<s>[INST] Why multiply a matrix with its transpose? This might be a very stupid question, but I do not seem to understand why I would multiple a matrix with its transpose. I am not a mathematician, but I am very interested in understanding the practical usage of equations: 
Imagine I have three products sales Apple, Orange and Pear for the last 3 days in a matrix form called A: 
$$ A= 
\begin{bmatrix}
        Apple & Orange & Pear \\
        10 & 2 & 5 \\
        5 & 3 & 10 \\
        4 & 3 & 2 \\
        5 & 10 & 5 \\
        \end{bmatrix}$$
What will $AA^{\rm T}$ tell me? 
I have seen this long answer link: Is a matrix multiplied with its transpose something special?, but I did not get it at all. 
I see that a lot of equations use the product $AA^{\rm T}$ and I really hope that someone will give a very simple answer. 
 [/INST] It can be a part of a bigger question (context needed, for example, linear regression as mentioned by lhf in the linked post).
Assume that the vector (Q) of quantity apple sales and the vector (R) of revenues on four days are:
$$Q=\begin{pmatrix}10\\ 5\\ 4 \\ 8\end{pmatrix}; \ \ R=\begin{pmatrix}20\\ 10\\ 8\\ 16\end{pmatrix}$$
And now we want to find the linear revenue function $R=aQ+b$. Obviously, it is $R=2Q$:
$$R=\begin{pmatrix}R_1\\ R_2\\ R_3\\ R_4\end{pmatrix}=2\begin{pmatrix}10\\ 5\\ 4\\ 8\end{pmatrix}=\begin{pmatrix}20\\ 10\\ 8\\ 16\end{pmatrix}$$ 
Keeping in mind that the predicted function is not always perfectly linear fit and for demonstration purpose, assume that we want to find the linear revenue function $y=b_0+b_1x$ using linear regression, where $y=R,x=Q$ and $b_0,b_1$ are the parameters to be found. Then the linear function can be written in the matrix form as:
$$Y=\begin{pmatrix}y_1\\ y_2\\ y_3\\ y_4\end{pmatrix}=\begin{pmatrix}1&x_1\\ 1&x_2\\ 1&x_3\\ 1&x_4\end{pmatrix}\begin{pmatrix}b_0\\ b_1\end{pmatrix}=Xb$$
Now we need to solve the matrix equation:
$$Y=Xb \Rightarrow \\
X^TY=X^TXb \Rightarrow \\
b=(X^TX)^{-1}X^TY=\\
\left[\begin{pmatrix}1&1&1&1\\x_1&x_2&x_3&x_4\end{pmatrix}\begin{pmatrix}1&x_1\\ 1&x_2\\ 1&x_3\\ 1&x_4\end{pmatrix}\right]^{-1}\begin{pmatrix}1&1&1&1\\x_1&x_2&x_3&x_4\end{pmatrix}\begin{pmatrix}y_1\\ y_2\\ y_3\\ y_4\end{pmatrix}=\\
\left[\begin{pmatrix}1&1&1&1\\10&5&4&8\end{pmatrix}\begin{pmatrix}1&10\\ 1&5\\ 1&4\\ 1&8\end{pmatrix}\right]^{-1}\begin{pmatrix}1&1&1&1\\10&5&4&8\end{pmatrix}\begin{pmatrix}20\\ 10\\ 8\\ 16\end{pmatrix}=\\
\left[\begin{pmatrix}4&27\\27&205\end{pmatrix}\right]^{-1}\begin{pmatrix}54\\410\end{pmatrix}=\\
\frac{1}{91}\begin{pmatrix}205&-27\\ -27&4\end{pmatrix}\begin{pmatrix}54\\410\end{pmatrix}=\begin{pmatrix}0\\2\end{pmatrix}=\begin{pmatrix}b_0\\b_1\end{pmatrix}$$
as expected: $y=0+2x \iff R=2Q$.
</s>