Assignment 2: Trace / Determinant / Factorization

1. Problem set 1

(1) Showing \(A^TA \neq AA^T\) in General

Given \(A\) as an \(m \times n\) matrix, \(A^T\) is an \(n \times m\) matrix.

The product \(A^TA\) results in an \(n \times n\) matrix, while \(AA^T\) results in an \(m \times m\) matrix.

Proof Using an Example:

Let’s take a matrix \(A\):

\[ A = \begin{pmatrix} 1 & 2 \\ 3 & 4 \\ \end{pmatrix} \]

The transpose, \(A^T\), is:

\[ A^T = \begin{pmatrix} 1 & 3 \\ 2 & 4 \\ \end{pmatrix} \]

Calculating \(A^TA\):

\[ A^TA = \begin{pmatrix} 1 & 3 \\ 2 & 4 \\ \end{pmatrix} \begin{pmatrix} 1 & 2 \\ 3 & 4 \\ \end{pmatrix} = \begin{pmatrix} 10 & 14 \\ 14 & 20 \\ \end{pmatrix} \]

Calculating \(AA^T\):

\[ AA^T = \begin{pmatrix} 1 & 2 \\ 3 & 4 \\ \end{pmatrix} \begin{pmatrix} 1 & 3 \\ 2 & 4 \\ \end{pmatrix} = \begin{pmatrix} 5 & 11 \\ 11 & 25 \\ \end{pmatrix} \]

Thus, \(A^TA \neq AA^T\), demonstrating that in general, \(A^TA \neq AA^T\).

(2) Conditions for \(A^TA = AA^T\)

For the equality \(A^TA = AA^T\) to hold, the matrix \(A\) must satisfy certain conditions. This is true for:

• Symmetric Matrices: Where \(A = A^T\). For such matrices, it’s common that \(A^TA = AA^T\) due to their symmetry.
• Orthogonal Matrices: Which satisfy \(A^TA = I\), where \(I\) is the identity matrix. Since \(A^T = A^{-1}\) for orthogonal matrices, both \(A^TA\) and \(AA^T\) yield the identity matrix, indicating that \(A^TA = AA^T\) in this case.

2. Problem set 2

Here is R function for LU decomposition of a square matrix

luDecomposition <- function(A) {
  if (!is.matrix(A)) {
    stop("Input must be a matrix.")
  }
  
  n <- nrow(A)
  if (ncol(A) != n) {
    stop("Matrix must be square.")
  }
  
  L <- diag(1, n) # Initializing L as an identity matrix of size n
  U <- A # Copying A into U for transformation
  
  for (j in 1:(n-1)) {
    for (i in (j+1):n) {
      if (U[j, j] == 0) {
        stop("Zero pivot encountered.")
      }
      L[i, j] <- U[i, j] / U[j, j]
      U[i, ] <- U[i, ] - L[i, j] * U[j, ]
    }
  }
  
  return(list(L = L, U = U))
}

An example of Application using the LU function above:

A <- matrix(c(1, 2, 1, 3,
              3, 8, 1, 5,
              0, 4, 1, 9,
              4, 11, 3, 0), byrow = TRUE, nrow = 4)

result <- luDecomposition(A)

print("L Matrix:")

## [1] "L Matrix:"

print(result$L)

##      [,1] [,2] [,3] [,4]
## [1,]    1  0.0  0.0    0
## [2,]    3  1.0  0.0    0
## [3,]    0  2.0  1.0    0
## [4,]    4  1.5  0.4    1

print("U Matrix:")

## [1] "U Matrix:"

print(result$U)

##      [,1] [,2] [,3]  [,4]
## [1,]    1    2    1   3.0
## [2,]    0    2   -2  -4.0
## [3,]    0    0    5  17.0
## [4,]    0    0    0 -12.8

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