Let m be the number of rows in matrix \(\mathbf{A}\).
Let n be the number of columns in matrix \(\mathbf{A}\).
Let \(A_{m,n}\) = \(\left[ \begin{array}{ccc} a_{1,1} & a_{1,2} & a_{1,3} ..... & a_{1,m-1} & a_{1,n}\\ a_{2,1} & a_{2,2} & a_{2,3} ..... & a_{2,m-1} & a_{2,n}\\ ....\\ ....\\ ....\\ a_{m-1,1} & a_{m-2,2} & a_{m-1,3} ..... & a_{m-1,n-1} & a_{m-1,n}\\ a_{m,1} & a_{m,2} & a_{m,3} ..... & a_{m,n-1} & a_{m,n} \end{array} \right]\)
Let \(\mathbf{A}^T\) be the transpose of A.
By definition of transpose \(A_{m,n}\) is \(\mathbf{A}^T_{n,m}\).
Let \(\mathbf{A}^T_{n,m}\) = \(\mathbf{B}_{n,m}\) = \(\left[ \begin{array}{ccc} b_{1,1} & b_{1,2} & b_{1,3} ..... & b_{1,m-1} & b_{1,m}\\ b_{2,1} & b_{2,2} & b_{2,3} ..... & b_{2,m-1} & b_{2,m}\\ ....\\ ....\\ ....\\ b_{n-1,1} & b_{n-2,2} & b_{n-1,3} ..... & b_{m-1,n-1} & b_{n-1,m}\\ b_{n,1} & b_{n,2} & b_{n,3} ..... & b_{n,m-1} & b_{n,m} \end{array} \right]\)
Let \(\mathbf{A}^T_{n,m}\mathbf{A}_{m,n}\) = \(\mathbf{C}_{m,m}\).
Let \(\mathbf{A}_{m,n}\mathbf{A}^T_{n,m}\) = \(\mathbf{D}_{n,n}\).
Therefore \(\mathbf{C}_{m,m}\ne\mathbf{D}_{n,n}\) since m \(\ne\) n in general (except for square matrices).
For square matrices -
Let \(\mathbb{a}_{i,j}\) be the elements of square matrix \(\mathbf{A}\) where i is the row and j is the column and where i and j = 1 to m.
Let \(\mathbf{A}^T_{m,m}\) = \(\mathbf{B}_{m,m}\) be the elements of square matrix \(\mathbf{B}\) where j is the row and i is the column and where i and j = 1 to m.
Since -
\(\mathbf{B}_{m,m}\mathbf{A}_{m,m}\) = \(\mathbf{C}_{m,m}\)[\(\mathbb{a}_{i,j}\cdot\mathbb{b}_{j,i}\)] for each element of the product matrix and
\(\mathbf{A}_{m,m}\mathbf{B}_{m,m}\) = \(\mathbf{D}_{m,m}\)[\(\mathbb{b}_{j,i}\cdot\mathbb{a}_{i,j}\)] for each element of the product matrix.
Therefore -
\(\mathbf{C}_{m,m}\)[\(\mathbb{a}_{i,j}\cdot\mathbb{b}_{j,i}\)] \(\ne\) \(\mathbf{D}_{m,m}\)[\(\mathbb{b}_{i,j}\cdot\mathbb{a}_{j,i}\)] for any square matrix,
since \(\mathbb{a}_{i,j}\ne\mathbb{a}_{j,i}\) and \(\mathbb{b}_{j,i}\ne\mathbb{a}_{i,j}\) in general.
Let \(\mathbf{A}_{3,4}\) = \(\left[ \begin{array}{ccc} 3 & 7 & 4 & 11\\ 13 & 5 & 4 & 1\\ 1 & 0 & 0 & 5 \end{array} \right]\)
A = matrix(c(3,7,4,11,13, 5, 4, 1, 1, 0, 0, 5), nrow=3, byrow=TRUE)
A## [,1] [,2] [,3] [,4]
## [1,] 3 7 4 11
## [2,] 13 5 4 1
## [3,] 1 0 0 5\(\mathbf{A}^T_{4,3}\)
t(A)## [,1] [,2] [,3]
## [1,] 3 13 1
## [2,] 7 5 0
## [3,] 4 4 0
## [4,] 11 1 5\(\mathbf{A}^T\mathbf{A}\)=\(\mathbf{A}\mathbf{A}^T\)?
t(A)%*%A ## [,1] [,2] [,3] [,4]
## [1,] 179 86 64 51
## [2,] 86 74 48 82
## [3,] 64 48 32 48
## [4,] 51 82 48 147A%*%t(A)## [,1] [,2] [,3]
## [1,] 195 101 58
## [2,] 101 211 18
## [3,] 58 18 26Answer: no
For a square matrix, let \(\mathbf{A}_{4,4}\) = \(\left[ \begin{array}{ccc} 1 & 2 & 3 & 4\\ 4 & 3 & 2 & 1\\ 5 & 6 & 7 & 8\\ 8 & 7 & 6 & 5 \end{array} \right]\)
A = matrix(c(1,2,3,4,4,3,2,1,5,6,7,8,8,7,6,5), nrow=4, byrow=TRUE)
A## [,1] [,2] [,3] [,4]
