Minimum value trace of A*Atranspose For any matrix $A$, let $A^T$ denote its transpose matrix.
What is the minimum value of $\mathrm{trace}(AA^T)$ for an $n \times n$ non-singular matrix $A$ with integer entries?

You can directly calculate this trace.
Let $a_1^T,a_2^T \dots a_n^T$ denote rows of $A$.  
Then  on diagonal $AA^T$ you have $a_1^Ta_1,a_2^Ta_2 \dots a_n^Ta_n$.
So the sum of diagonal entries of $AA^T$ is just a sum of squares of vectors $a_i$ lengths.
More detailed picture of the situation:
$AA^T= \begin{bmatrix} a_1^T \\ a_2^T \\ \dots a_n^T \end{bmatrix}\begin{bmatrix} a_1  & a_2  & \dots a_n  \end{bmatrix}=\begin{bmatrix} \color{red}{a_1^Ta_1 } & a_1^Ta_2 & \dots & a_1^Ta_n\\   a_2^Ta_1  & \color{red}{a_2^Ta_2}  & \dots   & a_2^Ta_n\\ \dots &  \dots  & \color{red}\dots  & \dots  \\ a_n^Ta_1  & a_n^Ta_2  & \dots & \color{red}{a_n^Ta_n} \end{bmatrix}$
$\operatorname{tr} (AA^T)=\sum\limits_{i=1}^n \color{red}{a_i^Ta_i}$.
Note additionally that vectors $a_i$ can't be zero vectors as in such situation the matrix would not be invertible. Minimal length of $a_i$ vector is therefore for integer components equal to $1$.
We have $n$ row vectors which every can have minimal length $1$. 
It's easy to find such $n$ vectors with length $1$ that they are independent linearly (in this case $A$ as $A^T$ are full rank matrices).
So the  sum of squares of lengths has minimal value $n$.
