# [R] an easy way to construct this special matirx

De-Jian,ZHAO zhaodj at ioz.ac.cn
Thu Aug 16 15:26:23 CEST 2007

```Hi Gabor,

question in an easy way. I replied to this question in a more

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

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, shaowenhua at comcast.net <shaowenhua at comcast.net> 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
>
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> and provide commented, minimal, self-contained, reproducible code.
>

--
De-Jian Zhao
Institute of Zoology,Chinese Academy of Sciences
+86-10-64807217
zhaodj at ioz.ac.cn

```