Show that \[A^{T}A \neq AA^{T}\]
Proof:
In general, AB != BA. Given a square matrix:
\[ A = \left[\begin{array}{cc} a & b \\ c & d \end{array}\right] \, A^T = \left[\begin{array}{cc} a & c \\ b & d \end{array}\right]\]
\[ A^{T}A = \left[\begin{array}{cc} a & c \\ b & d \end{array}\right] \,\left[\begin{array}{cc} a & b \\ c & d \end{array}\right] \,= \left[\begin{array}{cc} a^{2} + c^{2} & ab + cd \\ ab + dc & b^{2} + d^{2} \end{array}\right]\]
\[ AA^{T} = \left[\begin{array}{cc} a & b \\ c & d \end{array}\right] \,\left[\begin{array}{cc} a & c \\ b & d \end{array}\right] \,= \left[\begin{array}{cc} a^{2} + b^{2} & ac + bd \\ ac + bd & c^{2} + d^{2} \end{array}\right]\]
In order for this to be true, \[c^{2} = b^{2}, ac + bd = ab + dc\]
This is clearly not always the case, so we know that \[A^{T}A \neq AA^{T}\].
Demonstration with a 3x3 matrix:
\[ A = \left[\begin{array}{cc}
1 & 4 & 7\\
2 & 5 & 8\\
3 & 6 & 9
\end{array}\right]
\, A^T = \left[\begin{array}{cc}
1 & 2 & 3\\
4 & 5 & 6\\
7 & 8 & 9
\end{array}\right]\]
\[ A^{T}A = \left[\begin{array}{cc} 1+4+9 & 4+10+18 & 7+16+27\\ 4+10+18 & 16+25+36 & 28+40+54\\ 7+16+27 & 28+40+54 & 49+64+81 \end{array}\right] \]
\[ A^{T}A = \left[\begin{array}{cc} 14 & 32 & 50\\ 32 & 77 & 122\\ 50 & 122 & 194 \end{array}\right] \]
\[ AA^{T} = \left[\begin{array}{cc} 1+16+49 & 2+20+56 & 3+24+63\\ 2+20+56 & 4+25+64 & 6+30+72\\ 3+24+63 & 6+30+72 & 9+36+81 \end{array}\right] \]
\[ AA^{T} = \left[\begin{array}{cc} 66 & 78 & 90\\ 78 & 93 & 108\\ 90 & 108 & 126 \end{array}\right] \]
\[ \left[\begin{array}{cc} 14 & 32 & 50\\ 32 & 77 & 122\\ 50 & 122 & 194 \end{array}\right] != \left[\begin{array}{cc} 66 & 78 & 90\\ 78 & 93 & 108\\ 90 & 108 & 126 \end{array}\right] \]
We can see from the above example that they do not match. We can confirm this in R:
A = matrix(c(1,2,3,4,5,6,7,8,9),
ncol=3)
At = t(A)
AtA = A %*% At
AAt = At %*% A
all(AtA == AAt)## [1] FALSEFor 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).
In order for this to be true, \[A = A^{T}\].