Problem Set 1

Problem 1 - Prove \(A^{T}A\neq AA^{T}\)

Proof by induction:

Assume: \(A^{T}A = AA^{T}\)

Then for rows i in A and columns j in \(A^{T}\), \(A_{i1}\cdot A^{T}_{1j} = A^{T}_{i1}\cdot A_{1j}\). This is true if and only if \(A_{i1} = A^{T}_{1j}\), such that \(A = A^{T}\). Therefore, the assumption is false outside of the rare case of identical matricies. Thus, \(A^{T}A\neq AA^{T}\).

Example:

Consider the following; \(A =\begin{bmatrix}1 & 1 & 1\\ 2 &2 & 2 \\ 3 & 3 & 3\end{bmatrix}\) and \(A^{T} =\begin{bmatrix} 1 & 2 & 3\\ 1 &2 & 3 \\ 1 & 2 & 3\end{bmatrix}\)

Here if we take the inner product of the first columns of \(A\) and the first row of \(A^{T}\) we get \(A_{i1} = 14\). Similarly, we take the first inner product of \(A^{T}\) and \(A\) we get \(A^{T}_{1j} = 3\), such at \(A_{i1} \neq A^{T}_{1j}\) and \(A^{T}A\neq AA^{T}\).

Problem 2 -Under what circumstances could Problem 1 be true?

As mentioned in the proof, this is true for any matrix \(A\) that is elementwise identical to its transpose \(A^{T}\). That is, \(A = A^{T}\).

Problem Set 2

Matrix Factorization

Write a function that factorizes any square matrix less than dimension 5x5. No permuting rows necessary.

matrix.factorize<- function(input_mat){
  mat_L<-diag(dim(input_mat)[1])
  row_idx<-1
  for(j in 1:(dim(input_mat)[2]-1)){
    for(i in 1:(dim(input_mat)[1]-row_idx)){
      gcm<-(input_mat[i+row_idx,j]/input_mat[j,j])*(-1)
      augment_vec<- (gcm*input_mat[row_idx,])+input_mat[i+row_idx,]
      input_mat[i+row_idx,]<-augment_vec
      mat_L[i+row_idx,j]<-gcm*(-1)
    }
    row_idx<-row_idx+1
  }
  return(list(input_mat, mat_L))
}

Test the known example from the video posted in “Weekly Materials”.

a<-matrix(c(2,4,-4,1,-4,3,-6,-9,5), byrow = T, nrow=3)
soln<-matrix.factorize(a)
soln[[2]]%*%soln[[1]]
##      [,1] [,2] [,3]
## [1,]    2    4   -4
## [2,]    1   -4    3
## [3,]   -6   -9    5

Returns the input matrix under multiplication such that, \(A = LU\).

Let’s try one more from the video.

b<-matrix(c(1,4,-3,-2,8,5,3,4,7), byrow = T, nrow=3)
soln_2<-matrix.factorize(b)
soln_2[[2]]%*%soln_2[[1]]
##      [,1] [,2] [,3]
## [1,]    1    4   -3
## [2,]   -2    8    5
## [3,]    3    4    7

Solid.