Show that \(A^{T}A \neq AA^{T}\) in
general. (Proof and demonstration).
Solution
For the expression \(A^{T}A = AA^{T}\) to be true in general it must be true for all A where A is a matrix. So, if we can get at least one A that does not satisfy the expression, then the expression is not true in general for all A. Suppose a matrix \(A\) is an \(m \times n\) matrix, its transpose \(A^{T}\) will be an \(n \times m\) matrix
Let \(A\) be a \(2 \times 2\) square matrix: \[ A = \begin{bmatrix} u & v \\ x & y \\ \end{bmatrix} \] The transpose of A will be \[ A^T = \begin{bmatrix} u & x \\ v & y \\ \end{bmatrix} \]
The multiplication of \(A^T\) by \(A\), \(A^TA\) is given by \[ A^TA = \begin{bmatrix} u^2+x^2 & uv+xy \\ uv+xy & v^2+y^2 \\ \end{bmatrix} \] The multiplication of \(A\) by \(A^T\), \(AA^T\) is given by
\[
AA^T =
\begin{bmatrix}
u^2+v^2 & ux+vy \\
ux+vy & x^2+y^2 \\
\end{bmatrix}
\] We can see that unless x = v, the two would not be
equal.
Therefore,
\[ \begin{bmatrix} u^2+x^2 & uv+xy \\ uv+xy & v^2+y^2 \\ \end{bmatrix} \neq \begin{bmatrix} u^2+v^2 & ux+vy \\ ux+vy & x^2+y^2 \\ \end{bmatrix} \] Therefore, \(A^{T}A \neq AA^{T}\) in general.
Demonstration with 2 x 2 numeric valued matrices
# Let A be a 2 x 2 matrix
A = matrix(c(2,4,6,8), 2, 2)
A_T = t(A) # A transpose
A_TA = A_T %*% A # A transpose multiplied by A
AA_T = A %*% A_T # A multiplied by A transposeA_TA## [,1] [,2]
## [1,] 20 44
## [2,] 44 100AA_T## [,1] [,2]
## [1,] 40 56
## [2,] 56 80# check if the two matrices are equal
AA_T == A_TA## [,1] [,2]
## [1,] FALSE FALSE
## [2,] FALSE FALSE# check if the two matrices are equal
identical(A_TA, AA_T)## [1] FALSEDemonstration with 3 x 3 numeric valued matrices
A = matrix(c(1,2,3,4,5,6,7,8,9), 3, 3)
A_T = t(A)
A_TA = A_T %*% A
AA_T = A %*% A_TA_TA## [,1] [,2] [,3]
## [1,] 14 32 50
## [2,] 32 77 122
## [3,] 50 122 194AA_T## [,1] [,2] [,3]
## [1,] 66 78 90
## [2,] 78 93 108
## [3,] 90 108 126# check if the two matrices are equal
AA_T == A_TA## [,1] [,2] [,3]
## [1,] FALSE FALSE FALSE
## [2,] FALSE FALSE FALSE
## [3,] FALSE FALSE FALSE# check if the two matrices are identical
identical(A_TA, AA_T)## [1] FALSEFor a special type of square matrix A, we get \(A^{T}A = AA^{T}\). Under what conditions
could this be true? (Hint: The Identity matrix I is an example of such a
matrix).
Solution
The condition for which we get the above is for a symmetric matrix.
# Let A be a symmetric matrix
A = matrix(c(3,-2,4,-2,6,2,4,2,3), 3, 3)
A## [,1] [,2] [,3]
## [1,] 3 -2 4
## [2,] -2 6 2
## [3,] 4 2 3# check if the matrix is symmetric
isSymmetric.matrix(A)## [1] TRUEA_T = t(A)
A_TA = A_T %*% A
AA_T = A %*% A_TA_TA## [,1] [,2] [,3]
## [1,] 29 -10 20
## [2,] -10 44 10
## [3,] 20 10 29AA_T## [,1] [,2] [,3]
## [1,] 29 -10 20
## [2,] -10 44 10
## [3,] 20 10 29# check if the two matrices are equal
AA_T == A_TA## [,1] [,2] [,3]
## [1,] TRUE TRUE TRUE
## [2,] TRUE TRUE TRUE
## [3,] TRUE TRUE TRUE# check if the two matrices are identical
identical(A_TA, AA_T)## [1] TRUEWrite an R function to factorize a square matrix A into LU or LDU.
