Transpose

1.Show that A\(^{T}\)A \(\neq\) AA\(^{T}\)

For any given m\(\times\)n matrix A, A\(^{T}\) is a n\(\times\)m matrix where each element a\(_{ij}\) in A\(^{T}\) is equal to a\(_{ji}\) in matrix A.

A = \(\begin{bmatrix}a_{11} & a_{12} & a_{13}\\a_{21} & a_{22} & a_{23}\\a_{31} & a_{32} & a_{33}\end{bmatrix}\) and A\(^{T}\) = \(\begin{bmatrix}a_{11} & a_{21} & a_{31}\\a_{12} & a_{22} & a_{32}\\a_{13} & a_{23} & a_{33}\end{bmatrix}\)

\(\because\) All elements in A\(^{T}\) don’t eaqual to corresponding elements in A except elements on the diagonal. \(\therefore\) A\(\neq\)A\(^{T}\)

Based on Properties of matrix multiplication, \(\because\) AB\(\neq\)BA for any given matrix A, B where A\(\neq\)B \(\therefore\) A\(^{T}\)A \(\neq\) AA\(^{T}\)

2.For a special type of square matrix A, we get A\(^{T}\)A = AA\(^{T}\). Under what conditions could this be true?

To satisfy A\(^{T}\)A = AA\(^{T}\), A must satisfy A = A\(^{T}\) so that AA\(^{T}\) = AA = A\(^{T}\)A\(^{T}\) = A\(^{T}\)A and therefore, for each a\(_{ij}\) in A must equal to a\(_{ji}\).

For example:
A = \(\begin{bmatrix}1 & 2 & 3\\2 & 1 & 4\\3 & 4 & 1\end{bmatrix}\) and A\(^{T}\) = \(\begin{bmatrix}1 & 2 & 3\\2 & 1 & 4\\3 & 4 & 1\end{bmatrix}\) = A

I = \(\begin{bmatrix}1 & 0 & 0\\0 & 1 & 0\\0 & 0 & 1\end{bmatrix}\) and I\(^{T}\) = \(\begin{bmatrix}1 & 0 & 0\\0 & 1 & 0\\0 & 0 & 1\end{bmatrix}\) = I

LU Decomposition

Write an R function to factorize a square matrix A into LU

LUDecomposition <- function(matrixA){
  r <- nrow(matrixA)
  matrixL <- diag(r)#create an identity matrix of the same size as A
  matrixU <- matrixA
  i <- 2
  j <- 1
  k <- 1
  while(k <= r) {
    while (i <= r) {
      if (matrixU[i,j]!=0) {
        multiplier <- (-matrixU[i,j]/matrixU[k,j])
        matrixL[i,j] <- -multiplier
            while (j <= r) {
              matrixU[i,j] <- matrixU[i,j] + matrixU[k,j]*multiplier
              j <- j+1
            }
        }
          i <- i+1
          j <- 1
      }
    k <- k+1
    i <- k+1
    j <- k
  }
  matrixList <- list(matrixL,matrixU)
  return (matrixList)
}

Test the function with a 3 x 3 matrix

testA <- matrix(c(2,1,-6,4,-4,-9,-4,3,5),nrow = 3)
testL <- LUDecomposition(testA)[[1]]
testU <- LUDecomposition(testA)[[2]]
testA

##      [,1] [,2] [,3]
## [1,]    2    4   -4
## [2,]    1   -4    3
## [3,]   -6   -9    5

testL

##      [,1] [,2] [,3]
## [1,]  1.0  0.0    0
## [2,]  0.5  1.0    0
## [3,] -3.0 -0.5    1

testU

##      [,1] [,2] [,3]
## [1,]    2    4 -4.0
## [2,]    0   -6  5.0
## [3,]    0    0 -4.5

testL%*%testU

##      [,1] [,2] [,3]
## [1,]    2    4   -4
## [2,]    1   -4    3
## [3,]   -6   -9    5