Source files: [https://github.com/djlofland/DATA605_S2020/tree/master/]
\[\text{Given: }A = \left[\begin{array}\\ a & b\\ c & d\\\end{array}\right]\ \text{then}\ A^{T} = \left[\begin{array}\\ a & c\\ b & d\\\end{array}\right]\\ \text{show} \ A^{T}A \ne AA^{T}\\ \left[\begin{array}\\ a & c\\ b & d\\\end{array}\right] \left[\begin{array}\\ a & b\\ c & d\\\end{array}\right] \ne \left[\begin{array}\\ a & b\\ c & d\\\end{array}\right] \left[\begin{array}\\ a & c\\ b & d\\\end{array}\right]\\ \left[\begin{array}\\ aa+cc & ca+cd\\ ba+dc & bb+dd\\\end{array}\right] \ne \left[\begin{array}\\ aa+bb & ac+bd\\ ca+db & cc+dd\\\end{array}\right]\\ (aa+cc) \ne (aa+bb), (ca+cd) \ne (ac+bd), (ba+dc) \ne (ca+db), (bb+dd) \ne (cc+dd)\\ \therefore \\ A^{T}A \ne AA^{T} \]
\[\text{Given: }A = \left[\begin{array}\\ a & b & c\\ d & e & f\\ g & h & i \end{array}\right]\ then\ A^{T} = \left[\begin{array}\\ a & d & g\\ b & e & h\\ c & f & i \end{array}\right]\]
\(A^{T}A = AA^{T}\) would hold true if \(b=d, c=g, f=h\) since \(A^{T}\) would then equal \(A\) and the problem simplifies to \(AA = AA\)
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 donโt 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.
factor_matrix <- function(A) {
size <- nrow(A)
max <- size - 1
# Build our base L matrix
L <- matrix(0, nrow=size, ncol=size)
for (i in 1:size) {
L[i,i] = 1L
}
# Copy A into U - we will be decompsing U as we go and keep our orig A
U <- matrix(0, nrow=size, ncol=size)
for (r in 1:size) {
for (c in 1:size) {
U[r,c] <- A[r,c]
}
}
for (c in 1:max) {
base_row <- c
next_row <- c+1
for (r in next_row:size) {
# print(c(c, r))
L[r, c] = U[r, c] / U[base_row, c]
U[r,] = U[r,] - U[base_row,] * L[r, c]
}
}
U <- round(U, 5)
L <- round(L, 5)
return(list("A"=A, "U"=U, "L"=L))
}
matrix_size <- 6
A <- matrix(sample(-100:100, matrix_size^2), byrow=T, nrow=matrix_size, ncol=matrix_size)
result <- factor_matrix(A)
print(result)## $A
## [,1] [,2] [,3] [,4] [,5] [,6]
## [1,] 35 44 -22 -57 61 -46
## [2,] -68 -95 -81 -11 -75 90
## [3,] 92 40 -20 49 -91 -70
## [4,] -21 55 -56 46 -17 -5
## [5,] 82 -87 -48 19 29 16
## [6,] 66 100 37 79 -100 34
##
## $U
## [,1] [,2] [,3] [,4] [,5] [,6]
## [1,] 35 44.00000 -22.0000 -57.0000 61.00000 -46.00000
## [2,] 0 -9.51429 -123.7429 -121.7429 43.51429 0.62857
## [3,] 0 0.00000 1021.8258 1166.9219 -597.36637 45.91592
## [4,] 0 0.00000 0.0000 258.2678 -267.48267 23.45966
## [5,] 0 0.00000 0.0000 0.0000 212.91718 21.99141
## [6,] 0 0.00000 0.0000 0.0000 0.00000 125.00501
##
## $L
## [,1] [,2] [,3] [,4] [,5] [,6]
## [1,] 1.00000 0.00000 0.00000 0.00000 0.00000 0
## [2,] -1.94286 1.00000 0.00000 0.00000 0.00000 0
## [3,] 2.62857 7.95195 1.00000 0.00000 0.00000 0
## [4,] -0.60000 -8.55556 -1.10380 1.00000 0.00000 0
## [5,] 2.34286 19.97898 2.42292 -0.93900 1.00000 0
## [6,] 1.88571 -1.78979 -0.13993 0.51065 -0.39522 1