HW2

1. Problem set 1

1.1 Show that AT A ̸= AAT in general. (Proof and demonstration.)

\[ A = \begin{bmatrix} a11 & a12 \\ a21 & a22 \end{bmatrix} , B = \begin{bmatrix} a11 & a21 \\ a12 & a22 \end{bmatrix} \]

Proof:

D11: (a11∗a11)+(a21∗a21)≠(a11∗a11)+(a12∗a12)

D21: (a12∗a11)+(a22∗a21)≠(a21∗a11)+(a22∗a12)

D12: (a11∗a12)+(a21∗a22)≠(a11∗a21)+(a12∗a22)

D22: (a12∗a12)+(a22∗a22)≠(a21∗a21)+(a22∗a22)

Demonstration:

A <- matrix(c(5, 8, 0, -33), nrow=2, ncol=2)
A

##      [,1] [,2]
## [1,]    5    0
## [2,]    8  -33

t(A)%*%A

##      [,1] [,2]
## [1,]   89 -264
## [2,] -264 1089

A%*%t(A)

##      [,1] [,2]
## [1,]   25   40
## [2,]   40 1153

1.2 For a special type of square matrix A, we get AT A = AAT . Under what conditions could this be true? (Hint: The Identity matrix I is an example of such a matrix).

ATA=AAT only if a matrix is symmetrical (aij=aji).

A <- matrix(c(4,-3,0,-3,8,12,0,12,-2), nrow=3, ncol=3)
A

##      [,1] [,2] [,3]
## [1,]    4   -3    0
## [2,]   -3    8   12
## [3,]    0   12   -2

t(A)%*%A

##      [,1] [,2] [,3]
## [1,]   25  -36  -36
## [2,]  -36  217   72
## [3,]  -36   72  148

A%*%t(A)

##      [,1] [,2] [,3]
## [1,]   25  -36  -36
## [2,]  -36  217   72
## [3,]  -36   72  148

2. Problem set 2

Write an R function to factorize a square matrix A into LU or LDU, whichever you prefer. You don’t have to worry about permuting rows of A and you can assume that A is less than 5x5, if you need to hard-code any variables in your code.

LU <- function(mtrx){

  L <- diag(nrow(mtrx))
  U <- mtrx

  for (i in 1:(nrow(mtrx) - 1)) {
    for (j in (i+1):nrow(mtrx)) {
      
      E =  U[j,i]/U[i,i]
      # updating lower triangular matrix
      L[j,i] <- E
     # updating upper triangular matrix
      U[j,] <- U[j,] - (U[i,] * (E))
      
  }
  }

  LU_list<- list("L" = L, "U" = U, "L*U" = L%*%U )
  
  return(LU_list)

}

Test

A<-matrix(c(-1,13,6,0,9,-5,3,7,5,-1,3,10,26,-12,0,17),ncol=4,nrow=4)
A

##      [,1] [,2] [,3] [,4]
## [1,]   -1    9    5   26
## [2,]   13   -5   -1  -12
## [3,]    6    3    3    0
## [4,]    0    7   10   17

LU(A)

## $L
##      [,1]      [,2] [,3] [,4]
## [1,]    1 0.0000000    0    0
## [2,]  -13 1.0000000    0    0
## [3,]   -6 0.5089286    1    0
## [4,]    0 0.0625000   14    1
## 
## $U
##      [,1] [,2]       [,3]       [,4]
## [1,]   -1    9  5.0000000  26.000000
## [2,]    0  112 64.0000000 326.000000
## [3,]    0    0  0.4285714  -9.910714
## [4,]    0    0  0.0000000 135.375000
## 
## $`L*U`
##      [,1] [,2] [,3] [,4]
## [1,]   -1    9    5   26
## [2,]   13   -5   -1  -12
## [3,]    6    3    3    0
## [4,]    0    7   10   17