Hi Gabor, I am glad to see your answer,which gives a hope to resove this question in an easy way. I replied to this question in a more complex way before seeing your answer.
However,I think your code needs some revision, because the original matrix is not a diagonal matrix. It has 4 rows and 5 columns.Looking forward to your revised codes. Best regards, On Thu, Aug 16, 2007 20:22, Gabor Grothendieck wrote: > 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. > -- De-Jian Zhao Institute of Zoology,Chinese Academy of Sciences +86-10-64807217 [EMAIL PROTECTED]
______________________________________________ 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.