矩阵中的R个独立列 [英] R independent columns in matrix

问题描述

我试图找到独立的列来求解线性方程组.这是我的简化示例:

I am trying to find independent columns to solve the system of linear equations. Here my simplified example:

> mat = matrix(c(1,0,0,0,-1,1,0,0,0,-1,1,0,0,0,-1,0,-1,0,0,1,0,0,1,-1), nrow=4, ncol=6, dimnames=list(c("A", "B", "C", "D"), paste("v", 1:6, sep="")))
> mat
  v1 v2 v3 v4 v5 v6
A  1 -1  0  0 -1  0
B  0  1 -1  0  0  0
C  0  0  1 -1  0  1
D  0  0  0  0  1 -1

矩阵为满分:

qr(mat)$rank

给我4，并且由于有6列，所以应该有6-4 = 2个独立的列，从中我可以计算其他列. 我知道列v4和v6是独立的...我的第一个问题是，如何找到这些列(也许使用qr(mat)$ pivot)?

gives me 4, and since there are 6 columns, there should be 6-4=2 independent columns from which I can calculate the others. I know that columns v4 and v6 are independent... My first question is, how can I find these columns (maybe with qr(mat)$pivot)?

通过在纸上重新排列线性方程，我看到了
[v1，v2，v3，v4，v5，v6] = [v4，v4-v6，v4-v6，v4，v4，v6，v6]

By rearranging the linear equations on paper, I see that
[v1, v2, v3, v4, v5, v6] = [v4, v4-v6, v4-v6, v4, v4, v6, v6]

因此，我可以通过将v4和v6乘以以下向量，从v4和v6的任意值中找到一个位于空空间中的向量:

and thus I can find from arbitrary values for v4 and v6 a vector that lies in the null space by multiplying v4 and v6 with the vectors below:

v4 * [1,1,1,1,0,0] + v6 * [0,-1,-1,0,1,1]

我的第二个问题是:如何找到这些向量，这意味着如何求解v4和v6的矩阵? 例如

My second question is: How do I find these vectors, meaning how do I solve the matrix for v4 and v6? For example

qr.solve(mat, cbind(c(0,0,0,0), c(0,0,0,0)))

给我两个长度为6的矢量，它们只有零.

gives me two vectors of length 6 with only zeros.

感谢您的帮助，在此先感谢您！

Any help is appreciated, many thanks in advance!

-H-

推荐答案

使用数据透视信息查找一组独立的列:

Use the pivot information to find a set of independent columns:

q <- qr(mat)

mmat <- mat[,q$pivot[seq(q$rank)]]

mmat
##   v1 v2 v3 v5
## A  1 -1  0 -1
## B  0  1 -1  0
## C  0  0  1  0
## D  0  0  0  1

qr(mmat)$rank
## [1] 4

为什么这样做? pivot的含义在?qr.Q带来的QR.Auxiliaries {base}中给出.特别是:

Why does this work? The meaning of pivot is given in QR.Auxiliaries {base} brought up with ?qr.Q. In particular:

qr.R returns R. This may be pivoted, e.g., if a <- qr(x) then x[, a$pivot] = QR.
The number of rows of R is either nrow(X) or ncol(X) (and may depend on whether
complete is TRUE or FALSE).

进行枢轴旋转以使特征值按递减的绝对值排序，以实现数值稳定性.这也意味着，任何0特征值都位于末尾，超出了q$pivot中的q$rank(并且在当前示例中不存在，其中Q是4x4正交矩阵).

Pivoting is done to order the eigenvalues in decreasing absolute value, for numerical stability. This also means that any 0 eigenvalues are at the end, beyond q$rank in q$pivot (and nonexistent in the current example, where Q is a 4x4 orthogonal matrix).

QR.Auxiliaries {base}中的最后几行显示了这种关系:

The final lines in the QR.Auxiliaries {base} show this relationship:

pivI <- sort.list(a$pivot) # the inverse permutation
stopifnot(
 all.equal(x[, a$pivot], qr.Q(a) %*% qr.R(a)),          # TRUE
 all.equal(x           , qr.Q(a) %*% qr.R(a)[, pivI]))  # TRUE too!

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