Show that: \[A^TA \neq AA^T \]
Proof:
One of the basic laws of matrices is:
\[AB \neq BA \]
This law shows that matrix multiplication is not commutative, therefore \[A^TA \neq AA^T \]
Demonstation:
Generate a random matrix A:
## [,1] [,2] [,3]
## [1,] 5 12 4
## [2,] 13 13 7
## [3,] 8 7 13## [,1] [,2] [,3]
## [1,] 5 13 8
## [2,] 12 13 7
## [3,] 4 7 13\[A^TA:\]
## [,1] [,2] [,3]
## [1,] 258 285 215
## [2,] 285 362 230
## [3,] 215 230 234\[AA^T: \]
## [,1] [,2] [,3]
## [1,] 185 249 176
## [2,] 249 387 286
## [3,] 176 286 282The example above shows that generally, the products of matrix multiplication are not equal when multiplication order is inversed.
For a special type of square matrix A, we get: \[A^TA = AA^T \] Under what conditions could this be true? (Hint: The Identity matrix I is an example of such a matrix).
This is true when the transpose of A is equal to A.
Demonstration:
## [,1] [,2] [,3]
## [1,] 1 0 1
## [2,] 0 1 0
## [3,] 1 0 1## [,1] [,2] [,3]
## [1,] TRUE TRUE TRUE
## [2,] TRUE TRUE TRUE
## [3,] TRUE TRUE TRUEMatrix factorization is a very important problem. There are supercomputers built just to do matrix factorizations. Every second you are on an airplane, matrices are being factorized. Radars that track flights use a technique called Kalman filtering. At the heart of Kalman Filtering is a Matrix Factorization operation. Kalman Filters are solving linear systems of equations when they track your flight using radars. Write an R function to factorize a square matrix A into LU or LDU, whichever you prefer.
To solve this question, I used an algorithm from this site.
lu <- function(matrix){
n = nrow(A)
lower = matrix((c(1:n)*0), nrow=n, ncol=n)
upper = matrix((c(1:n)*0), nrow=n, ncol=n)
for(i in (1:n)) {
for(k in (i:n)) {
sum = 0
for(j in (1:i)) {
sum = sum + (lower[i,j] * upper[j, k])
}
upper[i, k] = A[i, k] - sum
}
for(k in (i:n)) {
if(i == k) {
lower[i, i] = 1
} else {
sum = 0
for(j in (1:i)) {
sum = sum + (lower[k, j] * upper[j, i])
}
lower[k, i] = (A[k, i] - sum) / upper[i, i]
}
}
}
return(list(lower, upper))
}Generate a random 3x3 matrix:
## [,1] [,2] [,3]
## [1,] 3 16 6
## [2,] 3 14 16
## [3,] 3 4 19The function returns the following lower form matrix:
## [,1] [,2] [,3]
## [1,] 1 0 0
## [2,] 1 1 0
## [3,] 1 6 1The function returns the following upper form matrix:
## [,1] [,2] [,3]
## [1,] 3 16 6
## [2,] 0 -2 10
## [3,] 0 0 -47Test that the product of the lower and upper form matrices is equal to the original matrix:
## [,1] [,2] [,3]
## [1,] TRUE TRUE TRUE
## [2,] TRUE TRUE TRUE
## [3,] TRUE TRUE TRUE