Assignment2

Problem Set 1

1. Show that AT A 6= AAT in general. (Proof and demonstration.)

The following example will compute a Matrix A, and the transpose of that matrix and check for equality

library(matlib)

## Warning: package 'matlib' was built under R version 4.0.2

A<-matrix(c(1,7,4,63,6,7,2,9,67),nrow = 3)
A_T<-t(A)

A_T%*%A == A%*%A_T

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

The example shows that ATA is not equal to AAT

2. 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).

The condition will be true if the matrices are symstrical or identity matrices.

A<- matrix(c(3,1,2,1,3,1,2,1,3),nrow =3)
A

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

A_T<-t(A)


A_T%*%A == A%*%A_T

##      [,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 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.

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.

We create function and return an error if the matrix is too large (greater than 5x5) or the matrix is not a square matrix (nrow = ncol).

library(matrixcalc)

## 
## Attaching package: 'matrixcalc'

## The following object is masked from 'package:matlib':
## 
##     vec

library(Matrix)

## Warning: package 'Matrix' was built under R version 4.0.2

LU_function <- function(Mat){
  if(nrow(Mat)>= 5) {
    return("Matrix too large")
  }
  
  if(is.square.matrix(Mat) == FALSE) {
    return("Matrix not square")
  }
  
  n <- dim(Mat)[1]
  L <- diag(n)
  start <- 0
  
  while (start <= dim(Mat)[1]){
  start = start+1
  i = start
  j = start
  while (i <= dim(Mat)[1]-1){
    c = Mat[i+1,j]/Mat[j,j]
    Mat[i+1,] = Mat[i+1,]-c*Mat[j,]
    L[i+1,j] = c
    i = 1+i
  }

  }
  return(L)
}

In order to test the function:

testmatrix <- matrix(c(5,34,3,-32,4,66,3,1,9), nrow=3, byrow=TRUE)
(LU <- LU_function(testmatrix))

##      [,1]        [,2] [,3]
## [1,]  1.0  0.00000000    0
## [2,] -6.4  1.00000000    0
## [3,]  0.6 -0.08754513    1

We can compare this with R’s built in function lu.decomposition

lu.decomposition(testmatrix)[1]

## $L
##      [,1]        [,2] [,3]
## [1,]  1.0  0.00000000    0
## [2,] -6.4  1.00000000    0
## [3,]  0.6 -0.08754513    1

lu.decomposition(testmatrix)[[1]]==LU

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