eliminate t() and %*% using crossprod() and solve(A,b)
On Wed, 5 Oct 2005, Robin Hankin wrote:
I have a square matrix Ainv of size N-by-N where N ~ 1000
I have a rectangular matrix H of size N by n where n ~ 4.
I have a vector d of length N.
I need X = solve(t(H) %*% Ainv %*% H) %*% t(H) %*% Ainv %*% d
and
H %*% X.
It is possible to rewrite X in the recommended crossprod way:
X <- solve(quad.form(Ainv, H), crossprod(crossprod(Ainv, H), d))
where quad.form() is a little function that evaluates a quadratic form:
quad.form <- function (M, x){
jj <- crossprod(M, x)
return(drop(crossprod(jj, jj)))
}
That is not the same thing: it gives t(H) %*% Ainv %*% t(Ainv) %*% H .
QUESTION: how to calculate H %*% X in the recommended crossprod way? (I don't want to take a transpose because t() is expensive, and I know that %*% is slow).
Have you some data to support your claims? Here I find (for random matrices of the dimensions given on a machine with a fast BLAS)
system.time(for(i in 1:100) t(H) %*% Ainv)
[1] 2.19 0.01 2.21 0.00 0.00
system.time(for(i in 1:100) crossprod(H, Ainv))
[1] 1.33 0.00 1.33 0.00 0.00 so each is quite fast and the difference is not great. However, that is not comparing %*% with crossprod, but t & %*% with crossprod. I get
system.time(for(i in 1:1000) H %*% X)
[1] 0.05 0.01 0.06 0.00 0.00 which is hardly 'slow' (60 us for %*%), especially compared to forming X in
system.time({X = solve(t(H) %*% Ainv %*% H) %*% t(H) %*% Ainv %*% d})
[1] 0.04 0.00 0.04 0.00 0.00 I would probably have written
system.time({X <- solve(crossprod(H, Ainv %*% H), crossprod(crossprod(Ainv, H), d))})
1] 0.03 0.00 0.03 0.00 0.00 which is faster and does give the same answer. [BTW, I used 2.2.0-beta which defaults to gcFirst=TRUE.]
Brian D. Ripley, ripley at stats.ox.ac.uk Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/ University of Oxford, Tel: +44 1865 272861 (self) 1 South Parks Road, +44 1865 272866 (PA) Oxford OX1 3TG, UK Fax: +44 1865 272595