Question #1: Show that \(\
A^T A\neq AA^T\) in general. (Proof and
demonstration.)
To prove that \(\ A^T A\neq AA^T\)
Let A be a 3x3 matrix:
\(A=\begin{bmatrix} a & b & c\\ d & e & f\end{bmatrix}\)
And its transpose will be (interchanging the rows and columns of A):
\(A^T=\begin{bmatrix} a & d\\ b & e\\ c & f\end{bmatrix}\)
Now, let’s calculate \(A^TA\)
\(A^TA=\begin{bmatrix} a & d\\ b & e\\ c & f\end{bmatrix}\begin{bmatrix} a & b & c\\ d & e & f\end{bmatrix}\)
\(A^TA=\begin{bmatrix} a^2+d^2 & ab+de & ac+df\\ba+ed & b^2+e^2 & bc+ef\\ca+fd & cb+fe & c^2+f^2 \end{bmatrix}\)
The matrix \(A^TA\) is 3x3 matrix (it is the result of (3x2) by (2x3) matrices),
\(AA^T=\begin{bmatrix} a & b & c\\ d & e & f\end{bmatrix}\begin{bmatrix} a & d\\ b & e\\ c & f\end{bmatrix}\)
\(AA^T=\begin{bmatrix} a^2+b^2+c^2 & ad+be+cf\\ da+eb+fc & d^2+e^2+f^2\end{bmatrix}\)
while \(AA^T\) matrix is 2x2 matrix (the result of (2x3) by (3x2) matrices). And the elements on each result matrix are different, so
\(\ A^T A\neq AA^T\)
Question #2: For a special type of square matrix A, we get \(A^T A = AA^T\) . Under what conditions could this be true? (Hint: The Identity matrix I is an example of such a matrix).
Based on the hint, \(I^TI= II^T\) , what is special about the identity matrix that makes it have this property?
we know that the transpose of a matrix is obtained by interchanging the rows and the columns of the original matrix, so
Let:
\(I=\begin{bmatrix} 1 & 0 & 0\\ 0 & 1 & 0\\0 & 0 & 1\end{bmatrix}\)
the transpose of I is:
\(I^T=\begin{bmatrix} 1 & 0 & 0\\ 0 & 1 & 0\\0 & 0 & 1\end{bmatrix}\)
Which means the condition that is needed to exist in a matrix is symmetry: \(A=A^T\)
Let \(A=\begin{bmatrix} a & x_1 & x_2\\ x_1 & b & x_3\\ x_2 & x_3 & c\end{bmatrix}\)
The transpose of the above matrix is:
\(A^T=\begin{bmatrix} a & x_1 & x_2\\ x_1 & b & x_3\\ x_2 & x_3 & c\end{bmatrix}\)
Let’s calculate \(AA^T\)
\(AA^T=\begin{bmatrix} a & x_1 & x_2\\ x_1 & b & x_3\\ x_2 & x_3 & c\end{bmatrix}\begin{bmatrix} a & x_1 & x_2\\ x_1 & b & x_3\\ x_2 & x_3 & c\end{bmatrix}\)
\(AA^T=\begin{bmatrix} a^2+x_1^2+x_2^2 & ax_1+x_1b+x_2x_3 & ax_2+x_1x_3+x_2c\\ x_1a+bx_1+x_3^2 & x_1^2+b^2+x_3^2 & x_1x_2+bx_3+x_3c\\ x_2a+x_3x_1+cx_2 & x_2x_1+x_3b+cx_3 & x_2^2+x_3^2+c^2\end{bmatrix}\)
Now, let’s calculate \(A^TA\)
\(A^TA=\begin{bmatrix} a^2+x_1^2+x_2^2 & ax_1+x_1b+x_2x_3 & ax_2+x_1x_3+x_2c\\ x_1a+bx_1+x_3^2 & x_1^2+b^2+x_3^2 & x_1x_2+bx_3+x_3c\\ x_2a+x_3x_1+cx_2 & x_2x_1+x_3b+cx_3 & x_2^2+x_3^2+c^2\end{bmatrix}\)
Both \(AA^T\) and \(A^TA\) are 3x3 matrices, and comparing their elements, they are equal, so for symmetric matrices, \(AA^T=A^TA\)