Solution
LU_factorizer <- function(X) {
m <- dim(X)[1]
n <- dim(X)[2]
L <- diag(m)
U <- X
# Check to ensure that M is a square matrix
if (m != n) {
return(cat("This is a not a square matrix. \nThis function works for only square matrices"))
}
if (m > 3 | n > 3){
return("Please enter a square matrix of lower dimension")
}
# If dimension is 1, then U=A and L=[1]
if (n==1 && m==1) {
return(list(L,U))
}
# Check if zero is in diagonal of the provided matrix
if (0 %in% diag(X)){
return("This fuction does not work if zero is in the diagonal of the matrix")
}
# Loop through the lower triangle to determine the multiplier for each position:
for(i in 2:m) {
for(j in 1:(i-1)) {
multiplier <- -U[i,j] / U[j,j]
U[i, ] <- multiplier * U[j, ] + U[i, ]
L[i,j] <- -multiplier
}
}
return(list(L,U))
}Demonstration with a non-squared matrix
# Test for a non-square matrix
A_nonsquare = matrix(c(1, 2, 3, 4, 5, 6), nrow = 2, ncol = 3)
LU_factorizer(A_nonsquare)## This is a not a square matrix.
## This function works for only square matricesDemonstration with 3 x 3 numeric valued matrices
# Let's declare a random 5 x 5 matrix, Y
Y = matrix(1:9,ncol=3)
Y## [,1] [,2] [,3]
## [1,] 1 4 7
## [2,] 2 5 8
## [3,] 3 6 9Apply the LU_factorizer function on Y
LU = LU_factorizer(Y)
L = LU[[1]]
U = LU[[2]]Display the L and U decomposition
# display L
L## [,1] [,2] [,3]
## [1,] 1 0 0
## [2,] 2 1 0
## [3,] 3 2 1U## [,1] [,2] [,3]
## [1,] 1 4 7
## [2,] 0 -3 -6
## [3,] 0 0 0LU_result = L %*% U
LU_result## [,1] [,2] [,3]
## [1,] 1 4 7
## [2,] 2 5 8
## [3,] 3 6 9# check if Y = LU
Y == LU_result## [,1] [,2] [,3]
## [1,] TRUE TRUE TRUE
## [2,] TRUE TRUE TRUE
## [3,] TRUE TRUE TRUEDemonstration with a random 3x3 matrix
set.seed(94)
X <- matrix(sample(c(1:10), size = 9),nrow = 3)
X## [,1] [,2] [,3]
## [1,] 2 6 10
## [2,] 7 4 3
## [3,] 9 8 1Apply the LU_factorizer function on Y
LU = LU_factorizer(X)
L = LU[[1]]
U = LU[[2]]Display the L and U decomposition
# display L
L## [,1] [,2] [,3]
## [1,] 1.0 0.000000 0
## [2,] 3.5 1.000000 0
## [3,] 4.5 1.117647 1U## [,1] [,2] [,3]
## [1,] 2 6 10.000000
## [2,] 0 -17 -32.000000
## [3,] 0 0 -8.235294LU_result = L %*% U
LU_result## [,1] [,2] [,3]
## [1,] 2 6 10
## [2,] 7 4 3
## [3,] 9 8 1# check if X = LU
X == LU_result## [,1] [,2] [,3]
## [1,] TRUE TRUE TRUE
## [2,] TRUE TRUE TRUE
## [3,] TRUE TRUE TRUE