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. 
 A: Lets consider the matrix $A$ characterizing the values of some variables $a_{ij}$, $j=1...m$ with values at different times $i=1...n$, as in the OP example, but transposed.
If the variables are normalized in mean, the matrix $\frac 1m A^TA$ is the estimator of the covariances $s_{j_1j_2}=\mathbb{E}(a_{\cdot j_1}a_{\cdot j_2}) \approx \frac 1m \sum a_{j_1}a_{j_2}$ for the set of random variables $a_{\cdot j=1...m}$.
If the entries $a_{ij}$ of $A$ have units of $[a]$, then the entries of $AA^T$ will have units of $[a^2]$. This is consistent with the abovementioned.
When solving the problem $Ax=B$, the solution $x=(A^TA)^{-1}A^TB$ is the best estimator (LS), provided that the covariance as defined above, is enough variable to be invertible.
A: 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: $y=0+2x \iff R=2Q$.
A: In your case, $AA^T$ just sitting on a park bench doesn't tell you anything of great interest.
Hyprfrcb's answer talks about units. One of the elements has units of squared apples. Another has units of pear-oranges. Another has units of orange-pears! This by itself should be a red flag that values don't mean anything by themselves.
It can be a means to get to a least-squares solution if you were looking to model, say, how much of each fruit you'd expect to sell on a particular day. (This was one of the answers in the linked question.)
But by itself? It's just a fruit hybridization experiment gone terribly wrong.
