Assignment2

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\[\begin{equation} A=\begin{bmatrix} a & b \\ c & d \\ \end{bmatrix} \end{equation}\] \[\begin{equation} A^T=\begin{bmatrix} a & c \\ b & d \\ \end{bmatrix} \end{equation}\] \[\begin{equation} AA^T=\begin{bmatrix} a^2 + b^2 & ac + bd \\ ca + db & c^2 + d^2 \\ \end{bmatrix} A^TA=\begin{bmatrix} a^2 + c^2 & ab + cd \\ ba + dc & b^2 + d^2 \\ \end{bmatrix} \end{equation}\] \[\begin{equation} A^TA <> AA^T in general \end{equation}\]

To demostrate assume

\[\begin{equation} A=\begin{bmatrix} 1 & 2 \\ 3 & 4 \\ \end{bmatrix} \end{equation}\] \[\begin{equation} AA^T=\begin{bmatrix} 1 + 4 & 3 + 8 \\ 3 + 8 & 9 + 16 \\ \end{bmatrix}= \begin{bmatrix} 5 & 11 \\ 11 & 25 \\ \end{bmatrix} A^TA=\begin{bmatrix} 1 + 9 & 2 + 12 \\ 2 + 12 & 4 + 16 \\ \end{bmatrix}= \begin{bmatrix} 10 & 14 \\ 14 & 20 \\ \end{bmatrix} \end{equation}\] \[\begin{equation} AA^T <> A^TA \end{equation}\]

However under condition when b = c then replace c with b

\[\begin{equation} AA^T=\begin{bmatrix} a^2 + b^2 & ab + bd \\ ba + db & b^2 + d^2 \\ \end{bmatrix} A^TA=\begin{bmatrix} a^2 + b^2 & ab + bd \\ ba + db & b^2 + d^2 \\ \end{bmatrix} \end{equation}\] \[\begin{equation} So AA^T = A^TA when A = A^T \end{equation}\]

a = function(A)
{
 n = dim(A)[1] 
 if (n==2){
 L1 = diag(n)
 L1[2,1] = -(L1[2,2]*A[2,1])/A[1,1]
 U= L1%*%A
 L=solve(L1)
return(list(L,U,L%*%U))}
else if (n==3)
{
 L1 = diag(n) 
 L1[2,1] = -(L1[2,2]*A[2,1])/A[1,1]
 L1[3,1] = -(L1[3,3]*A[3,1])/A[1,1]
 A=L1%*%A
 L2=diag(n)
 L2[3,2] = 1
 L2[3,2] = -(L2[3,3]*A[3,2])/A[2,2]
 L=solve(L1)%*%solve(L2)
 U=L2%*%A
 return(list(L,U,L%*%U))
}
else
{
 L1 = diag(n) 
 L1[2,1] = -(L1[2,2]*A[2,1])/A[1,1]
 L1[3,1] = -(L1[3,3]*A[3,1])/A[1,1]
 L1[4,1] = -(L1[4,4]*A[4,1])/A[1,1]
 A= L1%*%A
 L2= diag(n) 
 L2[3,2] = -(L2[3,3]*A[3,2])/A[2,2]
 L2[4,2]=-(L2[4,4]*A[4,2])/A[2,2]
 A=L2%*%A
 L3=diag(n)
 L3[4,3]=-(L3[4,4]*A[4,3])/A[3,3]
 L=solve(L1)%*%solve(L2)%*%solve(L3)
 U=L3%*%A
 return(list(L,U,L%*%U))
 
}
}  

A=matrix(c(1,2,3,4),2)
a(A)

## [[1]]
##      [,1] [,2]
## [1,]    1    0
## [2,]    2    1
## 
## [[2]]
##      [,1] [,2]
## [1,]    1    3
## [2,]    0   -2
## 
## [[3]]
##      [,1] [,2]
## [1,]    1    3
## [2,]    2    4

A=matrix(c(2,2,3,4,5,6,7,8,9),3)
a(A)

## [[1]]
##      [,1] [,2] [,3]
## [1,]  1.0    0    0
## [2,]  1.0    1    0
## [3,]  1.5    0    1
## 
## [[2]]
##      [,1] [,2] [,3]
## [1,]    2    4  7.0
## [2,]    0    1  1.0
## [3,]    0    0 -1.5
## 
## [[3]]
##      [,1] [,2] [,3]
## [1,]    2    4    7
## [2,]    2    5    8
## [3,]    3    6    9

A=matrix(c(2,2,3,4,5,6,7,8,9,5,7,8,8,8,4,7),4)
a(A)

## [[1]]
##      [,1] [,2]     [,3] [,4]
## [1,]  1.0  0.0 0.000000    0
## [2,]  1.0  1.0 0.000000    0
## [3,]  1.5 -0.5 1.000000    0
## [4,]  2.0 -2.0 2.117647    1
## 
## [[2]]
##      [,1] [,2] [,3]      [,4]
## [1,]    2    5  9.0  8.000000
## [2,]    0    1 -4.0  0.000000
## [3,]    0    0 -8.5 -8.000000
## [4,]    0    0  0.0  7.941176
## 
## [[3]]
##      [,1] [,2] [,3] [,4]
## [1,]    2    5    9    8
## [2,]    2    6    5    8
## [3,]    3    7    7    4
## [4,]    4    8    8    7