## Friday, June 28, 2013

### Upper triangle of a matrix to a vector by row

Say we want to get a square matrix into a vector. We can do:

to.upper<-function(X) X[upper.tri(X,diag=TRUE)]
but this will give us our upper triangular matrix as a vector by column, i.e.:
> X
[,1]   [,2]   [,3]   [,4]   [,5]
[1,] -0.329 -0.825 -0.906  0.834  0.702
[2,]  1.195 -0.237 -1.132 -0.827  0.150
[3,] -0.097 -1.270 -1.397  0.450  0.791
[4,]  0.883 -0.574  1.538 -2.632  0.135
[5,]  1.993 -0.520 -0.071  0.094 -0.064
> to.upper(X)
 -0.329 -0.825 -0.237 -0.906 -1.132 -1.397  0.834 -0.827  0.450 -2.632  0.702  0.150  0.791  0.135 -0.064

If we want to get our upper triangular matrix as a vector by row (as I did), we can use:

to.upper<-function(X) t(X)[lower.tri(X,diag=TRUE)]
which works just as we'd hoped, i.e.:
> X
[,1]   [,2]   [,3]   [,4]   [,5]
[1,] -0.329 -0.825 -0.906  0.834  0.702
[2,]  1.195 -0.237 -1.132 -0.827  0.150
[3,] -0.097 -1.270 -1.397  0.450  0.791
[4,]  0.883 -0.574  1.538 -2.632  0.135
[5,]  1.993 -0.520 -0.071  0.094 -0.064
> to.upper(X)
 -0.329 -0.825 -0.906  0.834  0.702 -0.237 -1.132 -0.827  0.150 -1.397  0.450  0.791 -2.632  0.135 -0.064

(Note that if we have a symmetric matrix than the upper triangular matrix by row & the lower triangular matrix by column are the same. Here I used a non-symmetric square matrix so that we could tell these apart.)

1. Similarly, here's a function that puts a vector into the upper diagonal of a matrix - by row:

upper.diag<-function(x){
m<-(-1+sqrt(1+8*length(x)))/2
X<-lower.tri(matrix(NA,m,m),diag=TRUE)
X[X==TRUE]<-x
t(X)
}

1. Hi
Could you please explain why m<-(-1+sqrt(1+8*length(x)))/2
thanks

2. Hmmm. This is a very old post. Must be the solution for the number of elements in the upper diagonal of a square matrix of dimension m x m. In particular, x=m*(m-1)/2 solved for m.

2. Thank you so much for you post! I just need this!!!

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