605 HW4

Here, we can input the matrix \(A\) and use it to create to special matrices X and Y where \(X = AA^T\) and \(Y=A^TA\).

A <- matrix(c(1,-1,2,0,3,4), nrow=2, ncol=3) #Matrix A
A #manual check via console

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

X = A %*% t(A) # A*A^T
Y = t(A) %*% A #A A^T*A

Then, we can computer the eigenvectors of x and y as well as the svd of

eig_x = eigen(X) 
eig_x$vectors

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

svd(A)$u #left handed vector

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

eig_y = eigen(Y)
eig_y$vectors[0:1]

[1] -0.01856629

svd(A)$v #right handed one

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

We can also see that the eigenvalues of the svd (X), the significant ones for Y, and the singular square values of A are all the same.

svd(X)$d # eigenvalue of svd(X)

[1] 26.601802  4.398198

svd(Y)$d[1:2] # eigenvalue of svd(Y)

[1] 26.601802  4.398198

squares <- c(svd(A)$d)# single values of A
squares <- (squares^2) #squaring the squares vector
squares

[1] 26.601802  4.398198

Problem 2

eye <- diag(rep(1,3))
myinverse <- function(A){
  if (dim(A)[1]==dim(A)[2] & det(A)!=0 ){
    C <- A
    for(i in 1:dim(A)[1]){
      for(j in 1:dim(A)[1]){
        C[i,j] <- (det(A[-i,-j])*(-1)^(i+j))
      }
    }
    B  <- t(C)/det(A) 
    return(B)
  }
    else{
    return(-1)
    }
}
if (all(myinverse(eye)== eye)){ #identity matrix should equal it's own inverse. Otherwise, my function is broken.
  print('function passed!') 
}

[1] "function passed!"