homework2

Show that ATA != AAT in general. (Proof and demonstration.)

## creating a 3x3 matrix with numbers 1 through 9
A <- matrix(seq(from= 1, to=9), nrow=3, byrow = T)
A

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

creating the transpose of the first A and multiplying it by A and storing it in a variable

ATA <- t(A) %*% A
ATA

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

multiplying the A by the transpose of A

AAT <-A %*% (t(A))
AAT

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

We can see that they’re not equal but just to make sure lets use a function to check if they’re equal and it returns a Boolean

equal <- all(ATA == AAT)
print(paste('Are the Matrices equivalent?',equal))

## [1] "Are the Matrices equivalent? FALSE"

Conclusion we conclude that ATA != AAT

For a special type of square matrix A, we get AT A = AAT Under what conditions could this be true? To make AT A = AAT, the matrix has to be symmetric, meaning the transpose of the matrix has to be equal to the matrix itself

I <- diag(3) 

# Print the identity matrix
print("Identity matrix:")

## [1] "Identity matrix:"

print(I)

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

if (all(I %*% t(I) == t(I) %*% I)) {
  print("AT A = A AT")
} else {
  print("AT A != A AT")
}

## [1] "AT A = A AT"

Problem set 2

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

LU_factorization <- function(B) {
  n <- nrow(B)
  L <- matrix(0, nrow = n, ncol = n)
  U <- matrix(0, nrow = n, ncol = n)
  
  for (j in 1:n) {
    U[1, j] <- B[1, j]
  }
  
  for (i in 1:n) {
    L[i, i] <- 1
  }
  
  for (i in 2:n) {
    L[i, 1] <- B[i, 1] / U[1, 1]
  }
  
  for (i in 2:n) {
    for (j in 2:n) {
      if (j >= i) {
        U[i, j] <- B[i, j] - sum(L[i, 1:(i-1)] * U[1:(i-1), j])
      } else {
        L[j, i] <- (B[j, i] - sum(L[j, 1:(j-1)] * U[1:(j-1), i])) / U[j, j]
      }
    }
  }
  
  return(list(L = L, U = U))
}


B <- matrix(c(2, -1, -2, -4, 6, 3, -4, 1, 10), nrow = 3, byrow = TRUE)

# Factorize B into LU decomposition
LU <- LU_factorization(B)
L <- LU$L
U <- LU$U

print("Lower triangular matrix L:")

## [1] "Lower triangular matrix L:"

print(L)

##      [,1] [,2]  [,3]
## [1,]    1    0  0.00
## [2,]   -2    1 -0.25
## [3,]   -2    0  1.00

print("Upper triangular matrix U:")

## [1] "Upper triangular matrix U:"

print(U)

##      [,1] [,2] [,3]
## [1,]    2   -1   -2
## [2,]    0    4   -1
## [3,]    0    0    6