## [1,] 1 2 3 4
## [2,] 4 3 2 1
## [3,] 5 6 7 8
## [4,] 8 7 6 5\(\mathbf{A}^T\mathbf{A}\)=\(\mathbf{A}\mathbf{A}^T\)?
t(A)%*%A ## [,1] [,2] [,3] [,4]
## [1,] 106 100 94 88
## [2,] 100 98 96 94
## [3,] 94 96 98 100
## [4,] 88 94 100 106A%*%t(A)## [,1] [,2] [,3] [,4]
## [1,] 30 20 70 60
## [2,] 20 30 60 70
## [3,] 70 60 174 164
## [4,] 60 70 164 174t(A)%*%A == A%*%t(A)## [,1] [,2] [,3] [,4]
## [1,] FALSE FALSE FALSE FALSE
## [2,] FALSE FALSE FALSE FALSE
## [3,] FALSE FALSE FALSE FALSE
## [4,] FALSE FALSE FALSE FALSEAnswer: no
Answer -
\(\mathbf{A}^T\mathbf{A}\)=\(\mathbf{A}\mathbf{A}^T\) is true for special cases where the row elements of the square matrix are identical to the corresponding column elements. That is,
\(\left[ \begin{array}{ccc} a_{i,j} & a_{i,j+1} & a_{i,j+2} ..... & a_{i,m-1} & a_{i,m} \end{array} \right]\) = \(\left[ \begin{array}{ccc} b_{i,j}\\ b_{i+1,j}\\ b_{i+2,j}\\ ....\\ ....\\ b_{m-1,j}\\ b_{m,j} \end{array} \right]\) where i,j values are from 1 to m of a \(\mathbf{A}_{m,m}\) square matrix
For a square matrix, let \(\mathbf{A}_{5,5}\) = \(\left[ \begin{array}{ccc} 1 & 2 & 3 & 4 & 5\\ 2 & 1 & 6 & 7 & 8\\ 3 & 6 & 1 & 9 & 10\\ 4 & 7 & 9 & 1 & 11\\ 5 & 8 & 10 & 11 & 1 \end{array} \right]\)
A = matrix(c(1,2,3,4,5,2,1,6,7,8,3,6,1,9,10,4,7,9,1,11,5,8,10,11,1), nrow=5, byrow=TRUE)
A## [,1] [,2] [,3] [,4] [,5]
## [1,] 1 2 3 4 5
## [2,] 2 1 6 7 8
## [3,] 3 6 1 9 10
## [4,] 4 7 9 1 11
## [5,] 5 8 10 11 1\(\mathbf{A}^T\mathbf{A}\)=\(\mathbf{A}\mathbf{A}^T\)?
t(A)%*%A ## [,1] [,2] [,3] [,4] [,5]
## [1,] 55 90 104 104 100
## [2,] 90 154 161 164 163
## [3,] 104 161 227 182 182
## [4,] 104 164 182 268 188
## [5,] 100 163 182 188 311A%*%t(A)## [,1] [,2] [,3] [,4] [,5]
## [1,] 55 90 104 104 100
## [2,] 90 154 161 164 163
## [3,] 104 161 227 182 182
## [4,] 104 164 182 268 188
## [5,] 100 163 182 188 311t(A)%*%A == A%*%t(A)## [,1] [,2] [,3] [,4] [,5]
## [1,] TRUE TRUE TRUE TRUE TRUE
## [2,] TRUE TRUE TRUE TRUE TRUE
## [3,] TRUE TRUE TRUE TRUE TRUE
## [4,] TRUE TRUE TRUE TRUE TRUE
## [5,] TRUE TRUE TRUE TRUE TRUEAnswer: yes
For a square matrix, let \(\mathbf{A}_{4,4}\) = \(\left[ \begin{array}{ccc} 18 & 2 & 3 & 4\\ 2 & 15 & 6 & 7\\ 3 & 6 & 13 & 9\\ 4 & 7 & 9 & 14 \end{array} \right]\)
A = matrix(c(1,2,3,4,2,15,6,7,3,6,13,9,4,7,9,14), nrow=4, byrow=TRUE)
A## [,1] [,2] [,3] [,4]
## [1,] 1 2 3 4
## [2,] 2 15 6 7
## [3,] 3 6 13 9
## [4,] 4 7 9 14\(\mathbf{A}^T\mathbf{A}\)=\(\mathbf{A}\mathbf{A}^T\)?
t(A)%*%A ## [,1] [,2] [,3] [,4]
## [1,] 30 78 90 101
## [2,] 78 314 237 265
## [3,] 90 237 295 297
## [4,] 101 265 297 342A%*%t(A)## [,1] [,2] [,3] [,4]
## [1,] 30 78 90 101
## [2,] 78 314 237 265
## [3,] 90 237 295 297
## [4,] 101 265 297 342t(A)%*%A == A%*%t(A)## [,1] [,2] [,3] [,4]
## [1,] TRUE TRUE TRUE TRUE
## [2,] TRUE TRUE TRUE TRUE
## [3,] TRUE TRUE TRUE TRUE
## [4,] TRUE TRUE TRUE TRUEAnswer: yes
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 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.
Matrix Factorization Using LU Decomposition
LUD <- function (M)
{
r = dim(M)[1]
U = M
L = matrix(0,r,r)
for (x in 1:r)
{
L[x,x] = 1
}
M
U
L
for (p in 1:(r-1))
{
for (i in p:(r-1)) #row
{
mult = ((U[i+1,p])/(U[p,p]))
L[i+1,p] = mult
print (mult)
for (j in p:r) #column
{
U[i+1,j] = U[i+1,j] - U[p,j]*mult
}
}
}
U
L
df = data.frame(U,L)
return(df)
}
#M = matrix(c(1, 2, -1, 1, -1, -2, 3, 5, 4), nrow=3)
#M
#s <- LUD(M)
#print(s)
M = matrix(c(2, 1, -6, 4, -4, -9, -4, 3, 5), ncol=3)
#M
s <- LUD(M)## [1] 0.5
## [1] -3
## [1] -0.5print(s)## X1 X2 X3 X1.1 X2.1 X3.1
## 1 2 4 -4.0 1.0 0.0 0
## 2 0 -6 5.0 0.5 1.0 0
## 3 0 0 -4.5 -3.0 -0.5 1