Demonstration:
\[ \begin{aligned} \text{Lets Consider the Matrix A = } \begin{bmatrix} 1 & 0 & 3 & \\2 & -4 & 2\\1 & -3 & 2\end{bmatrix} \text{We know that } A^T = \begin{bmatrix} 1 & 2 & 1 & \\0 & -4 & -3\\3 & 2 & 2\end{bmatrix} \end{aligned} \]
A <- matrix(c(1,0,3,2,-4,2,1,-3,2), nrow=3, byrow=TRUE)
At <- t(A)
A## [,1] [,2] [,3]
## [1,] 1 0 3
## [2,] 2 -4 2
## [3,] 1 -3 2At## [,1] [,2] [,3]
## [1,] 1 2 1
## [2,] 0 -4 -3
## [3,] 3 2 2# Multiplying A by its transpose we get 3x3
A %*% At## [,1] [,2] [,3]
## [1,] 10 8 7
## [2,] 8 24 18
## [3,] 7 18 14# Multiplying transpose At by A we get 3x3
At %*% A ## [,1] [,2] [,3]
## [1,] 6 -11 9
## [2,] -11 25 -14
## [3,] 9 -14 17From the above it is obvious that \(\begin{aligned} A^tA \neq AA^t \end{aligned}\)
Prof:
\[\begin{aligned} \text{Lets Consider the Matrix A with i rows and j columns where } \begin{aligned} i\neq j \end{aligned} = \begin{bmatrix} a_{1,1} & a_{1,2} &...... &a_{1,j} \\ a_{2,1} & a_{2,2} &...... & a_{2,j}\\ a_{3,1} & a_{3,2} &...... & a_{3,j}\\ . & . &...... & .\\ . & . &...... & .\\ a_{i,1} & a_{i,2} &...... & a_{i,j}\\ \end{bmatrix} \end{aligned}\] \[\begin{aligned} \text{Transpose of Matrix A with j rows and i columns where } \begin{aligned} j\neq i is A^T_{(ji)} \end{aligned} = \begin{bmatrix} a_{1,1} & a_{2,1} &...... &a_{i,1} \\ a_{1,2} & a_{2,2} &...... & a_{i,2}\\ a_{1,3} & a_{2,3} &...... & a_{i,3}\\ . & . &...... & .\\ . & . &...... & .\\ a_{1,j} & a_{2,j} &...... & a_{j,i}\\ \end{bmatrix} \end{aligned}\]\[ \begin{aligned} A{(ij)}A^t{(ji)} \end{aligned}\text{will result the matrix} \begin{aligned} AA^T{(ij)} \end{aligned} \] \[ \begin{aligned} A^t{(ji)} A{(ij)} \end{aligned}\text{will result the matrix} \begin{aligned} A^TA{(ji)} \end{aligned} \] It is obvious that => \(\begin{aligned} A{(ij)} A^t{(ji)} \neq A^t{(ji)}A{(ij)} \end{aligned}\)
A = matrix(c(1,3,2,1, 1,2,1,1, 1,1,0,1, 1,2,2,1, 3,4,1,3), nrow=4, ncol=5)
A## [,1] [,2] [,3] [,4] [,5]
## [1,] 1 1 1 1 3
## [2,] 3 2 1 2 4
## [3,] 2 1 0 2 1
## [4,] 1 1 1 1 3#transpose of matrix A, generates 5 X 4
tA = t(A)
tA## [,1] [,2] [,3] [,4]
## [1,] 1 3 2 1
## [2,] 1 2 1 1
## [3,] 1 1 0 1
## [4,] 1 2 2 1
## [5,] 3 4 1 3#Create 4 X 4 matrix
A%*%tA## [,1] [,2] [,3] [,4]
## [1,] 13 20 8 13
## [2,] 20 34 16 20
## [3,] 8 16 10 8
## [4,] 13 20 8 13#Create 5 X 5 matrix
tA%*%A## [,1] [,2] [,3] [,4] [,5]
## [1,] 15 10 5 12 20
## [2,] 10 7 4 8 15
## [3,] 5 4 3 4 10
## [4,] 12 8 4 10 16
## [5,] 20 15 10 16 35Prof:
# Symmetrical 4 X 4 matrix
A = matrix(c(9,1,3,6, 1,11,7,6, 3,7,4,1, 6,6,1,10), nrow=4, ncol=4)
A## [,1] [,2] [,3] [,4]
