[R] about the Choleski factorization
Duncan Murdoch
murdoch at stats.uwo.ca
Fri Mar 27 18:24:22 CET 2009
On 3/27/2009 11:46 AM, 93354504 wrote:
> Hi there,
>
> Given a positive definite symmetric matrix, I can use chol(x) to obtain U where U is upper triangular
> and x=U'U. For example,
>
> x=matrix(c(5,1,2,1,3,1,2,1,4),3,3)
> U=chol(x)
> U
> # [,1] [,2] [,3]
> #[1,] 2.236068 0.4472136 0.8944272
> #[2,] 0.000000 1.6733201 0.3585686
> #[3,] 0.000000 0.0000000 1.7525492
> t(U)%*%U # this is exactly x
>
> Does anyone know how to obtain L such that L is lower triangular and x=L'L? Thank you.
>
> Alex
>
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> and provide commented, minimal, self-contained, reproducible code.
> rev <- matrix(c(0,0,1,0,1,0,1,0,0),3,3)
> rev
[,1] [,2] [,3]
[1,] 0 0 1
[2,] 0 1 0
[3,] 1 0 0
(the matrix that reverses the row and column order when you pre and post
multiply it).
Then
L <- rev %*% chol(rev %*% x %*% rev) %*% rev
is what you want, i.e. you reverse the row and column order of the
Choleski square root of the reversed x:
> x
[,1] [,2] [,3]
[1,] 5 1 2
[2,] 1 3 1
[3,] 2 1 4
> L <- rev %*% chol(rev %*% x %*% rev) %*% rev
> L
[,1] [,2] [,3]
[1,] 1.9771421 0.000000 0
[2,] 0.3015113 1.658312 0
[3,] 1.0000000 0.500000 2
> t(L) %*% L
[,1] [,2] [,3]
[1,] 5 1 2
[2,] 1 3 1
[3,] 2 1 4
Duncan Murdoch
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