在正半定矩阵的Cholesky分解中正确使用枢轴 [英] Correct use of pivot in Cholesky decomposition of positive semi-definite matrix

问题描述

我不明白如何在R中使用chol函数来分解一个正半定矩阵. (或者我这样做，并且有一个错误.)

I don't understand how to use the chol function in R to factor a positive semi-definite matrix. (Or I do, and there's a bug.) The documentation states:

如果ivot = TRUE，则可以计算正半定x的Choleski分解. x的等级以attr(Q，"rank")的形式返回，这会受到数值误差的影响.枢轴返回为attr(Q，"pivot"). t(Q)％*％Q不再等于x.但是，设置枢轴<-attr(Q，"pivot")和oo <-order(pivot)，则确实是t(Q [，oo])％*％Q [，oo]等于x ...

If pivot = TRUE, then the Choleski decomposition of a positive semi-definite x can be computed. The rank of x is returned as attr(Q, "rank"), subject to numerical errors. The pivot is returned as attr(Q, "pivot"). It is no longer the case that t(Q) %*% Q equals x. However, setting pivot <- attr(Q, "pivot") and oo <- order(pivot), it is true that t(Q[, oo]) %*% Q[, oo] equals x ...

以下示例似乎掩盖了这一描述.

The following example seems to belie this description.

> x <- matrix(1, nrow=3, ncol=3)
> Q <- chol(x, pivot=TRUE)
> oo <- order(attr(Q, 'pivot'))
> t(Q[, oo]) %*% Q[, oo]
     [,1] [,2] [,3]
[1,]    1    1    1
[2,]    1    1    1
[3,]    1    1    3

结果不是x.我是否使用了不正确的枢轴?

The result is not x. Am I using the pivot incorrectly?

推荐答案

对于全秩输入，即正定矩阵x，我们需要

For full rank input, i.e., a positive-definite matrix x, we need

Q <- chol(x, TRUE)
oo <- order(attr(Q, 'pivot'))
unpivQ <- Q[, oo]
all.equal(crossprod(unpivQ), x)

对于有效秩不足的输入，即正半定矩阵x(具有负本征值的不确定矩阵是非法的，但未在chol中进行检查)，请记住零缺陷尾随对角线块:

For a valid rank-deficient input, i.e., a positive semi-definite matrix x (indefinite matrix with negative eigen values are illegal, but not checked in chol), remember to zero deficient trailing diagonal block:

Q <- chol(x, TRUE)
r <- attr(Q, 'rank')
if (r < nrow(x)) Q[(r+1):nrow(x), (r+1):nrow(x)] <- 0
oo <- order(attr(Q, 'pivot'))
unpivQ <- Q[, oo]
all.equal(crossprod(unpivQ), x)

有人将此称为chol的错误"，但实际上这是基础LAPACK例程dpstrf的功能.分解继续进行，直到第一个对角线元素低于公差为止，而尾随矩阵在出口处保持不变.

Some people call this a 'bug' of chol, but actually it is a feature of the underlying LAPACK routine dpstrf. The factorization proceeds till the first diagonal element which is below a tolerance, leaving the trailing matrix simply untouched on exit.

感谢伊恩(Ian)的以下观察:

Thanks to Ian for the following observation:

您可以在Q[-(1:r): -(1:r)] <- 0中使用R的负索引来跳过if语句.

You could use R's negative indexing in Q[-(1:r): -(1:r)] <- 0 to skip the if statement.

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