Problem Set 1

(1) Show that \(A^TA\) \(\ne\) \(AA^T\) in general. (Proof and demonstration.)

A <- matrix(c(1,1,1,2,0,-1,3,1,-2), nrow=3, byrow=TRUE)
AT <- t(A)
#Matrix A and its Transpose
A;
##      [,1] [,2] [,3]
## [1,]    1    1    1
## [2,]    2    0   -1
## [3,]    3    1   -2
AT
##      [,1] [,2] [,3]
## [1,]    1    2    3
## [2,]    1    0    1
## [3,]    1   -1   -2
#Multiplying A by its Transpose

A %*% AT
##      [,1] [,2] [,3]
## [1,]    3    1    2
## [2,]    1    5    8
## [3,]    2    8   14
#Multiplying A Transpose by A

AT %*% A
##      [,1] [,2] [,3]
## [1,]   14    4   -7
## [2,]    4    2   -1
## [3,]   -7   -1    6

Above calculation proves that \(A^TA\) \(\ne\) \(AA^T\). We can check this further more by using logical comparision in R

A %*% AT == AT %*% A
##       [,1]  [,2]  [,3]
## [1,] FALSE FALSE FALSE
## [2,] FALSE FALSE FALSE
## [3,] FALSE FALSE FALSE

(2) 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).

#Square matrix A
A <- matrix(c(2,7,3,7,9,4,3,4,7), nrow=3, byrow=TRUE)
A
##      [,1] [,2] [,3]
## [1,]    2    7    3
## [2,]    7    9    4
## [3,]    3    4    7
#Transpose of square matrix A
AT <- t(A)
AT
##      [,1] [,2] [,3]
## [1,]    2    7    3
## [2,]    7    9    4
## [3,]    3    4    7
#Multiplying square matrix A by its Transpose
A %*% AT
##      [,1] [,2] [,3]
## [1,]   62   89   55
## [2,]   89  146   85
## [3,]   55   85   74
#Multiplying square matrix A Transpose by square matrix A
AT %*% A
##      [,1] [,2] [,3]
## [1,]   62   89   55
## [2,]   89  146   85
## [3,]   55   85   74
#logical comparision
A %*% AT == AT %*% A
##      [,1] [,2] [,3]
## [1,] TRUE TRUE TRUE
## [2,] TRUE TRUE TRUE
## [3,] TRUE TRUE TRUE

Above calculations prove that if matrix A is symmetric/square, then \(A = A^T\) and \(A^TA\) \(=\) \(AA^T\).