On 1/11/2007, at 9:13 AM, Michael Gormley wrote: > Given a matrix B, where B=A'A, how can I find A? > In other words, if I have a matrix B which I know is another matrix > A times > its transpose, can I find matrix A?
You can't, because A is not unique. You can easily find ***a*** solution. E.g. A1 <- matrix(1:4,ncol=2) B <- t(A1)%*%A1 A2 <- msqrt(B) A2 != A1 (A2 is symmetric), yet t(A2)%*%A2 == B. The function msqrt() above is a simple-minded calculation of the square root of a positive semi-definite real matrix, the code of which I just cribbed from an old posting by Prof. Brian Ripley: msqrt <- function(X) { e <- eigen(X) V <- e$vectors V%*%diag(sqrt(e$values))%*%t(V) } The problem of finding ***all possible*** solutions A to A'A = B (for B symmetric positive semi-definite) is likely to be hard, but may have been solved by the linear algebraists. I dunno. cheers, Rolf Turner ###################################################################### Attention:\ This e-mail message is privileged and confid...{{dropped:9}} ______________________________________________ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.