^{-1} = \left[\begin{matrix}1 & 0 \\ 4 & 1\end{matrix}\right] L=E−1=[1401]
D = [ 2 0 0 3 ] , U = [ 1 1 2 0 1 ] , L = E − 1 = [ 1 0 4 1 ] D = \left[\begin{matrix}2 & 0\\ 0 & 3\end{matrix}\right], U = \left[\begin{matrix}1 & \frac{1}{2}\\ 0 & 1\end{matrix}\right], L = E^{-1} = \left[\begin{matrix}1 & 0 \\ 4 & 1\end{matrix}\right] D=[2003],U=[10211],L=E−1=[1401]
A = L D U A = LDU A=LDU
例2: 3 × 3 3 \times 3 3×3矩阵
E 3 , 2 E 3 , 1 E 2 , 1 A = U E_{3,2}E_{3,1}E_{2,1}A = U E3,2E3,1E2,1A=U
A = E 2 , 1 − 1 E 3 , 1 − 1 E 3 , 2 − 1 U A = E_{2,1}^{-1}E_{3,1}^{-1}E_{3,2}^{-1}U A=E2,1−1E3,1−1E3,2−1U
L = E 2 , 1 − 1 E 3 , 1 − 1 E 3 , 2 − 1 L = E_{2,1}^{-1}E_{3,1}^{-1}E_{3,2}^{-1} L=E2,1−1E3,1−1E3,2−1
E 2 , 1 = [ 1 0 0 − 2 1 0 0 0 1 ] , E 3 , 2 = [ 1 0 0 0 1 0 0 − 5 1 ] , E 3 , 1 = [ 1 0 0 0 1 0 0 0 1 ] E_{2,1} = \left[\begin{matrix}1 & 0 & 0\\ -2 & 1 & 0\\0 & 0 & 1\end{matrix}\right], E_{3,2} = \left[\begin{matrix}1 & 0 & 0\\ 0 & 1 & 0\\0 & -5& 1\end{matrix}\right], E_{3,1} = \left[\begin{matrix}1 & 0 & 0\\ 0 & 1 & 0\\0 & 0 & 1\end{matrix}\right] E2,1=⎣⎡1−20010001⎦⎤,E3,2=⎣⎡10001−5001⎦⎤,E3,1=⎣⎡100010001⎦⎤
L = ( E 3 , 2 E 3 , 1 E 2 , 1 ) − 1 = [ 1 0 0 2 1 0 0 5 1 ] L = (E_{3,2}E_{3,1}E_{2,1})^{-1} = \left[\begin{matrix}1 & 0 & 0\\ 2 & 1 & 0\\0 & 5& 1\end{matrix}\right] L=(E3,2E3,1E2,1)−1=⎣⎡120015001⎦⎤
置换矩阵 P : 3 × 3 P : 3 \times 3 P:3×3
P 0 = [ 1 0 0 0 1 0 0 0 1 ] , P 1 = [ 1 0 0 0 0 1 0 1 0 ] , P 2 = [ 0 1 0 1 0 0 0 0 1 ] , P 3 = [ 0 1 0 0 0 1 1 0 0 ] , P 4 = [ 0 0 1 1 0 0 0 1 0 ] , P 5 = [ 0 0 3 0 1 0 1 0 0 ] , P_0 = \left[\begin{matrix}1 & 0 & 0\\ 0 & 1 & 0\\0 & 0 & 1\end{matrix}\right], P_1 = \left[\begin{matrix}1 & 0 & 0\\ 0 & 0 & 1\\0 & 1 & 0\end{matrix}\right], P_2 = \left[\begin{matrix}0 & 1 & 0\\ 1 & 0 & 0\\0 & 0 & 1\end{matrix}\right], P_3 = \left[\begin{matrix}0 & 1 & 0\\ 0 & 0 & 1\\1 & 0 & 0\end{matrix}\right], P_4 = \left[\begin{matrix}0 & 0 & 1\\ 1 & 0 & 0\\0 & 1 & 0\end{matrix}\right], P_5 = \left[\begin{matrix}0 & 0 & 3\\ 0 & 1 & 0\\1 & 0 & 0\end{matrix}\right], P0=⎣⎡100010001⎦⎤,P1=⎣⎡100001010⎦⎤,P2=⎣⎡010100001⎦⎤,P3=⎣⎡001100010⎦⎤,P4=⎣⎡010001100⎦⎤,P5=⎣⎡001010300⎦⎤,
P 0 ∗ P 1 = P 1 P_0*P_1 = P_1 P0∗P1=P1
P 1 T = P 3 P_1^T = P_3 P1T=P3
P 2 T = P 4 P_2^T = P_4 P2T=P4
P 1 ∗ P 2 = P 3 P_1 * P_2 = P_3 P1∗P2=P3
P − 1 = P T P^{-1} = P^T P−1=PT
P 1 ∗ P 4 = I , P 1 T = P 4 = P 1 − 1 P_1 * P_4 = I, P_1^T = P_4 = P_1^{-1} P1∗P4=I,P1T=P4=P1−1