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#' Calculate the Pseudo-Determinant of a Matrix
#'
#' When a given square matrix \eqn{A} is rank deficient, determinant is zero.
#' Still, we can compute the pseudo-determinant by multiplying all non-zero
#' eigenvalues. Since thresholding to determine near-zero eigenvalues is subjective,
#' we implemented the function as of original limit problem. When matrix is
#' non-singular, it coincides with traditional determinant.
#'
#' @param A a square matrix whose pseudo-determinant be computed.
#'
#' @return a scalar value for computed pseudo-determinant.
#'
#' @examples
#' ## show the convergence of pseudo-determinant
#' # settings
#' n = 10
#' A = cov(matrix(rnorm(5*n),ncol=n)) # (n x n) matrix
#' k = as.double(Matrix::rankMatrix(A)) # rank of A
#'
#' # iterative computation
#' ntry = 11
#' del.vec = exp(-(1:ntry))
#' det.vec = rep(0,ntry)
#' for (i in 1:ntry){
#' del = del.vec[i]
#' det.vec[i] = det(A+del*diag(n))/(del^(n-k))
#' }
#'
#' # visualize the results
#' opar <- par(no.readonly=TRUE)
#' plot(1:ntry, det.vec, main=paste("true rank is ",k," out of ",n,sep=""),"b", xlab="iterations")
#' abline(h=pdeterminant(A),col="red",lwd=1.2)
#' par(opar)
#'
#' @references
#' \insertRef{holbrook_differentiating_2018}{maotai}
#'
#' @export
pdeterminant <- function(A){ ## wrapped-up function
if (!is.matrix(A)){
stop("* pdeterminant : input 'A' should be a matrix.")
}
if (nrow(A)!=ncol(A)){
stop("* pdeterminant : input 'A' should be a square matrix.")
}
n = nrow(A)
k = as.double(Matrix::rankMatrix(A))
if (k==n){
return(base::determinant(A))
} else {
multccc = 0.9
del.old = 1
det.old = det(A+del.old*diag(n))/(del.old^(n-k))
for (i in 1:496){
# print(paste("iteration ",i," initiated...",sep=""))
del.new = del.old*multccc
det.new = det(A+del.new*diag(n))/(del.new^(n-k))
if ((abs(det.new-det.old)/abs(det.old))<1e-5){
return(det.new)
} else {
del.old = del.new
det.old = det.new
}
}
return(det.old)
}
}
# # # personal tests ----------------------------------------------------------
# n = 10
# A = cov(matrix(rnorm(5*n),ncol=n))
# k = as.double(Matrix::rankMatrix(A))
#
# ntry = 20
# del.vec = exp(-(1:ntry))
# det.vec = rep(0,ntry)
# for (i in 1:ntry){
# del = del.vec[i]
# det.vec[i] = det(A+del*diag(n))/(del^(n-k))
# }
# plot(1:ntry, det.vec, main=paste("true rank is ",k,"/",n,sep=""),"b")
# abline(h=pdeterminant(A),col="red",lwd=1.2)
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