Problem Set 1

Verify using R that SVD and Eigenvalues are related

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

Compute the left-singular, singular values, and right-singular vectors of A using the svd command.

The vector d contains the singular values The matrix u contains the left singular vectors (rows). The matrix v contains the right singular vectors (columns).

A_svd <- svd(A, nu=nrow(A), nv=ncol(A))
singular_values <- as.matrix(diag(A_svd$d, nrow=nrow(A), ncol=ncol(A)))
left_singluar_vectors <- as.matrix(A_svd$u)
right_singluar_vectors <- as.matrix(A_svd$v)

Singular values from SVD

singular_values
##          [,1]     [,2] [,3]
## [1,] 5.157693 0.000000    0
## [2,] 0.000000 2.097188    0

The left singular vectors from SVD function

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

The right singular vectors from SVD function

right_singluar_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
  1. Examine the two sets of singular vectors and show that they are indeed eigenvectors of X and Y.

Compare left singular vector by computing AAT and taking its eigenvector:

X <- A %*% t(A)
X_eigen <- eigen(X) #AAT
X_left_singluar_vectors <- as.matrix(X_eigen$vectors)
X_eigenvalues <- X_eigen$values

let us compare the two

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

The numbers are the same but there are differences in signs.

The square root of the eigenvalues of X match the singular values of the SVD.

singular_values
##          [,1]     [,2] [,3]
## [1,] 5.157693 0.000000    0
## [2,] 0.000000 2.097188    0
diag(sqrt(X_eigenvalues), nrow=nrow(A), ncol=ncol(A))
##          [,1]     [,2] [,3]
## [1,] 5.157693 0.000000    0
## [2,] 0.000000 2.097188    0

Let us now compare right singular vector by computing ATA and taking its eigenvector

Y <- t(A)%*% A #ATA
Y_eigen <- eigen(Y)
Y_right_singluar_vectors <- as.matrix(Y_eigen$vectors)
Y_eigenvalues <- Y_eigen$values

The SVD right singular vectors and ATA eigenvectors match but there are sign differences

right_singluar_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
Y_right_singluar_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

The eigenvalues of X and Y match for nonzero values

round(X_eigenvalues,3)
## [1] 26.602  4.398
round(Y_eigenvalues,3)
## [1] 26.602  4.398  0.000

Problem Set 2

#The function to get cofactors
getcofactors <- function(M) {
  cofactors <- M
  for(i in 1:nrow(M)){
    for(j in 1:ncol(M)){
      cofactors[i,j] <- (det(M[-i,-j])*(-1)^(i+j)) 
    }
  }
  return(cofactors) # output of cofactors matrix
}

#The function to get the inverse of a square matrix that is invertible
myinverse <- function(M){
  M_cofactors <- getcofactors(M)
  M_cofactors_transpose <- t(M_cofactors)
  M_det <- det(M)
  return(M_cofactors_transpose/M_det)
}

Testing the function

mat<- matrix(c(4,3,7,8,6,2,2,1,6,1,2,1,2,3,6,7), nrow=4, ncol=4, byrow=TRUE)
mat
##      [,1] [,2] [,3] [,4]
## [1,]    4    3    7    8
## [2,]    6    2    2    1
## [3,]    6    1    2    1
## [4,]    2    3    6    7
mat_inverse<-myinverse(mat)
mat_inverse
##      [,1] [,2]   [,3]   [,4]
## [1,]    1  0.5 -0.625 -1.125
## [2,]    0  1.0 -1.000  0.000
## [3,]   -5 -3.0  4.500  5.500
## [4,]    4  2.0 -3.250 -4.250

Check if mat*mat_inverse=Identity

print(mat%*%mat_inverse)
##               [,1]          [,2]         [,3]         [,4]
## [1,]  1.000000e+00  7.105427e-15 7.105427e-15 7.105427e-15
## [2,] -3.552714e-15  1.000000e+00 1.065814e-14 1.776357e-15
## [3,] -3.552714e-15 -4.440892e-16 1.000000e+00 1.776357e-15
## [4,]  0.000000e+00  1.776357e-15 1.421085e-14 1.000000e+00