Is a matrix multiplied with its transpose something special? In my math lectures, we talked about the Gram-Determinant where a matrix times its transpose are multiplied together.

Is $A A^\mathrm T$ something special for any matrix $A$?
The main thing is presumably that $AA^T$ is symmetric. Indeed $(AA^T)^T=(A^T)^TA^T=AA^T$. For symmetric matrices one has the Spectral Theorem which says that we have a basis of eigenvectors and every eigenvalue is real.

Moreover if $A$ is invertible, then $AA^T$ is also positive definite, since $$x^TAA^Tx=(A^Tx)^T(A^Tx)&gt; 0$$

Then we have: A matrix is positive definite if and only if it's the Gram matrix of a linear independent set of vectors.

Last but not least if one is interested in how much the linear map represented by $A$ changes the norm of a vector one can compute

$$\sqrt{\left&lt;Ax,Ax\right&gt;}=\sqrt{\left&lt;A^TAx,x\right&gt;}$$ 

which simplifies for eigenvectors $x$ to the eigenvalue $\lambda$ to

$$\sqrt{\left&lt;Ax,Ax\right&gt;}=\sqrt \lambda\sqrt{\left&lt;x,x\right&gt;},$$

The determinant is just the product of these eigenvalues.
The matrix $A$ represents a rotation if $A^TA=I$ and Det A=1 (where $A^T$ means $A$ transpose)