Solution:
Proof:
Suppose we have A that is a m x n matrix. The transpose of A, \({ A }^{ T }\), would be a n x m matrix. Using the definition of matrix multiplication, we can multiply \({ A }^{ T }\) by A which would result in a n x n matrix; a square matrix. If we multiply A by \({ A }^{ T }\) this would result in a m x m matrix; also square matrix. Since you end up with two different square matrices, one that is n x n and the other that is m x m, \({ A }^{ T } A \neq A{ A }^{ T }\).
Demonstration:
Let A be a 2 x 3 matrix. Then the transpose of A is \({ A }^{ T }\), a 3 x 2 matrix. Then \({ A }^{ T }A\) would be a 3 x 3 matrix and \(A{ A }^{ T }\) would be a 2 x 2 matrix.
A = \(\begin{bmatrix} 1 & 2\quad 3 \\ 3 & 2\quad 1 \end{bmatrix}\)
\({ A }^{ T }\) = \(\begin{bmatrix} 1 & 3 \\ 2 & 2 \\ 3 & 1 \end{bmatrix}\)
\({ A }^{ T }A\) = \(\begin{bmatrix} 1 & 3 \\ 2 & 2 \\ 3 & 1 \end{bmatrix}\) * \(\begin{bmatrix} 1 & 2\quad 3 \\ 3 & 2\quad 1 \end{bmatrix}\)
= \(\begin{bmatrix} 10 & 8 & 6 \\ 8 & 8 & 8 \\ 6 & 8 & 10 \end{bmatrix}\)
\(A{ A }^{ T }\) = \(\begin{bmatrix} 1 & 2\quad 3 \\ 3 & 2\quad 1 \end{bmatrix}\) * \(\begin{bmatrix} 1 & 3 \\ 2 & 2 \\ 3 & 1 \end{bmatrix}\)
= \(\begin{bmatrix} 14 & 10 \\ 10 & 14 \end{bmatrix}\)
As you can see above, \({ A }^{ T }A \neq A{ A }^{ T }\).
Solution:
The conditions that could make \({ A }^{ T }A = A{ A }^{ T }\) for a special type of square matrix A is if A and \({ A }^{ T }\) are equal to each other. Since A and \({ A }^{ T }\) are equal to each other, this would mean the special type of square matrix would be a symmetric matrix. For example the Identity matrix is a special type of square matrix that shows this since it is a square matrix whose transpose would be equal to itself. In conclusion, if a square matrix A is a symmetric matrix, meaning that A and \({ A }^{ T }\) are equal to each other, then \({ A }^{ T }A = A{ A }^{ T }\).
Matrix_Factorization <- function(A) {
x <- dim(A)[1]
D <- matrix(diag(x), nrow = x, ncol = x)
L <- D
U <- A
for (b in 1:(x - 1)) {
for (a in (b + 1):x) {
L[a,b] <- U[a,b] / U[b,b]
U[a,] <- U[a,] - U[b,] * L[a,b]
}
}
diag(D) <- diag(U)
for (c in 1:x) {
U[c,] <- U[c,] / U[c,c]
}
LDU = list("Lower" = L, "Diagonal" = D, "Upper" = U)
return(LDU)
}Test to see if function works:
A <- matrix(c(1, 3, 4, 2, 3, 1, 4, 1, 3), nrow = 3, byrow = TRUE)
Matrix_Factorization(A)## $Lower
## [,1] [,2] [,3]
## [1,] 1 0.000000 0
## [2,] 2 1.000000 0
## [3,] 4 3.666667 1
##
## $Diagonal
## [,1] [,2] [,3]
## [1,] 1 0 0.00000
## [2,] 0 -3 0.00000
## [3,] 0 0 12.66667
##
## $Upper
## [,1] [,2] [,3]
## [1,] 1 3 4.000000
## [2,] 0 1 2.333333
## [3,] 0 0 1.000000