Elina Azrilyan

September 5th, 2019

Problem set 1

  1. Show that A x t(A) does not equal t(A) x A in general. (Proof and demonstration.)
Proof:

One of the basic laws of matrixes is that AB != BA, matrix multiplication is not commutative — order matters. This is generally true with some exceptions (see question 2 in this problem set).

Demonstration for nxn matrix:
A <- matrix(c(2, 2, 1, 4, 2, 3, 1, 0, 1),nrow=3)
B <- t(A)

A %*% B == B %*% A
##       [,1]  [,2]  [,3]
## [1,] FALSE FALSE FALSE
## [2,] FALSE FALSE FALSE
## [3,] FALSE FALSE FALSE
Demonstration for matrixes other than nxn - we can’t even calculate a product for those.
A <- matrix(c(2, 2, 1, 4, 2, 3, 1, 0, 1, 2, 3, 1),nrow=3)
B <- t(A)

#We get an error since the dimansions are incompatible

#A %*% B == B %*% A

#Error in B * A : non-conformable arrays
#Error in A * B : non-conformable arrays
  1. For a special type of square matrix A, we get AT A = AAT . Under what conditions could this be true? (Hint: The Identity matrix I is an example of such a matrix).
Answer:

t(A) x A and A x t(A) are equal when the matrix is symetric. A symmetric matrix is a square matrix that is equal to its transpose.

Example:
A <- matrix(c(2, 1, 1, 1, 2, 1, 1, 1, 2),nrow=3)
B <- t(A)

A
##      [,1] [,2] [,3]
## [1,]    2    1    1
## [2,]    1    2    1
## [3,]    1    1    2
B
##      [,1] [,2] [,3]
## [1,]    2    1    1
## [2,]    1    2    1
## [3,]    1    1    2
(A %*% B) == (B %*% A)
##      [,1] [,2] [,3]
## [1,] TRUE TRUE TRUE
## [2,] TRUE TRUE TRUE
## [3,] TRUE TRUE TRUE

Problem set 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.

Write an R function to factorize a square matrix A into LU or LDU, whichever you prefer. Please submit your response in an R Markdown document using our class naming convention, E.g. LFulton_Assignment2_PS2.png

You don’t have to worry about permuting rows of A and you can assume that A is less than 5x5, if you need to hard-code any variables in your code. If you doing the entire assignment in R, then please submit only one markdown document for both the problems.

The code below creates the function:
factorization <- function(A) {
  #Let's assign our matrix to 'upper' so that we can calculate U from that
  
  upper <- A 
  #Let's create an identity matrix of the same dimension and call it 'lower' so that we can use that as a starting point for L calculation
  
  lower <- diag(nrow = dim(A)[1])

  #The following loops move through the matrix and calculate the Upper and Lower Triangular Matrixes:
  
  for (i in 2:dim(A)[1]){ #
    for (j in 1:(i-1)){
      lower[i,j] <- upper[i,j] / upper[j,j]
      upper[i, ] <- -lower[i,j] * upper[j, ] + upper[i, ]
    }
  }
  
  print("Upper Triangular Martix:")
  print(upper)
  print("Lower Triangular Martix:")
  print(lower)
  print("Original Matrix A:")
  print(A)
  print("LU:")
  print(lower%*%upper)
  print("Comparing A and LU")
  print(A==lower%*%upper)
  
  #If we want to return the U and L rather than print - we would use the following code:
  #return(list(lower,upper))
}
We can test out the function with some examples:
A <- matrix(c(2, 4, -4, 1, -4, 3, -6, -9, 5), nrow=3, ncol=3, byrow = TRUE) 
factorization(A)
## [1] "Upper Triangular Martix:"
##      [,1] [,2] [,3]
## [1,]    2    4 -4.0
## [2,]    0   -6  5.0
## [3,]    0    0 -4.5
## [1] "Lower Triangular Martix:"
##      [,1] [,2] [,3]
## [1,]  1.0  0.0    0
## [2,]  0.5  1.0    0
## [3,] -3.0 -0.5    1
## [1] "Original Matrix A:"
##      [,1] [,2] [,3]
## [1,]    2    4   -4
## [2,]    1   -4    3
## [3,]   -6   -9    5
## [1] "LU:"
##      [,1] [,2] [,3]
## [1,]    2    4   -4
## [2,]    1   -4    3
## [3,]   -6   -9    5
## [1] "Comparing A and LU"
##      [,1] [,2] [,3]
## [1,] TRUE TRUE TRUE
## [2,] TRUE TRUE TRUE
## [3,] TRUE TRUE TRUE
B <- matrix(c(-1, 3, 4, 2, 4, 2, -9, 3, -1), nrow=3, ncol=3, byrow = TRUE) 
factorization(B)
## [1] "Upper Triangular Martix:"
##      [,1] [,2] [,3]
## [1,]   -1    3    4
## [2,]    0   10   10
## [3,]    0    0  -13
## [1] "Lower Triangular Martix:"
##      [,1] [,2] [,3]
## [1,]    1  0.0    0
## [2,]   -2  1.0    0
## [3,]    9 -2.4    1
## [1] "Original Matrix A:"
##      [,1] [,2] [,3]
## [1,]   -1    3    4
## [2,]    2    4    2
## [3,]   -9    3   -1
## [1] "LU:"
##      [,1] [,2] [,3]
## [1,]   -1    3    4
## [2,]    2    4    2
## [3,]   -9    3   -1
## [1] "Comparing A and LU"
##      [,1] [,2] [,3]
## [1,] TRUE TRUE TRUE
## [2,] TRUE TRUE TRUE
## [3,] TRUE TRUE TRUE