Solution found...
Sorry for not having known this,...
Apparently, what I was after is called a "Choleski factorization".
The solution pops right out of R, as follows:
> M<-matrix(c(0.6098601, 0.2557882, 0.1857773,
+ 0.2557882, 0.5127065, -0.1384238,
+ 0.1857773, -0.1384238, 0.9351089 ),
+ nrow=3, ncol=3, byrow=TRUE)
> chol(M)
[,1] [,2] [,3]
[1,] 0.7809354 0.3275408 0.2378907
[2,] 0.0000000 0.6367288 -0.3397722
[3,] 0.0000000 0.0000000 0.8735398
>
Thanks again for all your help!
/shawn
B %*% t(B) = R , then solve for B
2 messages · Shawn Koppenhoefer, Doran, Harold
1 day later
Correct. In the example I gave you yesterday, I used a different matrix, but showed this solution because it also answered the other question you had about doing it on non-square matrices. Of course, Spencer Graves also gave a very useful answer suggesting QR decomposition. I also gave you the example in the context of linear regression because that is commonly why statisticians will use these factorizations. If your matrix is small, chol() works fine. If your matrix is big and sparse, see Cholesky() in the matrix package (a package that I often refer to as a God-send)
-----Original Message----- From: r-help-bounces at r-project.org [mailto:r-help-bounces at r-project.org] On Behalf Of Shawn Koppenhoefer Sent: Tuesday, April 12, 2011 12:33 PM To: r-help at r-project.org Subject: Re: [R] B %*% t(B) = R , then solve for B Importance: High Solution found... Sorry for not having known this,... Apparently, what I was after is called a "Choleski factorization". The solution pops right out of R, as follows:
> M<-matrix(c(0.6098601, 0.2557882, 0.1857773,
+ 0.2557882, 0.5127065, -0.1384238, + 0.1857773, -0.1384238, 0.9351089 ), + nrow=3, ncol=3, byrow=TRUE)
> chol(M)
[,1] [,2] [,3] [1,] 0.7809354 0.3275408 0.2378907 [2,] 0.0000000 0.6367288 -0.3397722 [3,] 0.0000000 0.0000000 0.8735398
>
Thanks again for all your help! /shawn
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