Problem Set 1

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

Proof follows:

\[ A=\begin{bmatrix} a & b \\ c & d \end{bmatrix} \]

\[ A^T=\begin{bmatrix} a & c \\ b & d \end{bmatrix} \]

\[ A^TA= \begin{bmatrix} a & c \\ b & d \end{bmatrix} \times \begin{bmatrix} a & b \\ c & d \end{bmatrix} = \begin{bmatrix} a \times a + c \times c & c \times a + c \times c \\ b \times b + d \times d & b \times d + d \times d \end{bmatrix} = \begin{bmatrix} a^2 + c^2 & ac + c^2 \\ b^2 + d^2 & bd + d^2 \end{bmatrix} \]

\[ AA^T=\begin{bmatrix} a & b \\ c & d \end{bmatrix} \times \begin{bmatrix} a & c \\ b & d \end{bmatrix} = \begin{bmatrix} a \times a + b \times b & a \times c + b \times d \\ c \times a + d \times d & c \times c + d \times d \end{bmatrix} = \begin{bmatrix} a^2 + b^2 & ac + bd \\ ac + d^2 & c^2 + d^2 \end{bmatrix} \]

\[ \begin{bmatrix} a^2 + c^2 & ac + c^2 \\ b^2 + d^2 & bd + d^2 \end{bmatrix} \neq \begin{bmatrix} a^2 + b^2 & ac + bd \\ ac + d^2 & c^2 + d^2 \end{bmatrix} \]

Demonstration follows:

##      [,1] [,2]
## [1,]    1    2
## [2,]    3    4
##      [,1] [,2]
## [1,]    1    3
## [2,]    2    4
##      [,1] [,2]
## [1,]    5   11
## [2,]   11   25
##      [,1] [,2]
## [1,]   10   14
## [2,]   14   20

The following show that \(A^TA \neq AA^T\)

##       [,1]  [,2]
## [1,] FALSE FALSE
## [2,] FALSE FALSE

2. For a special type(s) of square matrix A, we get \(A^TA = AA^T\). Under what conditions could this be true?:

Special cases where \(A^TA = AA^T\) are:

  • one matrix is the Identity matrix I
  • one matrix is the Zero matrix
  • both matrices are 2 x 2 ratation matrices
  • both matrices are Diagonal matrices