Data 605-Homework #2

Problem set 1

(1) Show that ATA???AAT in general.(Proof and demonstration.)

A <- matrix(c(1, 4, 7, 2, 5, 8, 3, 6, 9), nrow=3, ncol=3) 
AT <- matrix(c(1, 2, 3, 4, 5, 6, 7, 8, 9), nrow=3, ncol=3)

A%*%AT

##      [,1] [,2] [,3]
## [1,]   14   32   50
## [2,]   32   77  122
## [3,]   50  122  194

AT%*%A

##      [,1] [,2] [,3]
## [1,]   66   78   90
## [2,]   78   93  108
## [3,]   90  108  126

(2) For a special type of square matrix A, we get AT A = AAT. Under what conditions could this be true? (Hint: The Identity matrix I is an example of such a matrix).

A <- matrix(c(1, 0, 0, 0, 1, 0, 0, 0, 1), nrow=3, ncol=3) 
AT <- matrix(c(1, 0, 0, 0, 1, 0, 0, 0, 1), nrow=3, ncol=3) 

A%*%AT

##      [,1] [,2] [,3]
## [1,]    1    0    0
## [2,]    0    1    0
## [3,]    0    0    1

AT%*%A

##      [,1] [,2] [,3]
## [1,]    1    0    0
## [2,]    0    1    0
## [3,]    0    0    1

Problem set 2

(1) Write an R function to factorize a square matrix A into LU or LDU, whichever you prefer.

factorize_square_Matrix <- function(A, diagonalMatrix, index) {

    columns = ncol(A)
    rows    = nrow(A)
  
    if (index >= columns) {
      return(list(diagonalMatrix, A))    
    } else {
      for(i in (index+1):rows) {
        multiplier <- (-A[i, index] / A[index, index])
        diagonalMatrix[i, index] <- round(-multiplier, 2)
        A[i, ] <- A[i, ] + (multiplier * A[index, ])
      }
      print (A)
    }
    factorize_square_Matrix(A, diagonalMatrix, index + 1)
}

A <- matrix(c(2, 4, -4, 1, -4, 3, -6, -9, 5), nrow=3, ncol=3, byrow = TRUE)
lowerMatrix <- factorize_square_Matrix(A, diag(nrow(A)), 1)

##      [,1] [,2] [,3]
## [1,]    2    4   -4
## [2,]    0   -6    5
## [3,]    0    3   -7
##      [,1] [,2] [,3]
## [1,]    2    4 -4.0
## [2,]    0   -6  5.0
## [3,]    0    0 -4.5

L<- lowerMatrix[[1]]
U <- lowerMatrix[[2]]

A == L %*% U

##      [,1] [,2] [,3]
## [1,] TRUE TRUE TRUE
## [2,] TRUE TRUE TRUE
## [3,] TRUE TRUE TRUE