## [1,] 9 1 3 6
## [2,] 1 11 7 6
## [3,] 3 7 4 1
## [4,] 6 6 1 10#transpose of matrix A is similar to A
tA = t(A)
tA## [,1] [,2] [,3] [,4]
## [1,] 9 1 3 6
## [2,] 1 11 7 6
## [3,] 3 7 4 1
## [4,] 6 6 1 10#matrix multiplication
A%*%tA## [,1] [,2] [,3] [,4]
## [1,] 127 77 52 123
## [2,] 77 207 114 139
## [3,] 52 114 75 74
## [4,] 123 139 74 173#matrix multiplication
tA%*%A## [,1] [,2] [,3] [,4]
## [1,] 127 77 52 123
## [2,] 77 207 114 139
## [3,] 52 114 75 74
## [4,] 123 139 74 173It is obvious that for Symmetrical matrix => \(\begin{aligned} A{(ii)} A^t{(ii)} = A^t{(ii)}A{(ii)} \end{aligned}\)
Prof:
# Identical 4 X 4 matrix
A = diag(4)
A## [,1] [,2] [,3] [,4]
## [1,] 1 0 0 0
## [2,] 0 1 0 0
## [3,] 0 0 1 0
## [4,] 0 0 0 1# Identical 4 X 4 matrix
At = t(A)
At## [,1] [,2] [,3] [,4]
## [1,] 1 0 0 0
## [2,] 0 1 0 0
## [3,] 0 0 1 0
## [4,] 0 0 0 1#matrix multiplication
A%*%At## [,1] [,2] [,3] [,4]
## [1,] 1 0 0 0
## [2,] 0 1 0 0
## [3,] 0 0 1 0
## [4,] 0 0 0 1#matrix multiplication
At%*%A## [,1] [,2] [,3] [,4]
## [1,] 1 0 0 0
## [2,] 0 1 0 0
## [3,] 0 0 1 0
## [4,] 0 0 0 1It is obvious that for Identical matrix => \(\begin{aligned} A{(ij)} A^t{(ji)} = A^t{(ji)}A{(ij)} \end{aligned}\)
Matrix 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 ights 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 Fight using radars.
Write an R function to factorize a square matrix A into LU or LDU, whichever you prefer. The function factorize checks for the size of a square matrix, and then factorizes it into LU.
factorizeLU <- function(A) {
# Check that A is square
if (dim(A)[1]!=dim(A)[2]) {
return(NA)
}
U <- A
n <- dim(A)[1]
L <- diag(n)
# If dimension is 1, then U=A and L=[1]
if (n==1) {
return(list(L,U))
}
# Loop through the lower triangle (by rows and columns)
# Determine multiplier for each position and add it to L
for(i in 2:n) {
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))
}A <- matrix(c(1,4,-3,-2,8,5,3,4,7), nrow=3, byrow=TRUE)
LU <- factorizeLU(A)
L<-LU[[1]]
U<-LU[[2]]
A## [,1] [,2] [,3]
## [1,] 1 4 -3
## [2,] -2 8 5
## [3,] 3 4 7L## [,1] [,2] [,3]
## [1,] 1 0.0 0
## [2,] -2 1.0 0
## [3,] 3 -0.5 1U## [,1] [,2] [,3]
## [1,] 1 4 -3.0
## [2,] 0 16 -1.0
## [3,] 0 0 15.5L %*% U## [,1] [,2] [,3]
## [1,] 1 4 -3
## [2,] -2 8 5
## [3,] 3 4 7A == L %*% U## [,1] [,2] [,3]
## [1,] TRUE TRUE TRUE
## [2,] TRUE TRUE TRUE
## [3,] TRUE TRUE TRUEA <- matrix(c(2,1,6,8),nrow=2, byrow = T)
LU <- factorizeLU(A)
L<-LU[[1]]
U<-LU[[2]]
A## [,1] [,2]
## [1,] 2 1
## [2,] 6 8L## [,1] [,2]
## [1,] 1 0
## [2,] 3 1U## [,1] [,2]
## [1,] 2 1
## [2,] 0 5A == L %*% U## [,1] [,2]
## [1,] TRUE TRUE
## [2,] TRUE TRUE