DATA_605_Assignment_2_Thonn

** DATA_605_Assignment_2_Thonn **

** Problem Set 1 **

1-1). Show that \(A^T \times A \ne A \times A^T\)

\(A = \left[\begin{array}{rrrr} a & b \\ c & d \\ \end{array} \right]\)

\(A^T = \left[\begin{array}{rrrr} a & c \\ b & d \\ \end{array} \right]\)

\(A^T \times A = \left[\begin{array}{rrrr} a^2 + c^2 + ab+cd \\ ba+dc + b^2+d^2 \\ \end{array} \right]\)

\(A \times A^T = \left[\begin{array}{rrrr} ca + db + c^2+d^2 \\ ca+db + c^2+d^2 \\ \end{array} \right]\)

therefore: \(A^T \times A \ne A \times A^T\)

Demonstration \(A^T \times A \ne A \times A^T\)

A <- matrix (c (1,2,3,4), nrow = 2, ncol=2 , byrow = TRUE)
A_T <- t(A)

A

##      [,1] [,2]
## [1,]    1    2
## [2,]    3    4

A_T

##      [,1] [,2]
## [1,]    1    3
## [2,]    2    4

#A * A^T
A %*% A_T

##      [,1] [,2]
## [1,]    5   11
## [2,]   11   25

#A^T * A
 A_T %*% A 

##      [,1] [,2]
## [1,]   10   14
## [2,]   14   20

 # conclusion-1a: shows the outcome of these products is two different matrices

1-2)For a special type of square matrix A, we get \(A^T \times A = AA^T\) . Under what conditions could this be true?

This is true if the matrix is symetrical.

\(A = \left[\begin{array}{rrrr} a & c \\ c & a \\ \end{array} \right]\)

\(A^T = \left[\begin{array}{rrrr} a & c \\ c & a \\ \end{array} \right]\)

Therfore: \(A^T \times A = A \times A^T = A^T\)

A <- matrix (c (1,2,2,1), nrow = 2, ncol=2 , byrow = TRUE)
A_T <- t(A)

A

##      [,1] [,2]
## [1,]    1    2
## [2,]    2    1

A_T

##      [,1] [,2]
## [1,]    1    2
## [2,]    2    1

#A * A^T
A %*% A_T

##      [,1] [,2]
## [1,]    5    4
## [2,]    4    5

#A^T * A
 A_T %*% A 

##      [,1] [,2]
## [1,]    5    4
## [2,]    4    5

 # conclusion-1b: shows the outcome of these products is the same matrix

** Problem Set 2 ** 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

#Reference: LU Decomposition Steps: https://math.stackexchange.com/questions/404419/lu-decomposition-steps
#Method: Gaussian Elimination
#Factorize: obtain the lower and upper factors of a matrix
Factorize <- function(x) { 
intMatrices <- list()
i<-0
# Obtain upper matrix
  # iterate through rows of matrix x
  for ( row in 2: nrow(x)) {
    # iterate through columns of matrix x
    for(col in 1: (row -1 ))  {
       i <- i + 1 
      dnom <- x[col,col]
      num <- x[row ,col]
      ratio <- num / dnom 
      
      subx <- diag(nrow(x))
      subx[row,col] <- -1 * ratio

      x <- subx %*% x
      
      intMatrices[[i]] <- subx
    }
  }
  # Obtain lower matrix
  y <- diag(nrow(x))
  for (i in 1:length(intMatrices))  {
    y <- y %*% solve(intMatrices[[i]])
  }
  
  LU <- list ( Lower = y , Upper = x )
  LU
  
  }

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

##      [,1] [,2] [,3]
## [1,]    1    2    3
## [2,]    4    5    6
## [3,]    7    8    9

LU <- lapply(Factorize(A),round,1)
# LU Matrix with Low and Up factors
LU

## $Lower
##      [,1] [,2] [,3]
## [1,]    1    0    0
## [2,]    4    1    0
## [3,]    7    2    1
## 
## $Upper
##      [,1] [,2] [,3]
## [1,]    1    2    3
## [2,]    0   -3   -6
## [3,]    0    0    0

# Test: Multiply Lower and Upper matrixes to obtain the original matrix
LU[["Lower"]] %*% LU[["Upper"]] 

##      [,1] [,2] [,3]
## [1,]    1    2    3
## [2,]    4    5    6
## [3,]    7    8    9

# conclusion-2:

END