SOrtegaCruz-Assign2
Sergio Ortega Cruz
February 7, 2019
Assignment 2
Problem set 1
Answer 1
knitr::include_graphics("C:\\Users\\sergioor\\Desktop\\Old PC\\CUNY\\DATA605\\hw1ex1.jpg")
Demonstration:
\(Consider\quad a\quad matrix\quad A\quad i,j:\\ \\ \begin{bmatrix} { a }_{ 1,1 } & { a }_{ 1,2 } & { a }_{ 1,3 } & . & . & . & { a }_{ 1,j } \\ { a }_{ 2,1 } & { a }_{ 2,2 } & { a }_{ 2,3 } & . & . & . & . \\ { a }_{ 3,1 } & { a }_{ 3,2 } & { a }_{ 3,3 } & . & . & . & . \\ . & . & . & . & . & . & . \\ . & . & . & . & . & . & . \\ . & . & . & . & . & . & . \\ { a }_{ i,1 } & { a }_{ i,2 } & { a }_{ i,3 } & . & . & . & { a }_{ i,j } \end{bmatrix}\)
Assuming \(i\neq j\)
\({ A }_{ (j,i) }^{ T }\) will result in j rows and i columns
\(Transpose\quad of\quad A,{ A }^{ T\quad }=\quad \begin{bmatrix} { a }_{ 1,1 } & { a }_{ 2,1 } & { a }_{ 3,1 } & . & . & . & { a }_{ i,1 } \\ { a }_{ 1,2 } & { a }_{ 2,2 } & { a }_{ 3,2 } & . & . & . & { a }_{ 2,j } \\ { a }_{ 1,3 } & { a }_{ 2,3 } & { a }_{ 3,3 } & . & . & . & { a }_{ 3,j } \\ . & . & . & . & . & . & . \\ . & . & . & . & . & . & . \\ . & . & . & . & . & . & . \\ { a }_{ 1,j } & { a }_{ 2,j } & { a }_{ 3,j } & . & . & . & { a }_{ j,i } \end{bmatrix}\)
\({ A }_{ (i,j) }.{ A }_{ (j,i) }^{ T }\quad will\quad result\quad in\quad matrix\quad of\quad i\quad rows\quad and\quad i\quad columns,\quad A.{ A }_{ (i,i) }^{ T }\\ { A }_{ (j,i) }.{ A }_{ (i,j) }^{ T }\quad will\quad result\quad in\quad matrix\quad of\quad j\quad rows\quad and\quad j\quad columns,\quad { A }_{ }^{ T }{ A }_{ (j,j) }\\ Hence,\quad { A }_{ (i,j) }{ A }_{ (j,i) }^{ T }\quad \neq \quad { A }_{ (j,i) }^{ T }{ A }_{ (i,j) }\quad \\ A{ A }_{ (i,i) }^{ T }\quad \neq \quad { A }_{ }^{ T }{ A }_{ (j,j) }\)
Consider the 4x5 matrix
A = matrix(c(1,3,2,2, 3,2,1,1, 2,1,0,1, 1,-1,2,1, 1,4,1,3), nrow=4, ncol=5)
A## [,1] [,2] [,3] [,4] [,5]
## [1,] 1 3 2 1 1
## [2,] 3 2 1 -1 4
## [3,] 2 1 0 2 1
## [4,] 2 1 1 1 3Generate traspose of A:
tA = t(A)
tA## [,1] [,2] [,3] [,4]
## [1,] 1 3 2 2
## [2,] 3 2 1 1
## [3,] 2 1 0 1
## [4,] 1 -1 2 1
## [5,] 1 4 1 3Transpose now is a 5x4 matrix
proving original \(\quad \\ A{ A }_{ (i,i) }^{ T }\quad \neq \quad { A }_{ }^{ T }{ A }_{ (j,j) }\):
A%*%tA## [,1] [,2] [,3] [,4]
## [1,] 16 14 8 11
## [2,] 14 31 10 20
## [3,] 8 10 10 10
## [4,] 11 20 10 16#geenrates 4x4 matrixtA%*%A## [,1] [,2] [,3] [,4] [,5]
## [1,] 18 13 7 4 21
## [2,] 13 15 9 4 15
## [3,] 7 9 6 2 9
## [4,] 4 4 2 7 2
## [5,] 21 15 9 2 27#geenrates 5x5 matrixHence \(\quad \\ A{ A }_{ (i,i) }^{ T }\quad \neq \quad { A }_{ }^{ T }{ A }_{ (j,j) }\) is correct.
