以下函数不使用行旋转进行LU分解.R中是否存在使用行枢轴进行 LU分解的现有函数?
> require(Matrix)
> expand(lu(matrix(rnorm(16),4,4)))
$L
4 x 4 Matrix of class "dtrMatrix"
[,1] [,2] [,3] [,4]
[1,] 1.00000000 . . .
[2,] 0.13812836 1.00000000 . .
[3,] 0.27704442 0.39877260 1.00000000 .
[4,] -0.08512341 -0.24699820 0.04347201 1.00000000
$U
4 x 4 Matrix of class "dtrMatrix"
[,1] [,2] [,3] [,4]
[1,] 1.5759031 -0.2074224 -1.5334082 -0.5959756
[2,] . -1.3096874 -0.6301727 1.1953838
[3,] . . 1.6316292 0.6256619
[4,] . . . 0.8078140
$P
4 x 4 sparse Matrix of class "pMatrix"
[1,] | . . .
[2,] . | . .
[3,] . . . |
[4,] . . | .
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luR中的功能使用部分(行)枢轴。您没有在示例中提供原始矩阵,因此我将创建一个新示例进行演示。
luR中的函数是计算A = PLU,这等效于计算矩阵A的LU分解,矩阵A的行由置换矩阵P -1:P -1 A = LU置换。有关更多信息,请参见Matrix软件包文档。
> A <- matrix(c(1, 1, 1, 1, 1, 1, -1, -1, 1, -1, -1, 1, 1, -1, 1, -1), 4)
> A
[,1] [,2] [,3] [,4]
[1,] 1 1 1 1
[2,] 1 1 -1 -1
[3,] 1 -1 -1 1
[4,] 1 -1 1 -1
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这是L因素:
> luDec <- lu(A)
> L <- expand(luDec)$L
> L
4 x 4 Matrix of class "dtrMatrix" (unitriangular)
[,1] [,2] [,3] [,4]
[1,] 1 . . .
[2,] 1 1 . .
[3,] 1 0 1 .
[4,] 1 1 -1 1
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这是U因素:
> U <- expand(luDec)$U
> U
4 x 4 Matrix of class "dtrMatrix"
[,1] [,2] [,3] [,4]
[1,] 1 1 1 1
[2,] . -2 -2 0
[3,] . . -2 -2
[4,] . . . -4
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这是置换矩阵:
> P <- expand(luDec)$P
> P
4 x 4 sparse Matrix of class "pMatrix"
[1,] | . . .
[2,] . . | .
[3,] . | . .
[4,] . . . |
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我们可以看到这LU是行的排列版本A:
> L %*% U
4 x 4 Matrix of class "dgeMatrix"
[,1] [,2] [,3] [,4]
[1,] 1 1 1 1
[2,] 1 -1 -1 1
[3,] 1 1 -1 -1
[4,] 1 -1 1 -1
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回到原始身份A = PLU,我们可以恢复A(与A上面比较):
> P %*% L %*% U
4 x 4 Matrix of class "dgeMatrix"
[,1] [,2] [,3] [,4]
[1,] 1 1 1 1
[2,] 1 1 -1 -1
[3,] 1 -1 -1 1
[4,] 1 -1 1 -1
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