\(A^TA\neq AA^T\)
Prof: For a non-squire matrix, \(n \times m\), \(A^TA\) produces \(m \times m\) matrix, and \(AA^T\) produces \(n \times n\) matrix.
See an example below.
a <- matrix(1:12, ncol = 3, nrow = 4)
at<-t(a)
aat<-a%*%at
aat
## [,1] [,2] [,3] [,4]
## [1,] 107 122 137 152
## [2,] 122 140 158 176
## [3,] 137 158 179 200
## [4,] 152 176 200 224
ata<-at%*%a
ata
## [,1] [,2] [,3]
## [1,] 30 70 110
## [2,] 70 174 278
## [3,] 110 278 446
For a squire matrix, the \(A^TA= AA^T\) only if a matrix is symmetrical ( \(a_{ij} =a_{ji}\)).
See, example below.
a <- matrix(c(1,2,3,2,4,5,3,5,6), ncol = 3, nrow = 3)
a
## [,1] [,2] [,3]
## [1,] 1 2 3
## [2,] 2 4 5
## [3,] 3 5 6
at<-t(a)
aat<-a%*%at
aat
## [,1] [,2] [,3]
## [1,] 14 25 31
## [2,] 25 45 56
## [3,] 31 56 70
ata<-at%*%a
ata
## [,1] [,2] [,3]
## [1,] 14 25 31
## [2,] 25 45 56
## [3,] 31 56 70
It a squire matrix is not symetrical, then equality won’t hold.
See example below.
a <- matrix(1:9, ncol = 3, nrow = 3)
at<-t(a)
aat<-a%*%at
aat
## [,1] [,2] [,3]
## [1,] 66 78 90
## [2,] 78 93 108
## [3,] 90 108 126
ata<-at%*%a
ata
## [,1] [,2] [,3]
## [1,] 14 32 50
## [2,] 32 77 122
## [3,] 50 122 194
A<-matrix(c(1,7,6,2,9,5,3,7,5,89,3,67,67,45,3,5),ncol=4,nrow=4)
A
## [,1] [,2] [,3] [,4]
## [1,] 1 9 5 67
## [2,] 7 5 89 45
## [3,] 6 3 3 3
## [4,] 2 7 67 5
LU<-function(C){
D<-diag(nrow(C))
for (i in 2:(nrow(C)))
{for (j in i:(nrow(C)))
{
temp<-C[j,i-1]/C[i-1,i-1]
C[j,]<-C[j,]-(C[j,i-1]/C[i-1,i-1])*C[i-1,]
D[j,i-1]<-temp
}
}
print(C)
print(D)
return(D%*%C)
}
D<-LU(A)
## [,1] [,2] [,3] [,4]
## [1,] 1 9 5.00000 67.00000
## [2,] 0 -58 54.00000 -424.00000
## [3,] 0 0 -74.48276 -26.17241
## [4,] 0 0 0.00000 -65.01667
## [,1] [,2] [,3] [,4]
## [1,] 1 0.0000000 0.0000000 0
## [2,] 7 1.0000000 0.0000000 0
## [3,] 6 0.8793103 1.0000000 0
## [4,] 2 0.1896552 -0.6277778 1
D
## [,1] [,2] [,3] [,4]
## [1,] 1 9 5 67
## [2,] 7 5 89 45
## [3,] 6 3 3 3
## [4,] 2 7 67 5