data-605-week2-Assignment

Problem set 1

1. Show that A^TA ≠ AA^T in general. (Proof and demonstration.) You can use R-markdown to submit your responses to this problem set. If you decide to do it in paper, then please either scan it or take a picture using a smartphone and attach that picture. Please make sure that the picture is legible before submitting.

# Declare a matrix
A <- matrix(c(1,2,3,4,5,6,7,8,9), nrow = 3, byrow = T)
A

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

# Calculate Transpose of matrix A
t(A)

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

# Transposed of matrix A multiplied by matrix A 
ATA <- t(A) %*% A
ATA

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

# Matrix A multiplied by transposed of matrix A
AAT <- A %*% t(A)
AAT

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

#Checking

ATA == AAT

##       [,1]  [,2]  [,3]
## [1,] FALSE FALSE FALSE
## [2,] FALSE FALSE FALSE
## [3,] FALSE FALSE FALSE

Hence, A^TA ≠ AA^T

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). Please typeset your response using LaTeX mode in RStudio. If you do it in paper, please either scan or take a picture of the work and submit it. Please ensure that your image is legible and that your submissions are named using your first initial, last name, assignment and problem set within the assignment. E.g. LFulton_Assignment2_PS1.png

# Declare an identity matrix
A <- matrix(c(1,0,0,0,1,0,0,0,1), nrow = 3, byrow = T)
A

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

# Calculate Transpose of matrix A
t(A)

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

# Transposed of matrix A multiplied bymatrix A 
ATA <- t(A) %*% A
ATA

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

# Matrix A multiplied by transposed matrix of A
AAT <- A %*% t(A)
AAT

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

#Checking

ATA == AAT

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

Hence, A^TA = AA^T

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 lights 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 light 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.

funclu <- function(A){
  if((nrow(A) != ncol(A)) || (nrow(A) >= 5) || (ncol(A) >= 5)){
    stop("Input needs to be a square matrix and less than 5x5. Please try again with valid input matrix.")
  }
    else {
    r <- nrow(A)
    L <- diag(r)
    U <- A
      for (i in 1:(r - 1)) {
      for (j in (i + 1):r) {
        L[j, i] <- U[j, i] / U[i, i]
        U[j, ]  <- U[j, ] - L[j, i] * U[i, ]
      }
    }
    return(list("Lower:-"= L, "Upper:-" = U))
    }
}

(A <- matrix(c(1:9), nrow=3))

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

(funclu(A))

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