RE: (HH^t)^{-1}

[Faheem Mitha <faheem@email.unc.edu>]

On Thu, 11 Oct 2001, Mikael Adlers wrote:

> The problem of computing \mu is a least squares problem. What you
> have written is the normal equations to
>
>  min || H\mu - h||_2
>
> Another way is to solve the problem by QR factorization, H = Q*R,
> ||H\mu -h||_2 = ||R \mu - Q^Th||_2. To solve this you solve R \mu = Q^Tb
> This method has much better error bounds and should be preferred.

Dear Mikael Adlers,

Thanks, this is a really good suggestion. I was handed these expressions
by someone else (I think they come from Bayesian computations) but I
haven't checked them myself. Like a ninny, I didn't realise /mu was a
solution to a least squares problem. This way of handling it is certainly
much better than the clumsy things I was doing. Thanks a lot for pointing
it out.

                              Sincerely, Faheem Mitha.