PROBLEM SET 1

Show that ATA != AAT. (Proof and demonstration)

# create as sample matrix 
M <- matrix(c(7,1,5,8,2,3,9,0,4),ncol =3)

# Display the sample matrix
M

##      [,1] [,2] [,3]
## [1,]    7    8    9
## [2,]    1    2    0
## [3,]    5    3    4

# calculate as sample matrix transpose
Mt <- t(M)

# Display the sample matrix transpose
Mt

##      [,1] [,2] [,3]
## [1,]    7    1    5
## [2,]    8    2    3
## [3,]    9    0    4

# I will now multiply M by its transpose t(M) and display the product
Mt %*% M

##      [,1] [,2] [,3]
## [1,]   75   73   83
## [2,]   73   77   84
## [3,]   83   84   97

# I will now perform a check (prove correctness), that in fact ATA != AAT

Mt %*% M == M %*% Mt

##       [,1]  [,2]  [,3]
## [1,] FALSE FALSE FALSE
## [2,] FALSE FALSE FALSE
## [3,] FALSE FALSE FALSE

We can see from the result above that in fact ATA != AAT is true.

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). Please typeset your response using LaTeX mode in RStudio. If you do it in paper, please either scan or take a picture of the work and submit it. Please ensure that your image is legible and that your submissions are named using your first initial, last name, assignment and problem set within the assignment. E.g. LFulton_Assignment2_PS1.pn

This holds true under the conditions that a particular matrix is symmetrical on the diagonal. Symmetry along the diagonal would indicate that a particular matrix and its transpose are in fact equal. Below is a depiction of this:

# Create a sample Matrix M
M <- matrix(c(5,2,0,2,5,0,0,0,1),ncol =3)

# Display the sample matrix
M

##      [,1] [,2] [,3]
## [1,]    5    2    0
## [2,]    2    5    0
## [3,]    0    0    1

# calculate as sample matrix transpose
Mt <- t(M)

# Display the sample matrix transpose
Mt

##      [,1] [,2] [,3]
## [1,]    5    2    0
## [2,]    2    5    0
## [3,]    0    0    1

# Prove correctness that this matrix is symmetrical on the diagonal by showing that this matrix is equal to it s transpose

# Since, we assume a symmetric matrix, then: A == t(A). This substitution, will give us the following:

(M %*% t(M)) == (t(M) %*% M)

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

We can see from the result above that in fact ATA == AAT under the condition of a matrix being symmetrical on diagonal.

PROBLEM SET 2

Matrix factorization is a very important problem. There are supercomputers built just to do matrix factorization. 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.

#sample square matrix A:
A <- matrix (c(6, 7, 8, 1, 4, 5, 5, 2, 1), ncol = 3)


#Function which factorizes a square matrix into Lower and Upper decomposition
SM_Factorize <- function(A){
  
#Number of rows in matrix A
number_of_rows <- nrow(A)
#Lower triangular matrix represents diagonal matrix (where all elements above and below main diagonal are 0)
Lower <- diag(number_of_rows)
#Upper triangular matrix represents the matrix itself
Upper<- A

#looping through two vectors - 1 representing the rows, 2 representing the columns
for (z in c(1:number_of_rows )){
    for(x in c(2:number_of_rows )){
      
     #algorithm starts when column > row, meaning, in A 3X3 square matrix we look at the row of elements in       positions:A12,A13,A23
      if(x > z){
        # current row we  look at is the first row
        current_row <- Upper[z,];
        #retrieve the multiplier by dividing the appropriate elements by the current row
        multiplier <- Upper[x, z] / current_row[z]
        #have multiplier swap rows to ensure that a triangle of zeros is formed below the diagonal
        Upper[x,] <- Upper[x,] - (multiplier * current_row)
        #have multiplier swap rows to ensure that a triangle of zeros is formed above the diagonal
        Lower[x,z] <- multiplier
      }
    }
  }
  
  #printing the results

  #upper triangular matrix
  print(Upper)
  print (Lower)
  print (A)
  
}

#Run the function
SM_Factorize(A)

##      [,1]     [,2]       [,3]
## [1,]    6 1.000000  5.0000000
## [2,]    0 2.833333 -3.8333333
## [3,]    0 0.000000 -0.7058824
##          [,1]     [,2] [,3]
## [1,] 1.000000 0.000000    0
## [2,] 1.166667 1.000000    0
## [3,] 1.333333 1.294118    1
##      [,1] [,2] [,3]
## [1,]    6    1    5
## [2,]    7    4    2
## [3,]    8    5    1

Assignment # 2

Vladimir Nimchenko

2023-01-28

PROBLEM SET 1

PROBLEM SET 2