用 R 求解非平方线性系统 [英] Solving non-square linear system with R

问题描述

如何用 R 求解非平方线性系统:A X = B ?

How to solve a non-square linear system with R : A X = B ?

(在系统无解或有无穷多个解的情况下)

(in the case the system has no solution or infinitely many solutions)

示例:

A=matrix(c(0,1,-2,3,5,-3,1,-2,5,-2,-1,1),3,4,T)
B=matrix(c(-17,28,11),3,1,T)

A
     [,1] [,2] [,3] [,4]
[1,]    0    1   -2    3
[2,]    5   -3    1   -2
[3,]    5   -2   -1    1


B
     [,1]
[1,]  -17
[2,]   28
[3,]   11

推荐答案

如果矩阵 A 的行数多于列数，则应使用最小二乘法拟合.

If the matrix A has more rows than columns, then you should use least squares fit.

如果矩阵 A 的行数少于列数，则应执行奇异值分解.每个算法都尽其所能，通过使用假设为您提供解决方案.

If the matrix A has fewer rows than columns, then you should perform singular value decomposition. Each algorithm does the best it can to give you a solution by using assumptions.

这是一个显示如何使用 SVD 作为求解器的链接:

Here's a link that shows how to use SVD as a solver:

http://www.ecse.rpi.edu/~qji/简历/svd_review.pdf

让我们将其应用于您的问题，看看它是否有效:

Let's apply it to your problem and see if it works:

你的输入矩阵A和已知的RHS向量B:

Your input matrix A and known RHS vector B:

> A=matrix(c(0,1,-2,3,5,-3,1,-2,5,-2,-1,1),3,4,T)
> B=matrix(c(-17,28,11),3,1,T)
> A
     [,1] [,2] [,3] [,4]
[1,]    0    1   -2    3
[2,]    5   -3    1   -2
[3,]    5   -2   -1    1
> B
     [,1]
[1,]  -17
[2,]   28
[3,]   11

让我们分解你的 A 矩阵:

Let's decompose your A matrix:

> asvd = svd(A)
> asvd
$d
[1] 8.007081e+00 4.459446e+00 4.022656e-16

$u
           [,1]       [,2]       [,3]
[1,] -0.1295469 -0.8061540  0.5773503
[2,]  0.7629233  0.2908861  0.5773503
[3,]  0.6333764 -0.5152679 -0.5773503

$v
            [,1]       [,2]       [,3]
[1,]  0.87191556 -0.2515803 -0.1764323
[2,] -0.46022634 -0.1453716 -0.4694190
[3,]  0.04853711  0.5423235  0.6394484
[4,] -0.15999723 -0.7883272  0.5827720

> adiag = diag(1/asvd$d)
> adiag
          [,1]      [,2]        [,3]
[1,] 0.1248895 0.0000000 0.00000e+00
[2,] 0.0000000 0.2242431 0.00000e+00
[3,] 0.0000000 0.0000000 2.48592e+15

这里是关键:d 中的第三个特征值非常小；相反，adiag 中的对角线元素非常大.在求解之前，将其设置为零:

Here's the key: the third eigenvalue in d is very small; conversely, the diagonal element in adiag is very large. Before solving, set it equal to zero:

> adiag[3,3] = 0
> adiag
          [,1]      [,2] [,3]
[1,] 0.1248895 0.0000000    0
[2,] 0.0000000 0.2242431    0
[3,] 0.0000000 0.0000000    0

现在让我们计算解决方案(请参阅上面我提供给您的链接中的幻灯片 16):

Now let's compute the solution (see slide 16 in the link I gave you above):

> solution = asvd$v %*% adiag %*% t(asvd$u) %*% B
> solution
          [,1]
[1,]  2.411765
[2,] -2.282353
[3,]  2.152941
[4,] -3.470588

既然我们有了一个解决方案，让我们把它替换回去看看它是否给了我们相同的B:

Now that we have a solution, let's substitute it back to see if it gives us the same B:

> check = A %*% solution
> check
     [,1]
[1,]  -17
[2,]   28
[3,]   11

那是你开始的 B 方面，所以我认为我们很好.

That's the B side you started with, so I think we're good.

这是来自 AMS 的另一个精彩的 SVD 讨论:

Here's another nice SVD discussion from AMS:

http://www.ams.org/samplings/feature-column/fcarc-svd

这篇关于用 R 求解非平方线性系统的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！