Matrix factorization is a very important problem. There are supercomputers built just to do matrix factorizations. Every second you are on an airplane, matrices are being factorized. Radars that track flights use a technique called Kalman filtering. At the heart of Kalman Filtering is a Matrix Factorization operation. Kalman Filters are solving linear systems of equations when they track your flight using radars. Write an R function to factorize a square matrix A into LU or LDU, whichever you prefer. Please submit your response in an R Markdown document using our class naming convention, E.g. LFulton_Assignment2_PS2.png You don’t have to worry about permuting rows of A and you can assume that A is less than 5x5, if you need to hard-code any variables in your code. If you doing the entire assignment in R, then please submit only one markdown document for both the problems.
LU = function(A){
# Get matrix dimensions
rows <- columns <- dim(A)[1]
# A = LDU
# L = Lower triangle matrix
# D = Diagonal matrix
# U = Upper triangle matrix
U <- A
L <- D <- diag(rows)
#Column Loop
for (j in 1:(columns-1)){
#row loop
for (i in (j+1):rows){
#elimination
L[i,j] <- (U[i,j]/U[j,j])
U[i,] <- U[i,]-(U[j,]*L[i,j])
}
}
#transfer the middle diagonal from upper triangular matrix
diag(D) <- diag(U)
for (l in 1:rows){
U[l,] <- U[l,]/U[l,l]
}
LDU = list("Lower"=L,"Diagonal"=D,"Upper"=U)
return(LDU)
}
#Matrix example
A <- matrix(c(1,1,3,2,-1,5,-1,-2,4), nrow=3, ncol=3)
LU(A)## $Lower
## [,1] [,2] [,3]
## [1,] 1 0.0000000 0
## [2,] 1 1.0000000 0
## [3,] 3 0.3333333 1
##
## $Diagonal
## [,1] [,2] [,3]
## [1,] 1 0 0.000000
## [2,] 0 -3 0.000000
## [3,] 0 0 7.333333
##
## $Upper
## [,1] [,2] [,3]
## [1,] 1 2 -1.0000000
## [2,] 0 1 0.3333333
## [3,] 0 0 1.0000000