行分解的LU分解 [英] LU decomposition with row pivot

问题描述

以下函数不使用行透视进行LU分解. R中是否存在使用行透视分解LU的现有函数?

The following function does not use row pivoting for LU decomposition. Is there an existing function in R that does LU decomposition with row pivot?

> 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,] . . | .

推荐答案

R中的lu函数正在使用部分(行)旋转.您没有在示例中提供原始矩阵，所以我将创建一个新示例进行演示.

The lu function in R is using partial (row) pivoting. You did not give the original matrix with your example, so I will create a new example to demonstrate.

函数lu正在计算 A = PLU ，这等效于计算矩阵 A 的LU分解，其行由置换矩阵 P ^-1 : P ^-1 A = LU .有关更多信息，请参见 Matrix软件包文档. /p>

示例

Function lu in R is computing A = PLU, which is equivalent to computing the LU decomposition of matrix A with its rows permuted by the permutation matrix P^-1: P^-1A = LU. See the Matrix package documentation for more information.

> 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

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

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

这是置换矩阵:

> P <- expand(luDec)$P
> P
4 x 4 sparse Matrix of class "pMatrix"

[1,] | . . .
[2,] . . | .
[3,] . | . .
[4,] . . . |

我们可以看到LU是A的行排列版本:

We can see that LU is a row-permuted version of 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

回到原始身份 A = PLU ，我们可以恢复A(与上面的A比较):

Going back to the original identity A = PLU we can recover A (compare with A above):

> 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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