矩阵A可以LU分解的充要条件

矩阵A可以LU分解的充要条件是
∀ k ∈ { 1 , ⋯   , n } , r a n k A 1 : k , 1 : k + k ⩾ r a n k A 1 : k , 1 : n + r a n k A 1 : n , 1 : k \forall k \in \{1,\cdots,n\}, \mathrm{rank} A_{1:k,1:k} + k \geqslant \mathrm{rank} A_{1:k,1:n} + \mathrm{rank} A_{1:n,1:k} k{1,,n},rankA1:k,1:k+krankA1:k,1:n+rankA1:n,1:k.

如果矩阵A可以LU分解, 分解的唯一性需要通过下式分析
L y = f Ly = f Ly=f, U x = y Ux = y Ux=y.

下面给出一个分解算法.

  1. INPUT A ∈ M n × n ( F ) A \in \mathcal{M}_{n \times n}(\mathcal{F}) AMn×n(F)
  2. FOR k = 1 , ⋯   , k k = 1, \cdots, k k=1,,k DO
    1. IF A = 0 A = 0 A=0 THEN
      1. L : , k ← 0 L_{:,k} \gets 0 L:,k0
      2. U k , : ← 0 U_{k,:} \gets 0 Uk,:0
    2. ELSE
      1. FIND i , j , i,j, i,j, WHICH MINIMIZE min ⁡ { i , j } \min\{i,j\} min{i,j} SUBJECT TO a i j ≠ 0 a_{ij} \neq 0 aij=0
      2. L : , k ← a i j − 1 A : , j L_{:,k} \gets a_{ij}^{-1} A_{:,j} L:,kaij1A:,j
      3. U k , : ← A i , : U_{k,:} \gets A_{i,:} Uk,:Ai,:
      4. A ← A − L : , k U k , : A \gets A - L_{:,k} U_{k,:} AAL:,kUk,:
    3. END IF
  3. END FOR
  4. OUTPUT L L L, U U U,

请添加图片描述

参考文献
Pavel Okunev, Necessary And Sufficient Conditions For Existence of the LU Factorization of an Arbitrary Matrix.

书上的定理7.3是
矩阵A可以LU分解的充分条件是
A A A的顺序主子式 D i ≠ 0 D_i \neq 0 Di=0( i = 1 , 2 , ⋯   , n − 1 i = 1,2, \cdots, n-1 i=1,2,,n1)
此时分解是唯一的.