Objective: Show that $ATA ≠ \(AAT\)in general and discuss conditions under which \(ATA = AAT\)
Matrix multiplication is non-commutative, meaning that the order of multiplication matters. For instance, given a matrix A:
A * A^T = \[\begin{bmatrix} a_{11} & a_{12} & \dots & a_{1n} \\ a_{21} & a_{22} & \dots & a_{2n} \\ \vdots & \vdots & \ddots & \vdots \\ a_{m1} & a_{m2} & \dots & a_{mn} \end{bmatrix} \cdot \begin{bmatrix} a_{11} & a_{21} & \dots & a_{m1} \\ a_{12} & a_{22} & \dots & a_{m2} \\ \vdots & \vdots & \ddots & \vdots \\ a_{1n} & a_{2n} & \dots & a_{mn} \end{bmatrix}\]\((A^T A)_{ij} = \sum_{k=1}^{n} a_{ik} a_{jk}\)
It is evident that if \(n \neq m\) then \(A^TA\)and \(AA^T\) cannot be equal due to their different dimensions.
Symmetric matrices include diagonal square matrices, such as the identity matrix I. For instance:
\[I = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{bmatrix}\]
For example, suppose we have \(A\) and \(A^T\) as follows:
Let \(B\) be a matrix given by: \[ B = \begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \end{bmatrix} \]
The transpose of \(B\), denoted as \(B^T\), is: \[ B^T = \begin{bmatrix} 1 & 4 \\ 2 & 5 \\ 3 & 6 \end{bmatrix} \]
Now, let’s find the product \(BB^T\): \[ BB^T = \begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \end{bmatrix} \begin{bmatrix} 1 & 4 \\ 2 & 5 \\ 3 & 6 \end{bmatrix} \]
The element at position \((i, j)\) in \(BB^T\) is given by: \[ (BB^T)_{ij} = \sum_{k=1}^{3} b_{ik} \cdot (b_{jk})^T \]
After performing the calculations, the result is a \(2 \times 2\) matrix: \[ BB^T = \begin{bmatrix} 14 & 32 \\ 32 & 77 \end{bmatrix} \]
Next, let’s find the product \(B^TB\): \[ B^TB = \begin{bmatrix} 1 & 4 \\ 2 & 5 \\ 3 & 6 \end{bmatrix} \begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \end{bmatrix} \]
The element at position \((i, j)\) in \(B^TB\) is given by: \[ (B^TB)_{ij} = \sum_{k=1}^{3} (b_{ik})^T \cdot b_{kj} \]
After performing the calculations, the result is a \(3 \times 3\) matrix: \[ B^TB = \begin{bmatrix} 17 & 22 & 27 \\ 22 & 29 & 36 \\ 27 & 36 & 45 \end{bmatrix} \]
As in the previous example, \(BB^T\) and \(B^TB\) are not equal. Therefore, in general, \(BB^T \neq B^TB\) as \(B^T \neq B\). ***
For a special type of square matrix we get ATA=AAT. Under what conditions could this be true? (Hint: The Identity matrix I is an example of such a matrix)
Let \(C\) be a square matrix such that \(C^T = C\). This means that \(C\) is a symmetric matrix.
\[ C = \begin{bmatrix} c_{11} & c_{12} & \dots & c_{1n} \\ c_{21} & c_{22} & \dots & c_{2n} \\ \vdots & \vdots & \ddots & \vdots \\ c_{n1} & c_{n2} & \dots & c_{nn} \end{bmatrix} \]
Now, let’s find \(C^TC\) and \(CC^T\):
The element at position \((i, j)\) in \(C^TC\) is given by: \[ (C^TC)_{ij} = \sum_{k=1}^{n} c_{ki} \cdot c_{kj} \]
The element at position \((i, j)\) in \(CC^T\) is given by: \[ (CC^T)_{ij} = \sum_{k=1}^{n} c_{ik} \cdot c_{jk} \]
Since \(C^T = C\), the elements of \(C^TC\) and \(CC^T\) will be the same. Therefore, in this case, \(C^TC = CC^T\) for any symmetric square matrix \(C\). This includes diagonal matrices and, in particular, the identity matrix \(I\).
Define the LU or LDU Factorization Function
luOrLduFactorize <- function(A, ldu = FALSE) {
n <- nrow(A)
U <- A
L <- diag(n)
D <- diag(n)
for (i in 1:(n - 1)) {
for (j in (i + 1):n) {
L[j, i] = U[j, i] / U[i, i]
U[j, ] = U[j, ] - L[j, i] * U[i, ]
if (ldu) {
D[j, j] = U[j, j]
U[j, j] = 1 # Set diagonal elements of U to 1 in LDU factorization
}
}
}
if (ldu) {
result <- list("L" = L, "D" = D, "U" = U)
} else {
result <- list("L" = L, "U" = U)
}
return(result)
}Factorize a Square Matrix A into LU
A <- matrix(c(1, 2, 3, 1, 1, 1, 2, 0, 1), nrow = 3)
luResult <- luOrLduFactorize(A)
# Check if LU factorization is correct
if (all(luResult$L %*% luResult$U == A)) {
cat("LU Factorization is correct:\n")
print("Original Matrix A:")
print(A)
print("Lower Triangular Matrix L:")
print(luResult$L)
print("Upper Triangular Matrix U:")
print(luResult$U)
} else {
cat("LU Factorization is incorrect.\n")
}## LU Factorization is correct:
## [1] "Original Matrix A:"
## [,1] [,2] [,3]
## [1,] 1 1 2
## [2,] 2 1 0
## [3,] 3 1 1
## [1] "Lower Triangular Matrix L:"
## [,1] [,2] [,3]
## [1,] 1 0 0
## [2,] 2 1 0
## [3,] 3 2 1
## [1] "Upper Triangular Matrix U:"
## [,1] [,2] [,3]
## [1,] 1 1 2
## [2,] 0 -1 -4
## [3,] 0 0 3# Factorize a Square Matrix A into LDU
A <- matrix(c(1, 2, 3, 1, 1, 1, 2, 0, 1), nrow = 3)
lduResult <- luOrLduFactorize(A, ldu = TRUE)Check if LDU factorization is correct
if (all(lduResult$L %*% lduResult$D %*% lduResult$U == A)) {
cat("LDU Factorization is correct:\n")
print("Original Matrix A:")
print(A)
print("Lower Triangular Matrix L:")
print(lduResult$L)
print("Diagonal Matrix D:")
print(lduResult$D)
print("Upper Triangular Matrix U:")
print(lduResult$U)
} else {
cat("LDU Factorization is incorrect.\n")
}## LDU Factorization is incorrect.