Answer number 2
When a matrix is symmetrical square matrix or identity matrix then \({ A }_{ (i,j) }{ A }_{ (j,i) }^{ T }\quad =\quad { A }_{ (j,i) }^{ T }{ A }_{ (i,j) }\)
#4 X 4 matrix
A = matrix(c(9,1,3,6, 1,11,7,6, 3,7,4,1, 6,6,1,10), nrow=4, ncol=4)
A## [,1] [,2] [,3] [,4]
## [1,] 9 1 3 6
## [2,] 1 11 7 6
## [3,] 3 7 4 1
## [4,] 6 6 1 10#transpose of matrix A
#it is similar to A
tA = t(A)
tA## [,1] [,2] [,3] [,4]
## [1,] 9 1 3 6
## [2,] 1 11 7 6
## [3,] 3 7 4 1
## [4,] 6 6 1 10#matrix multiplication
A%*%tA## [,1] [,2] [,3] [,4]
## [1,] 127 77 52 123
## [2,] 77 207 114 139
## [3,] 52 114 75 74
## [4,] 123 139 74 173tA%*%A## [,1] [,2] [,3] [,4]
## [1,] 127 77 52 123
## [2,] 77 207 114 139
## [3,] 52 114 75 74
## [4,] 123 139 74 173Symmetric Square Matrix is one exception to the rule, where (i = j) \({ A }_{ (i,j) }{ A }_{ (j,i) }^{ T }\quad =\quad { A }_{ (j,i) }^{ T }{ A }_{ (i,j) }\quad =\quad A{ A }_{ (i,i) }^{ T }\quad =\quad { A }^{ T }{ A }_{ (j,j) }\)
Similarly for identity matrix
#4 X 4 identity matrix
A = diag(4)
A## [,1] [,2] [,3] [,4]
## [1,] 1 0 0 0
## [2,] 0 1 0 0
## [3,] 0 0 1 0
## [4,] 0 0 0 1#transpose of matrix A
#it is similar to A
tA = t(A)
tA## [,1] [,2] [,3] [,4]
## [1,] 1 0 0 0
## [2,] 0 1 0 0
## [3,] 0 0 1 0
## [4,] 0 0 0 1#matrix multiplication
A%*%tA## [,1] [,2] [,3] [,4]
## [1,] 1 0 0 0
## [2,] 0 1 0 0
## [3,] 0 0 1 0
## [4,] 0 0 0 1tA%*%A## [,1] [,2] [,3] [,4]
## [1,] 1 0 0 0
## [2,] 0 1 0 0
## [3,] 0 0 1 0
## [4,] 0 0 0 1Identity matrix is another exception to the rule, where (i = j),
\({ A }_{ (i,j) }{ A }_{ (j,i) }^{ T }\quad =\quad { A }_{ (j,i) }^{ T }{ A }_{ (i,j) }\quad =\quad A{ A }_{ (i,i) }^{ T }\quad =\quad { A }^{ T }{ A }_{ (j,j) }\)
Problem set 2
knitr::include_graphics("C:\\Users\\sergioor\\Desktop\\Old PC\\CUNY\\DATA605\\hw2ex2.jpg")
Following function accepts a 5 X 5 square matrix and diagonal value to create identity matrix. It returns two matrices lower(L) and upper(U) decomposition
lud_function <- function(a,diagonal){
#initiate identity matrix
l = diag(diagonal)
u = a
#check if a[1,1] equal zero move it following row
for (i in 1:nrow(u)){
if (u[1,1] == 0){
if (i+1 <= nrow(u)){
swap = u[i+1,]
u[i+1,] = u[1,]
u[1,] = swap
if (u[1,1] != 0){
break
}
}
}
}
#Loop through every column and every row.
#While working with column number 1 row number should be 2 (row > column)
#For the first row, leave values as is
#For the second row make first value zero
#For the third row make first and second column values zero
#Basically, we are doing row reduction echelon form, converting numbers on the lower side of diagonal to zero.
#And saving factor value on lower side of identity matrix.
k = 0
#read columns
for (j in 1:ncol(a)){
k = k + 1
#read rows
for (i in 1:nrow(u)){
#if row > column and value for the row is not equal to zero
if (i > j){
if (u[i,j] != 0){
#initiate the sign of the factor
s = -1
if (u[i,j] < 0){
if (u[k,j] < 0){
#when both numbers are negative keep sign positive
s = 1
}
}
#calculate the factor
f = s * u[i,j]/u[k,j]
#multiply upper row with the factor and sign
swap1 = s * (u[k,] * u[i,j])/u[k,j]
#get current row
swap2 = u[i,]
#add modified upper row to current row.
#this should make row,column value to zero when addition is performed
u[i,] = swap1 + swap2
#record the factor
l[i,j] = -1*f
}
}
}
}
return(list(l=l,u=u))
}
#get matrix
#a = matrix(c(1,-2,3, 4,8,4, -3,5,7), nrow=3, ncol=3)
#l = diag(3)
#a = matrix(c(1,2,0,0, 5,12,4,0, 0,5,13,6, 0,0,5,11), nrow=4, ncol=4)
#l = diag(4)
a = matrix(c(2,6,6,4,2, 3,1,3,2,1, 4,3,1,4,2, 1,1,2,7,4, 3,2,5,8,2), nrow=5, ncol=5)
m = lud_function(a,5)Lower Matrix
#Lower matrix
m$l## [,1] [,2] [,3] [,4] [,5]
## [1,] 1 0.00 0.0000000 0.000000 0
## [2,] 3 1.00 0.0000000 0.000000 0
## [3,] 3 -0.75 1.0000000 0.000000 0
## [4,] 2 -0.50 -0.4788732 1.000000 0
## [5,] 1 -0.25 -0.2394366 0.678392 1Upper Matric
#Upper matrix
m$u## [,1] [,2] [,3] [,4] [,5]
## [1,] 2 3.000000 4.000000 1.000000e+00 3.0000000
## [2,] 0 -8.000000 -9.000000 -2.000000e+00 -7.0000000
## [3,] 0 -12.000000 -17.750000 -2.500000e+00 -9.2500000
## [4,] 0 -13.746479 -17.000000 2.802817e+00 -5.9295775
## [5,] 0 2.452261 3.032663 -2.220446e-16 -0.9422111Original Matrix for proof
a## [,1] [,2] [,3] [,4] [,5]
## [1,] 2 3 4 1 3
## [2,] 6 1 3 1 2
## [3,] 6 3 1 2 5
## [4,] 4 2 4 7 8
## [5,] 2 1 2 4 2Multiplying L and U and we have the original Matrix back
#multipication of L and U
m$l%*%m$u## [,1] [,2] [,3] [,4] [,5]
## [1,] 2 3 4 1 3
## [2,] 6 1 3 1 2
## [3,] 6 3 1 2 5
## [4,] 4 2 4 7 8
## [5,] 2 1 2 4 2