I am trying to check if my matrix is singular using the eigenvalues approach (i.e. if one of the eigenvalues is zero then the matrix is singular). Here is the code:
z <- matrix(c(-3,2,1,4,-9,6,3,12,5,5,9,4),nrow=4,ncol=3)
eigen(t(z)%*%z)$values
I know the eigenvalues are sorted in descending order. Can someone please let me know if there is a way to find out what eigenvalue is associated to what column in the matrix? I need to remove the collinear columns.
It might be obvious in the example above but it is just an example intended to save you time from creating a new matrix.
Example:
z <- matrix(c(-3,2,1,4,-9,6,3,12,5,5,9,4),nrow=4,ncol=3)
m <- crossprod(z) ## slightly more efficient than t(z) %*% z
This tells you that the third eigenvector corresponds to the collinear combinations:
ee <- eigen(m)
(evals <- zapsmall(ee$values))
## [1] 322.7585 124.2415 0.0000
Now examine the corresponding eigenvectors, which are listed as columns corresponding to their respective eigenvalues:
(evecs <- zapsmall(ee$vectors))
## [1,] -0.2975496 -0.1070713 0.9486833
## [2,] -0.8926487 -0.3212138 -0.3162278
## [3,] -0.3385891 0.9409343 0.0000000
The third eigenvalue is zero; the first two elements of the third eigenvector (evecs[,3]
) are non-zero, which tells you that columns 1 and 2 are collinear.
Here's a way to automate this test:
testcols <- function(ee) {
## split eigenvector matrix into a list, by columns
evecs <- split(zapsmall(ee$vectors),col(ee$vectors))
## for non-zero eigenvalues, list non-zero evec components
mapply(function(val,vec) {
if (val!=0) NULL else which(vec!=0)
},zapsmall(ee$values),evecs)
}
testcols(ee)
## [[1]]
## NULL
## [[2]]
## NULL
## [[3]]
## [1] 1 2
You can use tmp <- svd(z)
to do a svd. The eigenvalues are then saved in tmp$d
as a diagonal matrix of eigenvalues. This works also with a non square matrix.
> diag(tmp$d)
[,1] [,2] [,3]
[1,] 17.96548 0.00000 0.000000e+00
[2,] 0.00000 11.14637 0.000000e+00
[3,] 0.00000 0.00000 8.787239e-16
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