Here are two solutions. In the first lo has TRUE on the lower diagonal and diagonal. Then we compute the exponents, multiplying by lo to zero out the upper triangle. In the second rn is a matrix of row numbers and rn >= t(rn) is the same as lo in the first solution.
r <- 2; n <- 5 # test data lo <- lower.tri(diag(n), diag = TRUE) lo * r ^ (row(lo) - col(lo) + 1) Here is another one: rn <- row(diag(n)) (rn >= t(rn)) * r ^ (rn - t(rn) + 1) On 8/15/07, [EMAIL PROTECTED] <[EMAIL PROTECTED]> wrote: > Hi, > Sorry if this is a repost. I searched but found no results. > I am wondering if it is an easy way to construct the following matrix: > > r 1 0 0 0 > r^2 r 1 0 0 > r^3 r^2 r 1 0 > r^4 r^3 r^2 r 1 > > where r could be any number. Thanks. > Wen ______________________________________________ R-help@stat.math.ethz.ch 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.
