HW2

Part 1:

1. Show that ATA =! AAT in general.
   Proof: \[ A^TA \ne AA^T \\ A^T_{kxn}A_{nxk} \ne A_{nxk} A^T_{kxn} \\ matrix_{kxk} \ne matrix_{nxn} \] Demonstration:

a <- matrix(c(1,4,2,5,3,6), nrow = 2, ncol = 3)
b <- t(a)
b%*%a

##      [,1] [,2] [,3]
## [1,]   17   22   27
## [2,]   22   29   36
## [3,]   27   36   45

a%*%b

##      [,1] [,2]
## [1,]   14   32
## [2,]   32   77

You can see that a does not equal to b.

2. For a special type of square matrix A, we get AT A = AAT. Under what conditions could this be true?
   When there is a symmetrical matrix.
   The following is a symmetrical matrix and it transpose equal to its inverse.

c <- matrix(c(1, 2, 3, 2, 1, 2, 3, 2 ,1), nrow = 3, ncol = 3)
d <- t(c)
c%*%d

##      [,1] [,2] [,3]
## [1,]   14   10   10
## [2,]   10    9   10
## [3,]   10   10   14

d%*%c

##      [,1] [,2] [,3]
## [1,]   14   10   10
## [2,]   10    9   10
## [3,]   10   10   14

Part 2:

Write an R function to factorize a square matrix A into LU or LDU.

LU factorize

lu_decomposition <- function(matrix, n) {
  # Copy of matrix to upper
  upper <- matrix
  # Diagonal matrix with 1 in diagonal 
  lower <- diag(n) 
  
  # Row Reduce matrix into Upper and Lower triangular matrices
  for(i in 1:(n-1)){
    p <- upper[i,i]
    
    for (j in range(i+1,n)){
      # Lower Triangular
      lower[j,i] <- upper[j,i]/p
      # Upper Triangular
      upper[j,] <- upper[j,]-lower[j,i]*upper[i,]
      
      if(i+1 == n) {break}
    }
  }
  
  triangle <- list(lower,upper)
  return (triangle)
}

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

print(lu_decomposition(matrix, 3))

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