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

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