R Notebook

Proof and Demonstration

We have a matrix size m * n. The product of their square \(A^TA\) noted as n * n and \(AA^T\) as matrix of m * n. To show \(A^TA \neq AA^T\) , it means that nn is not equal to mm unless n = m which mean \(A\) is a square. Therefore, for non-square matrices,\(A^TA \neq AA^T\) have different dimensions and cannot be equal.

\(A^T = \begin{bmatrix}a & c\\ b & d\end{bmatrix}\).

The product of \(AA^T\) will be: \(\begin{bmatrix}a\times a + b\times b & a\times c + b\times d\\ c\times a + d\times b & c\times c + d \times d\end{bmatrix}\).

The product of \(A^TA\) will be: \(\begin{bmatrix}a\times a + c\times c & a\times b + c\times d\\ a\times b + c\times d & b\times b + d \times d\end{bmatrix}\).

A <- matrix(seq(from = 2,8, by =2), nrow = 2)
A

##      [,1] [,2]
## [1,]    2    6
## [2,]    4    8

# creating AT 
A_T <- t(A)
A_T

##      [,1] [,2]
## [1,]    2    4
## [2,]    6    8

# Creaing A^TA
AT_A <- t(A)%*%A
AT_A

##      [,1] [,2]
## [1,]   20   44
## [2,]   44  100

#Creating AA^T
AA_T <- A%*%t(A)
AA_T

##      [,1] [,2]
## [1,]   40   56
## [2,]   56   80

Now we can argue that \(A^TA \neq AA^T\). \(A^TA = AA^T\) only when the matrix is orthogonal matrix and its transpose is equal to its inverse matrix. We have a matrix :

A = \(A^T = \begin{bmatrix}-1 & 0\\ 0 & 1\end{bmatrix}\).

\(A^T\) = \(\begin{bmatrix}-1 & 0\\ 0 & 1\end{bmatrix}\).

\(AA^T\) = \(\begin{bmatrix}(-1)\times (-1) + 0\times 0 & 0\times 0 + 0\times 1\\ 0\times (-1) + (1)\times 0 & 0\times 0 + 1 \times 1\end{bmatrix}\).

\(AA^T\) = \(\begin{bmatrix}-1 & 0\\ 0 & 1\end{bmatrix}\).

Problem #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.

Function that factorize a square matrix A LU and LDU:

# Function that factorize the matrix
lu_decomposition <- function(A){
  #Check for valid input
  if(nrow(A) != ncol(A)){
    stop("Input must be square matrix.")
  }
  else {
    n <- nrow(A)
    L <- diag(n)
    U <- A
    for (i in 1:(n - 1)) {
      for (j in (i + 1):n) {
        # get multipliers
        L[j, i] <- U[j, i] / U[i, i]
        # pivots and multiplication
        U[j, ]  <- U[j, ] - L[j, i] * U[i, ]
      }
    }
    
    #Get results
    LU <- list("L" = L, "U" = U)
  }
}

# Example usage:
b <- matrix(c(2, -1, -2, 4, 1, 2, -4, 6, 3), 3, 3)
result <- lu_decomposition(A)
 
result

## $L
##      [,1] [,2]
## [1,]    1    0
## [2,]    2    1
## 
## $U
##      [,1] [,2]
## [1,]    2    6
## [2,]    0   -4

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



print(lu_decomposition(b)$L %*% lu_decomposition(b)$U  == b)

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

R Notebook

Problem #1

Proof and Demonstration

Problem #2