Data 605 - Assignment #2

Problem Set 1

(1) Show that \(A^{T}A \neq AA^{T}\) in general. (Proof and demonstration.)

Answer: From textbook example MMNC (\(\xi MM\), page 151), we see that matrix multiplication is not communtative. In other words, for two matrices \(A\) and \(B\):

\(AB \neq BA\)

So, if \(B = A^{T}\) we get the similar result:

\(AA^{T} \neq A^{T}A\)

To show one case where they are not equal, we start with a matrix A:

\(A = \begin{bmatrix}8 & 4\\3 & 2\end{bmatrix}\) with \(A^{T} = \begin{bmatrix}8 & 3\\4 & 2\end{bmatrix}\)

\(AA^{T} = \left[\begin{array}{@{}c|c@{}} A\begin{bmatrix}8\\4\end{bmatrix} & A\begin{bmatrix}3\\2\end{bmatrix} \end{array}\right] = \left[\begin{array}{@{}c|c@{}} 8\begin{bmatrix}8\\4\end{bmatrix} + 4\begin{bmatrix}4\\2\end{bmatrix} & 3\begin{bmatrix}8\\4\end{bmatrix} + 2\begin{bmatrix}4\\2\end{bmatrix} \end{array}\right]\)

\(AA^{T} = \left[\begin{array}{@{}c|c@{}} \begin{bmatrix}64\\32\end{bmatrix} \begin{bmatrix}16\\8\end{bmatrix} & \begin{bmatrix}24\\12\end{bmatrix} \begin{bmatrix}8\\4\end{bmatrix} \end{array}\right] = \left[\begin{array}{@{}c|c@{}} \begin{bmatrix}82\\40\end{bmatrix} & \begin{bmatrix}32\\16\end{bmatrix} \end{array}\right] = \begin{bmatrix}82 & 32\\40 & 16\end{bmatrix}\)

Following similar steps, we generate \(A^{T}A = \begin{bmatrix}76 & 40\\30 & 16\end{bmatrix}\). So we can see that at least for one case, \(AA^{T} \ne A^{T}A\).

(2) For a special type of square matrix A, we get \(A^{T}A = AA^{T}\) . Under what conditions could this be true? (Hint: The Identity matrix I is an example of such a matrix).

Answer:

The above statement is true under two conditions:

Condition 1) \(A_{mn}\) is a square matrix (\(m = n\))

Condition 2) \(A_{ij} = A_{ji}\) for values of \(1 \ge i \ge m\) and \(1 \ge j \ge m\)

Problem Set 2

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

 find_lu_matrix <- function(m) {
  
  L <- diag(nrow(m))
  
  # for every row 
  for (start_row in 1:(nrow(m)-1)) {
    
    # start on columns from 'start_row'
    for (i in start_row:(nrow(m)-1)) {
      
      # Calculate the value to zero out a row
      x <- -(m[i+1,start_row] / m[start_row,start_row])
      
      # Save this value to the L matrix
      L[i+1,start_row] <- -x
      
      # Combine rows
      m[i+1, ] <- x * m[start_row,] + m[i+1,]
    }
  }
  
  # return a list of two matrices
  return_list <- list("U" = m, "L" = L)
  
  return (return_list);
 }

B <- matrix(c(1, -2, 3, 4, 8, 4, -3, 5, 7), nrow = 3)
# show the initial matrix
B

##      [,1] [,2] [,3]
## [1,]    1    4   -3
## [2,]   -2    8    5
## [3,]    3    4    7

my_matrices <- find_lu_matrix(B)

# display the lower and upper matrices
my_matrices$L

##      [,1] [,2] [,3]
## [1,]    1  0.0    0
## [2,]   -2  1.0    0
## [3,]    3 -0.5    1

my_matrices$U