Dat 605 HW 4 - SVD Decomposition Exercise

Prob Set 1

• SVD Decomposition Step by Step

library(matlib)
library(MASS)

Setting up A mxn Matrix

• X = AAT

• Y = ATA

A <- matrix(c(1, 2, 3, -1,0,4), nrow = 2, ncol = 3, byrow=T)

A

##      [,1] [,2] [,3]
## [1,]    1    2    3
## [2,]   -1    0    4

X <- A%*%t(A) #AAT

X

##      [,1] [,2]
## [1,]   14   11
## [2,]   11   17

Y <- t(A)%*%A #ATA

Y

##      [,1] [,2] [,3]
## [1,]    2    2   -1
## [2,]    2    4    6
## [3,]   -1    6   25

• Eigenvectors of ATA matrix

Y.e <- eigen(Y) #eigen of ATA

values.mat <-Y.e$values
values.mat #this will show that the last eignevalue is virtually zero

## [1] 2.660180e+01 4.398198e+00 1.058982e-16

vector.mat <- Y.e$vectors #eignen vectors of ATA

#vector.mat[,1:2] <- vector.mat[,1:2]*-1

vector.moat <- vector.mat[,1:2]

vector.moat

##             [,1]       [,2]
## [1,] -0.01856629 -0.6727903
## [2,]  0.25499937 -0.7184510
## [3,]  0.96676296  0.1765824

• Performing the same routine for AAT; X Matrix

X.e <- eigen(X) #eigen of AAT

uvalues.mat <- X.e$values #eigen values of AAT
uvector.mat <- X.e$vectors #eigen vectors of AAT

uvector.mat

##           [,1]       [,2]
## [1,] 0.6576043 -0.7533635
## [2,] 0.7533635  0.6576043

• The singular values r are the square roots of the non-zero eigenvalues of the AAT and ATA matrices

r <- sqrt(Y.e$values)

r <- r*diag(length(r))[,1:2]

r <-r[1:2,]

r

##          [,1]     [,2]
## [1,] 5.157693 0.000000
## [2,] 0.000000 2.097188

---

• The manual step by step process aligns with the output of the svd() function below.

• We can also show that the matrix A is indeed equal to the components resulting from singular value decomposition.

---

svd.matrix <- uvector.mat %*% r %*% t(vector.moat)
svd.matrix # this results the same as the original A matrix we began with

##      [,1]          [,2] [,3]
## [1,]    1  2.000000e+00    3
## [2,]   -1 -2.220446e-15    4

Confirmation with R built-in function below:

• The singular value decomposition of the matrix is computed using the R svd() function

A.svd <- svd(A)
A.svd

## $d
## [1] 5.157693 2.097188
## 
## $u
##            [,1]       [,2]
## [1,] -0.6576043 -0.7533635
## [2,] -0.7533635  0.6576043
## 
## $v
##             [,1]       [,2]
## [1,]  0.01856629 -0.6727903
## [2,] -0.25499937 -0.7184510
## [3,] -0.96676296  0.1765824

Prob Set 2

• Inverse of a Matrix using R built-in cofactor functions

myinverse <- function(A1,n){
  
# Pre-populate nxn matrix with zeros
  nrow <- n
  ncol <- n

#create place-holder C matrix of zeros
C <- matrix(0, nrow = n, ncol = n)
  
i <-1
#j<-1
while (i < n+1){
 for (j in 1:n) {
  C[i,j]<- cofactor(A1, i,j)
 }
  i<-i+1
}
#Matrix of Co-factors of original Matrix A
C
#transpose of Cofactor Matrix
print("Cofactors of Matrix A :")
Ctranspose <- t(C)

print(Ctranspose)

#Determinant of original Matrix A
print("The Determinant of Matrix A :")
Det_A1 <- det(A1)
print(Det_A1)

#Inverse of original Matrix A
print("The Inverse of A is :")

A_Inv <- Ctranspose/Det_A1
print(A_Inv)

print("To verify that AxA(inverse) = Identity :")
I <- A1 %*% A_Inv

return (I)  
}

Test the Function out; Driver Matrix A1

n<-3

A1 <- matrix(c(1,2,3,0,1,4,5,6,0),nrow=3,ncol=3, byrow=T)
print("Original Matrix of A :")

## [1] "Original Matrix of A :"

A1

##      [,1] [,2] [,3]
## [1,]    1    2    3
## [2,]    0    1    4
## [3,]    5    6    0

myinverse(A1,n)

## [1] "Cofactors of Matrix A :"
##      [,1] [,2] [,3]
## [1,]  -24   18    5
## [2,]   20  -15   -4
## [3,]   -5    4    1
## [1] "The Determinant of Matrix A :"
## [1] 1
## [1] "The Inverse of A is :"
##      [,1] [,2] [,3]
## [1,]  -24   18    5
## [2,]   20  -15   -4
## [3,]   -5    4    1
## [1] "To verify that AxA(inverse) = Identity :"

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