Assignment 4

1. Problem set 1

(1) In this problem, we’ll verify using R that SVD and Eigenvalues are related as worked out in the weekly module. Given a 3 × 2 matrix A write code in R to compute X = AA^T and Y = A^TA. Then, compute the eigenvalues and eigenvectors of X and Y using the built-in commans in R. Then, compute the left-singular, singular values, and right-singular vectors of A using the svd command. Examine the two sets of singular vectors and show that they are indeed eigenvectors of X and Y. In addition, the two non-zero eigenvalues (the 3rd value will be very close to zero, if not zero) of both X and Y are the same and are squares of the non-zero singular values of A.Your code should compute all these vectors and scalars and store them in variables.Please add enough comments in your code to show me how to interpret your steps.

a <- matrix(c(1,-1,2,0,3,4),nrow=2,ncol=3)
a

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

Using built in R function svd, which takes an arbitrary matrix argument and calculates the singular value decomposition. d= eigenvalues u= left-singular eigenvectors v= right-singular eigenvectors

svd(a)

## $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

Computing for x = aa^t

#x = aa^t
x = a %*% t(a)
x

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

svd(x)

## $d
## [1] 26.601802  4.398198
## 
## $u
##            [,1]       [,2]
## [1,] -0.6576043 -0.7533635
## [2,] -0.7533635  0.6576043
## 
## $v
##            [,1]       [,2]
## [1,] -0.6576043 -0.7533635
## [2,] -0.7533635  0.6576043

eigen(x)

## eigen() decomposition
## $values
## [1] 26.601802  4.398198
## 
## $vectors
##           [,1]       [,2]
## [1,] 0.6576043 -0.7533635
## [2,] 0.7533635  0.6576043

computing for y = a^ta

#y = a^ta
y = t(a) %*% a
y

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

svd(y)

## $d
## [1] 2.660180e+01 4.398198e+00 2.836098e-15
## 
## $u
##             [,1]       [,2]       [,3]
## [1,] -0.01856629 -0.6727903  0.7396003
## [2,]  0.25499937 -0.7184510 -0.6471502
## [3,]  0.96676296  0.1765824  0.1849001
## 
## $v
##             [,1]       [,2]       [,3]
## [1,] -0.01856629 -0.6727903 -0.7396003
## [2,]  0.25499937 -0.7184510  0.6471502
## [3,]  0.96676296  0.1765824 -0.1849001

eigen(y)

## eigen() decomposition
## $values
## [1] 2.660180e+01 4.398198e+00 1.058982e-16
## 
## $vectors
##             [,1]       [,2]       [,3]
## [1,] -0.01856629 -0.6727903  0.7396003
## [2,]  0.25499937 -0.7184510 -0.6471502
## [3,]  0.96676296  0.1765824  0.1849001

As, you can see the the left-singular eigenvectors of a equal the eigenvectors of x and right-singular eigenvectors of a are very close to eigenvectors of y.

svd(a)$u 

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

svd(x)$u

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

svd(x)$v

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

svd(a)$v

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

svd(y)$u

##             [,1]       [,2]       [,3]
## [1,] -0.01856629 -0.6727903  0.7396003
## [2,]  0.25499937 -0.7184510 -0.6471502
## [3,]  0.96676296  0.1765824  0.1849001

svd(y)$v

##             [,1]       [,2]       [,3]
## [1,] -0.01856629 -0.6727903 -0.7396003
## [2,]  0.25499937 -0.7184510  0.6471502
## [3,]  0.96676296  0.1765824 -0.1849001

2. Problem set 2

Using the procedure outlined in section 1 of the weekly handout, write a function to compute the inverse of a well-conditioned full-rank square matrix using co-factors. In order to compute the co-factors, you may use built-in commands to compute the determinant.Your function should have the following signature:B = myinverse(A) where A is a matrix and B is its inverse and A×B = I. The off-diagonal elements of I should be close to zero, if not zero. Likewise, the diagonal elements should be close to 1, if not 1. Small numerical precision errors are acceptable but the function myinverse should be correct and must use co-factors and determinant of A to compute the inverse. Please submit PS1 and PS2 in an R-markdown document with your first initial and last name.